Vector Calculus

UNIT

  

Vector Calculus  VECTOR DIFFERENTIAL CALCULUS The vector differential calculus extends the basic concepts of (ordinary) differential calculus to vector functions, by introducing derivative of a vector function and the new concepts of gradient, divergence and curl.

5.1 VECTOR FUNCTION If the vector r   varies corresponding to the variation of a scalar variable t that is its length and direction be known and determine as soon as a value of t  is given, then r   is called a vector function of t and written as r = f  (t)

and read it as r   equals a vector function of t. Any vector  f  (t) can be expressed in the component form   f  (t) = f 1(t) i  +  f 2 (t)  j  +  f 3 (t) k  Where  f 1(t),  f 2 (t),  f 3 (t) are three scalar functions of t.

For example, where

  r = 5t2 i  + t  j   – t3 k  2 f 1(t) = 5t ,  f 2(t) = t,  f 3 (t) = – t3.

5.2 VECTOR DIFFERENTIATION Let r  =  f  (t) be a single valued continuous vector point function of a scalar variable t. Let O be the origin of vectors. Let OP   represents the vector r   corresponding to a certain value t  to the scalar variable t. Then r = f  (t)

...( i) 333 33 3

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Let OQ  represents the vector r  + δ r  corresponding to the value t + δt of the scalar variable t, where δt is infinitesimally small. Then, Q

r  + δ r = f  (t + δt) Subracting ((ii) from (ii (ii)) δ r = f  (t + δt) –  f  (t) Dividing both sides by δt, we get

...(i) 

...(iii iii))

  f (t + δt ) − f (t ) δr = δt δt Taking the limit of both side as δt → 0.

  

  r

dr  — dt

     r   d  +

P 

r

O

lim δr = lim  f (t + δt) − f( t ) t →0 δt δt→0 δt

We obtain

d  f (t + δt) − f( t ) dr  f (t) = δlim  = t→0 dt δt dt

 

.

5.3 SOME RESULTS ON DIFFERENTIATION 

 d dr d s (r − s ) = − ⋅ dt dt dt

(a)



d(sr ) ds  dr = r+s dt dt dt

(b) (c)

 db d(a ⋅ b) da = a⋅ + ⋅b dt dt dt

(d)

d ( a × b) = da × b + a × db dt dt dt

(e)

 db     dc   d  da  a × (b × c) = × (b × c ) + a ×   + a ×  b ×  ⋅  dt × c   +   dt   dt dt



v = r  =

dr ⋅ dt

dv d2 r  ⋅  a = = dt dt 2   A particle moves along the curve x = t3 + 1, y 1,  y =  = t2, z  z =  = 2t 2 t + 5, where t is the time. t Find the component of its velocity and acceleration at   = 1 in the direction i +  j + 3k .  Velocity

dr d  = ( xi + y j + zk)  dt dt d 3 (t + 1)i + t 2 j + (2t + 5)  k   = dt = 3t 2 i + 2t j + 2 k  

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 , at t = 1. = 3i + 2 j + 2k   is Again unit vector in the direction of i + j + 3k 

=

 i + j + 3k 



12 + 12 + 32 

 =

 i + j + 3 k 

11

 is Therefore, the component of velocity at t  = 1 in the direction of i + j + 3k 

 3i + 2 j + 2k ⋅ i + j + 3k   11

=

3+2+6  = 11

11

        

d2 r d dr =  = 6ti + 2 j  = 6i + 2 j , at t = 1 2 dt dt dt

Acceleration

 is Therefore, the component of acceleration at t  = 1 in the direction i + j + 3 k 

6i + 2 j  ⋅ i + j + 3k    11

=

6+2 11

 =

8 11

·

  A particle moves along the curve x = 4 cos t,  y = 4 sin t, t, z  z =  = 6t. Find the velocity and acceleration at time t = 0 and t = π/2. Find also the magnitudes of the velocity and acceleration at any time t.   Let r = 4 cos t i   + 4 sin t j  + 6t k 



=

 4 j + 6 k 

v

=

 −4i + 6 k 

|v|

=

16 + 36  =

52  = 2 13

|v|

=

16 + 36  =

52  = 2 13 .

at

t = 0,

v

at

t =

at

π , 2 t = 0,

at

t = Again, acceleration,

a =

π , 2



d2 r   = – 4 cos t i  – 4 sin t  j dt 2

at

t = 0,

a

∴ at

t = 0,

|a|

at

t =

π , 2 π t = , 2

at

a

|a|

= – 4 i =

(−4)2 = 4

= – 4 j  =

(−4)2 = 4.

  If r  = a ent + b e –nt, where a, b are constant vectors, then prove that

d2 r dt 

2

− n2 r = 0

nt nt − nt r = a e  + b e dr be − nt ⋅ (− n) = ae nt ⋅ n + be dt

...(ii) ...(

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d2 r = ae nt ⋅ n2 + be be − nt ⋅ (− n)2 2 dt d2 r = n 2 ae ae nt + be − nt  = n2r 2 dt ⇒

d2 r – n 2 r = 0. dt 2 



[From ((ii)]

 



  If a , b , c   are constant vectors then show that r = at 2 + bt + c   is the path of a point moving with constant acceleration. 2 r = at + bt + c



dr = 2 at + b dt d2 r = 2 a   which is a constant vector. dt 2 Hence, acceleration of the moving point is a constant.  

EXERCISE 5.1  A particle moves along a curve whose parametric equations are x = e–t,  y  y   = 2 cos 3t 3 t,  z   = sin 3t.  z 3 t. Find the velocity and acceleration at t = 0.

[ r = xi + yj + zk  ] .

V e l. =

10 , acc. = 5 13

 A particle moves along the curve

x = 3 t2,  y  y =  = t2 – 2t,  z  z =  = t2. 

Find the velocity and acceleration at t = 1.

.

v = 2 10 , a = 2 11

 Find the angle between the directions of the velocity and acceleration vectors at time t of  2  arc cos t a particle with position vector r = (t 2 + 1)i − 2tj + (t 2 − 1)k .   2t 2 + 1

  

  

.

 A particle moves along the curve x = 2t2,  y  y =  = t2 – 4t,  z = 3t – 5 where t is the time. Find . the components of its velocity and acceleration at time t = 1 in the direction i − 3 j + 2 k 

  

.

8 14 14  ; −  7 7 

 If r  = (sec t) i  + (tan t) j   be the position vector of P. Find the velocity and acceleration

of P at t =

π . 6

 

.

 2 4  2 i + j, (5i + 4 j ) 3 3 3 3 

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VECTOR CALCULUS

5.4 SCALAR POINT FUNCTION If for each point P  of a region R, there corresponds a scalar denoted by  f (P), then  f   is called a ‘‘scalar point function’’ for the region R.   The temperature f (P) at any point P of a certain body occupying a certain region R is a scalar point function.   The distance of any point P(x,  y,  z) in space from a fixed point ( x0,  y0,  z0) is a scalar function. (x − x0 ) 2 + ( y − y0 ) 2 + ( z − z0 )2 ⋅

 f (P) =

  (U.P.T.U., 2001) Scalar field is a region in space such that for every point P in this region, the scalar function  f   associates a scalar  f (P).

5.5 VECTOR POINT FUNCTION If for each point P of a region R, there corresponds a vector  f (P) then  f   is called “vector point function” for the region R.  If the velocity of a particle at a point P, at any time t be  f  (P), then  f   is a vector point function for the region occupied by the particle at time t. If the coordinates of P be (x, y,  z ) then

 f  (P) = f 1 (x,  y,  z) i +  f 2 (x,  y,  z) j +  f 3 (x,  y,  z) k.   (U.P.T.U., 2001) Vector field is a region in space such that with every point P  in the region, the vector function  f  associates a vector  f (P).     The linear vector differential (Hamiltorian) operator ‘‘del’’ defined and ^

denoted as

∇ = i

∂ ^ ∂ ^ ∂ + j + k  ∂x ∂ y ∂ z

This operator also read nabla. It is not a vector but combines both differential and vectorial properties analogous to those of ordinary vectors.

5.6 GRADIENT OR SLOPE OF SCALAR POINT FUNCTION If f (x,  y, z) be a scalar point function and continuously differentiable then the vector  ∂f   ∂f   ∂f  ∇ f  = i ∂x +  j ∂ y + k ∂ z

is called the gradient of  f   and is written as grad  f . Thus

(U.P.T.U., 2006)

∂ f  ∂f  ∂f  grad  f  = i ∂x +  j ∂ y + k ∂ z = ∇ f 

It should be noted that ∇ f  is a vector whose three components are is a scalar point function, then ∇ f   is a vector point function.

∂ f  ∂ f  ∂ f  ⋅  Thus, if  f  , , ∂x ∂ y ∂ z

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5.7 GEOMETRICAL MEANING OF GRADIENT, NORMAL Consider any point P in a region throught which a scalar field

(U.P.T.U., 2001)

 f (x, y, z) = c defined. Suppose that ∇ f  ≠ 0  at P and that there is a  f   = const. surface S  through P  and a tangent plane T ; for instance, if  f   is a temperature field, then S is an isothermal surface (level surface). If n , at P, is choosen as any vector in the df  tangent plane T , then surely  must be zero. dS df  Since  = ∇ f  . n = 0 dS

Normal N at P Z

^

N

S

P n^

for every n  at P in the tangent plane, and both ∇ f   and n are non-zero, it follows that ∇ f    is normal to the tangent plane T  and hence to the surface S at P.

f = Constant

If letting n  be in the tangent plane, we learn that ∇ f   is normal to S, then to seek additional information about logical to let n   be along the normal line at P,. Then

∇ f    it seems

Y

X

 

df  df    = n  = , N     then ds dN  df    = ∇ f ⋅ 1 cos 0 = ∇ f  ⋅ = ∇ f ⋅ N  dN 

So that the magnitude of ∇ f   is the directional derivative of  f   along the normal line to S, in the direction of increasing  f . Hence, ‘‘The gradient

∇ f    of scalar field  f (x,  y,  z) at

P  is vector normal to the surface df   in that direction. dN 

 f   = const. and has a magnitude is equal to the directional derivative

5.8 DIRECTIONAL DERIVATIVE ∂ f  ∂ f  ∂ f  , ,   are the derivatives (rates of change) of  f  ∂x ∂ y ∂ z in the direction of the coordinate axes OX , OY , OZ respectively. This concept can be extended to

Let  f  =  f (x, y, z) then the partial derivatives

define a derivative of  f   in a "given" direction PQ . Let P be a point in space and b  be a unit vector from P  in the given direction. Let s  be the are length measured from P to another point Q along the ray C  in the direction of b . Now consider  f (s) = f (x, y,  z) =  f {x(s), y(s),  z(s)} Then

∂ f  dx ∂ f  dy ∂ f  dz df  = + + ds ∂x ds ∂ y ds ∂ z ds

s

P

^

b

Q

C

 

...(i)

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VECTOR CALCULUS

df    is called directional derivative of  f  at P in the direction b which gives the rate of  ds change of  f   in the direction of b. dx  dy  dz  i +  j + k  = b  = unit vector Since, ...(ii) ds ds ds Eqn. (i) can be rewritten as Here

 i ∂f  + j ∂f  + k  ∂f   ⋅   dx i + dy  j + dz k      ∂x ∂ y ∂ z     ds ds ds    i ∂ + y ∂ + k  ∂   f  ⋅ b = ∇f ⋅ b   δx δ y δ z      

df  = ds

df  = ...(iii) ds Thus the directional derivative of  f  at P is the component (dot product) of ∇ f  in the direction

of (with) unit vector b. Hence the directional derivative in the direction of any unit vector a is

  a   df  = ∇ f  ·    ds   a    

df  = ∇ f ⋅  n , where n   is the unit normal to the surface  f   = constant. dn

5.9 PROPERTIES OF GRADIENT   ( a ⋅ ∇) f   = a ⋅ (∇f ) 

L.H.S. = ( a ⋅ ∇) f  = =

a ⋅  i ∂ + j ∂ + k  ∂   f     ∂x ∂ y ∂ z      ( a ⋅ i) ∂ + ( a ⋅ j) ∂ + ( a ⋅ k  ) ∂  f   ∂x  ∂ y ∂ z  

= ( a ⋅ i)

∂f  ∂f  ∂f  ) + ( a ⋅ j) + ( a ⋅ k  ∂x ∂ y ∂ z

...(i)

R.H.S. = a ⋅  ∇f 

  ∂f  ∂f  ∂f   = a ⋅  i ∂x + j ∂ y + k ∂ z    

 

∂f  ∂f  ∂f  = ( a ⋅ i) ∂x + ( a ⋅ j) ∂ y + ( a ⋅ k ) ∂ z

From (i) and (ii),

 a ⋅∇  f 



 

= a ⋅ ∇ f   .

...(ii)

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       The necessary and sufficient condition that scalar point function φ  is a constant is that ∇φ = 0.  Let φ(x, y, z) = c ∂φ ∂φ ∂φ =  = =0 ∂ y ∂x ∂ z

Then,

∂φ ∂φ  ∂φ + j  + k  ∂ y ∂x ∂ z = i.0 +  j.0 + k .0 = 0. Hence, the condition is necessary. ∇φ = 0.  Let ∴

∇φ = i

∂φ ∂φ  ∂φ +  j  + k  = 0.i + 0.j + 0.k . ∂ y ∂x ∂ z Equating the coefficients of i,  j, k . ∂φ ∂φ ∂φ On both sides, we get = 0.  = 0, =0 ∂ y ∂x ∂ z ⇒ φ  is independent of x, y,  z ⇒ φ is constant. Hence, the condition is sufficient.

Then,

or

i

            If f   and  g are any two scalar point functions, then ∇ ( f  ±  g) = ∇ f  ± ∇ g grad ( f  ±  g) = grad  f   ± grad  g 

∇ ( f  ±  g) =

 i ∂ + j ∂ + k  ∂  ( f ± g)   ∂x ∂ y ∂ z   

∂ ∂ ∂ = i ∂x ( f ± g ) + j ∂ y ( f ± g) + k ∂ z ( f ± g)

 ∂f  ± ∂ g    +  j  ∂f  ± ∂ g   + k   ∂f  ± ∂ g    = i    ∂ y ∂ y     ∂ z ∂ z   ∂x ∂x   =

 i ∂f    ∂x

+  j

∂g ∂ y

+ k 

  ±  i ∂g      ∂x

∂f  ∂ z

+  j

∇( f  ± g) = grad f  ± grad g.

          If f  and g are two scalar point functions, then ∇( fg) = f ∇ g + g∇ f 

grad ( fg) = f  (grad g) + g (grad f ).

or 

∇( fg) =

 i ∂ + j ∂ + k  ∂  ( fg)   ∂x ∂ y ∂ z   

∂g ∂g + k  ∂ y ∂ z

    

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= i

∂ ∂ ∂ ( fg) + j ( fg) + k  ( fg) ∂x ∂ y ∂ z

  ∂g + g ∂f    + j  f  ∂g + g ∂f   + k   f  ∂g + g ∂f      ∂x ∂x     ∂ y ∂ y      ∂ z ∂ z     ∂g + j ∂g + k ∂g   + g  i ∂f  + j ∂f  + k ∂f   f i   ∂x ∂ y ∂ z      ∂x ∂ y ∂ z   

= i  f  =

∇( fg) = f ∇ g + g∇ f .

          If f  and g are two scalar point functions, then

  f          f   grad  g         f   ∇  g       ∇  g

or 

=

 g∇f − f∇g ,  g ≠ 0  g 2

=

 g(grad f ) − f (grad g ) , ≠ 0.  g 2

= =

 i ∂ + j ∂ + k  ∂     f     ∂x ∂ y ∂ z     g    ∂   f   ∂   f   ∂   f    +  j  + k     i       ∂x   g   ∂ y   g   ∂ z   g    g

= i

 

∂f  ∂g ∂f  ∂g ∂f  ∂g  g − f  − f  − f   g ∂ y ∂x ∂x + j ∂ y + k  ∂ z 2 ∂ z 2 2  g  g  g

 g i =

  f       

∇  g

=

 

 

∂f  ∂f  ∂f  ∂g ∂g ∂g +  j + k  −  f i +  j + k  ∂ z ∂ y ∂ z ∂x ∂ y ∂ z

 g 2  g∇f − f∇g .  g 2

  If r  = xi + yj + zk    then show that

(i)

∇( a ⋅ r) = a , where a  is a constant vector

(ii)

grad r =

(iii) grad (iv)

then

r r

1 r = – 3 r r

grad rn = nrn – 2 r , where r = r .

 (i) Let



 

a

=

a1i + a2 j + a 3 k, r = xi + yj + zk , 

a⋅r

=

( a1i + a2 j + a3 k) ⋅ ( xi + yj + zk)  = a1x + a2 y + a3 z.

(U.P.T.U., 2007)

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∴

∇( a ⋅ r) =

 i ∂ + j ∂ + k  ∂   (a x +   ∂x ∂ y ∂ z    1

a2 y + a3 z)



a1 i + a2 j + a3 k  =   a .  

=

∂ 2 ( x + y 2 + z 2 )1/2 ∂x x  x  =r = Σi 2 2 2 1/2  = Σi r (x + y + z )

grad r = ∆r = Σi

(ii)

  xi + yj + zk   r  = = r . r r  1   ∂  1  r  = −1 r = ∇     = ∂r  r   r r2 r

grad r =

Hence, (iii)

grad

 1    r  

= −

r r3

. 

(iv) Let r  = xi + yj + zk.  grad rn = ∇rn = Σi

Now,

∂ 2 ( x + y 2 + z2 )n/2 ∂x

= n (x2 +  y2 +  z2)n/2–1 = n(x2 +  y2 +  z2)(n–1)/2 = nr n−1

xi + yj + zk  xi + yj + zk  1/2 x 2 + y 2 + z 2 

r r

= nr n − 2 r .   If  f = 3x2 y –  y3 z2, find grad  f   at the point (1, –2, –1). 



grad  f  = ∇ f  =

(U.P.T.U., 2006)

 i ∂ +  j ∂ + k  ∂  ( 3x 2 y − y 3 z 2 )   ∂x ∂ y ∂ z   

= i

∂ ∂ ∂ 3x 2 y − y 3z 2 + j ( 3x 2 y − y 3 z 2 ) + k   ( 3x 2 y − y 3 z 2 ) ∂x ∂ y ∂ z





= i(6xy) +  j(3x2 – 3 y2 z2) + k (–2 y3 z) grad φ  at (1, –2, –1) = i(6)(1)(–2) +  j  [(3)(1) – 3(4)(1)] + k (–2)(–8)(–1) = –12i – 9 j – 16k.   Find the directional derivative of

 Here  f (x,  y,  z) =

Now

1  = r

1 2

2

x +y +z

2

1  in the direction r   where r  = xi +  yj +  zk . r (U.P.T.U., 2002, 2005)

= (x2 +  y2 +  z2)–1/2

  ∂ + j ∂ + k  ∂  x 2 + y 2 + z 2 −1/2   ∂x ∂ y ∂ z   

∇ f  = i

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VECTOR CALCULUS

= = =

∂ 2 x + y2 + z2 ∂x





−1/2

i+

∂ 2 2 2 −1/2 ∂ 2 k   ( x + y 2 + z 2 ) −1/2 j + ∂ z ( x + y + z ) ∂ y

− 1 (x 2 + y 2 + z 2 )−3/2 2xi + − 1 (x 2 + y 2 + z 2 )−3/2 2y   2   2  − xi + yj + zk   3/2 x 2 + y 2 + z 2 

 1   j + − ( x 2 + y 2 + z 2 ) −3/2 2 z  k   2 

and a   = unit vector in the direction of xi +  yj +  zk  =

xi + yj + zk  

As a =

x2 + y2 + z2

∴  Directional derivative = ∇ f ·  a = −

r r

xi + yj + zk   xi + yj + zk   ⋅ 2 2 3/2 (x + y + z ) x 2 + y 2 + z 2 1/2 2

(xi + yj + zk )  2   xi + yj + zk    ⋅ = −  x2 + y 2 + z 2    (x 2 + y 2 + z 2 ) 2 2

= −

  Find the directional derivative of φ = x2 yz + 4xz2 at (1, – 2, –1) in the direction 2i –  j – 2k . In what direction the directional derivative will be maximum and what is its magnitude? Also find a unit normal to the surface x2 yz + 4xz2  = 6 at the point (1, – 2, – 1). φ = x2 yz + 4xz2



∂φ = 2xyz + 4 z2 ∂x ∂φ 2 ∂ y = x  z,

∴

∂φ = x2 y + 8xz ∂ z ∂φ ∂φ ∂φ + j + k  grad φ = i ∂x ∂ y ∂ z

= (2xyz + 4 z2) i + (x2 z)  j + (x2 y + 8xz)k  = 8i –  j – 10k   at the point (1, – 2, – 1) Let a   be the unit vector in the given direction. Then ∴ Directional derivative

a =

2i − j − 2k  4+1+ 4

 =

2i − j − 2 k  3

dφ = ∇φ ⋅  a ds = (8i –  j – 10k ) · =

  2i − j − 2k      3  

37 16 + 1 + 20  = · 3 3

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A TEXTBOOK OF ENGINEERING MATHEMATICS—I

Again, we know that the directional derivative is maximum in the direction of normal which is the direction of grad φ. Hence, the directional derivative is maximum along grad φ = 8i –  j – 10k . Further, maximum value of the directional derivative = |grad φ| = |8i –  j – 10k | =

64 + 1 + 100  = Again, a unit vector normal to the surface =

165 .

grad φ |grad φ|

=

8i − j − 10k 

· 165   What is the greatest rate of increase of u = xyz2  at the point (1, 0, 3)?  u = xyz2 ∴

grad u = i

∂u ∂u ∂u + j + k  ∂x ∂ y ∂ z

= yz2 i + xz2  + 2xyz k  = 9 j at (1, 0, 3) point. Hence, the greatest rate of increase of u  at (1, 0, 3) = |grad u| at (1, 0, 3) point. = |9 j| = 9.   Find the directional derivative of  φ = (x2 +  y2 +  z2)–1/2 at the points (3, 1, 2) in the direction of the vector  yz i +  zx  j + xy k .  φ = (x2 +  y2 +  z2)–1/2 ∴

grad φ = i

∂ 2 ∂ ( x + y 2 + z 2 ) −1/2 + j ( x2 + y 2 + z 2 ) −1/2 ∂x ∂ y

+ k 

∂ 2 ( x + y 2 + z 2 ) −1/2 ∂ z

 1 2 2 2 −3/2 (2 x) + j − 1 ( x 2 + y 2 + z2 ) −3/2 (2 y) = i − ( x + y + z )   2   2 1 + k − ( x 2 + y 2 + z 2 ) −3/2 (2 z)  2 

= −

xi + yj + zk   ( x 2 + y 2 + z 2 )3/2

= −

3i + j + 2k    at (3, 1, 2) (9 + 1 + 4) 3/2

= −

3i + j + 2k  14 14

  at (3, 1, 2)

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VECTOR CALCULUS

Let a  be the unit vector in the given direction, then yzi + zxj + xyk  a =  y 2 z 2 + z 2 x2 + x 2 y 2 = Now,

2i + 6 j + 3 k    at (3, 1, 2) 7

dφ = a . grad φ ds  2i + 6 j + 3 k   ⋅   − 3 i + j + 2 k     =    7     14 14   = − = −

(2 )( 3 ) + (6 )(1) + (3 )(2 ) 7.14 14 18 9  = − · 7.14 14 49 14

  Find the directional derivative of the function φ  = x2 –  y2 + 2 z2  at the point P(1, 2, 3) in the direction of the line PQ, where Q is the point (5, 0, 4).  Here Position vector of P = i + 2 j + 3k  Position vector of Q = 5i + 0 j + 4k  ∴

PQ = Position vector of Q  – Position vector of P = (5i + 0 j + 4k ) – (i + 2 j + 3k ) = 4i – 2 j + k . Let a   be the unit vector along PQ, then 4i − 2 j + k 

a =

Also,

16 + 4 + 1

grad φ = i

 =

4i − 2 j + k  21

∂φ ∂φ ∂φ + j + k  ∂x ∂ y ∂ z

= 2x i – 2 y  j + 4 z k  = 2i – 4 j + 12k   at (1, 2, 3) Hence,

dφ = a . grad φ ds  4i − 2 j + k   ⋅ (2i − 4 j + 12 k ) =    21   = =

( 4)( 2 ) + ( −2)(−4) + (1 )(12 ) 21 28 21

⋅

  For the function φ =

 y x2 + y2

,

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A TEXTBOOK OF ENGINEERING MATHEMATICS—I

find the magnitude of the directional derivative making an angle 30° with the positive X -axis at the point (0, 1).  Here ∴

φ =

 y x2 + y2

∂φ 2 xy = – 2 ∂x ( x + y 2 )2 ∂φ x2 − y2 ∂φ =  and =0 2 2 2 ∂ y (x + y ) ∂ z

∴

grad φ = i = i =

∂φ ∂φ ∂φ + j + k  ∂x ∂ y ∂ z

 ∂φ = 0  ∂ z 

∂φ ∂φ + j ∂x ∂ y −2xy

i+

x2 − y2  j (x 2 + y 2 )2

( x 2 + y 2 )2 = –  j  at (0, 1). Let a  be the unit vector along the line making an angle 30° with the positive X -axis at the point (0, 1), then a = cos 30° i + sin 30°  j. Hence, the directional derivative is given by dφ = a . grad φ ds = (cos 30° i + sin 30°  j) · ( – j) 1 = – sin 30° = – · 2   Find the values of the constants a, b, c  so that the directional derivative of  φ = axy2 + byz + cz2x3 at (1, 2, –1) has a maximum magnitude 64 in the direction parallel to Z-axis. φ = axy2 + byz + cz2x3  ∴

grad φ =

∂φ  ∂φ  ∂φ i +  j + k  ∂x ∂ y ∂ z

= (ay2 + 3cz2x2) i + (2axy + bz)  j + (by + 2czx3) k  = (4a + 3c) i + (4a – b)  j + (2b – 2c) k . at (1, 2, –1). Now, we know that the directional derivative is maximum along the normal to the surface, i.e., along grad φ. But we are given that the directional derivative is maximum in the direction parallel to Z-axis, i.e., parallel to the vector k . Hence, the coefficients of i and  j in grad φ should vanish and the coefficient of k   should be positive. Thus 4a + 3c = 0 ...(i) 4a – b = 0 ...(ii) and 2b – 2c > 0 b > c ...(iii) Then grad φ = 2(b – c) k .

347

VECTOR CALCULUS

Also, maximum value of the directional derivative =|grad φ| ⇒ 64 = |2 (b – c) k | = 2 (b – c) b – c = 32. ⇒ Solving equations (i), (ii) and (iv), we obtain a = 6, b = 24, c  = – 8.

[ b > c] ...(iv)

  If the directional derivative of φ = ax2 y + by2 z + cz2x, at the point (1, 1, 1) has

maximum magnitude 15 in the direction parallel to the line of a, b and c.  Given

 y − 3 x −1  z  =  = , find the values 2 1 −2 (U.P.T.U., 2001)

φ = ax2 y + by2 z + cz2x

∇φ = i

∴

  ∂φ    +  j   ∂φ    + k   ∂φ     ∂x    ∂ y    ∂ z  

= i(2axy + cz2) +  j (ax2 + 2byz) + k  (by2 + 2czx) ∇φ   at the point (1, 1, 1) = i (2a + c) +  j (a + 2b) + k (b + 2c).

We know that the maximum value of the directional derivative is in the direction of ∇ f  i.e., |∇φ| = 15 ⇒ (2a + c)2 + (2b + a)2 + (2c + b)2  = (15)2 But, the directional derivative is given to be maximum parallel to the line. y−3 x −1 = −2 2 2a + c ⇒ 2 ⇒ 2a + c and 2b + a ⇒ 2b + a Solving (i) and (ii), we get

= = = = =

 z 1 2b + a 2c + b  = −2 1 – 2b – a ⇒ 3a + 2b + c = 0 – 2(2c + b) – 4c – 2b ⇒ a + 4b + 4c = 0

a b c =  =  = k   (say) −11 4 10 a = 4k , b  = – 11k , and c = 10k. ⇒ 2 Now, (2a + c) + (2b + a)2 + (2c + b)2 = (15)2 ⇒ (8k  + 10k )2  + (– 22k  + 4k )2  + (20k  – 11k )2 = (15)2 5 9 20 55 50 a = ± ,b= ±  and c = ± · 9 9 9

⇒

k  = ±

⇒

   

  Prove that ∇   f u du  =  f (u) ∇u.  Let

Then

   

 f u du  = F(u), a function of u so that

 



∇   f u du =

∂F =  f (u) ∂u

 i ∂ + j ∂ + k  ∂  F   ∂x ∂ y ∂ z   

(i) (ii)

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A TEXTBOOK OF ENGINEERING MATHEMATICS—I

= i

∂F ∂F ∂F  j  + ∂ y  + k  ∂x ∂ z

= i

∂F ∂u ∂F ∂u ∂F ∂u  +  j  + k  ∂u ∂x ∂u ∂ z ∂u ∂ y

   

    

∂F i ∂u +  j ∂u + k  ∂u ∂x ∂ y ∂ z ∂u = f  (u) ∇u.  

=

  Find the angle between the surfaces x2 +  y2 +  z2 = 9 and  z = x2 +  y2 – 3 at the point (2, – 1, 2) (U.P.T.U., 2002) 2 2 2  Let φ1 = x  +  y  +  z – 9 φ2 = x2 +  y2 –  z – 3 ∴

∇φ1 =

 i ∂ +  j ∂ + k  ∂   (x  +  y  +  z   – 9) = 2xi + 2 yj + 2 zk    ∂x ∂ y ∂ z    2

2

2

∇φ1  at the point (2, – 1, 2) = 4i – 2 j + 4k  

∇φ 2

=

(i)

 i ∂ +  j ∂ + k  ∂     ∂x ∂ y ∂ z   (x  +  y  –  z  – 3) = 2xi + 2 yj – 2

2

k 

∇φ2  at the point (2, – 1, 2) = 4i – 2 j – k  Let θ be the angle between normals ( i) and (ii)

(4i – 2 j + 4k ) · (4i – 2 j – k ) =

16 + 4 + 16

(ii) 16 + 4 + 1  cos θ

16 + 4 – 4 = 6 21  cos θ ⇒ 16 = 6 21   cos θ ⇒

cos θ =

8 8 ⇒ θ  = cos –1 · 3 21 3 21

  If u = x +  y +  z, v = x2 +  y2 +  z2, w =  yz +  zx + xy, prove that grad u, grad v and grad w  are coplanar vectors. (U.P.T.U., 2002)

grad u =



grad v = grad w =

 i ∂ +  j ∂ + k  ∂   (x +  y +  z) =  +  +  i j k    ∂x ∂ y ∂ z     i ∂ +  j ∂ + k  ∂   (x  +  y  +  z ) = 2x  + 2 y j    ∂x ∂ y ∂ z    i  i ∂ +  j ∂ + k  ∂  ( yz +  zx + xy)   ∂x ∂ y ∂ z    2

2

= i ( z + y ) + j ( z + x) + k (y + x ) Now,

1 1 1 2y 2z grad u  (grad v × grad w) = 2 x  z + y z + x y + x

2

 + 2 z k 

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VECTOR CALCULUS

1 1 1 y z = 2 x  z + y z + x y + x 1 = 2 x + z+ y  z + y

1 y + z+x z+x

1 z + y + x |Applying R2 → R2 + R3 y+x

1 1 1 = 2 (x +  y +  z) 1 1 1 =0  y + z z + x x + y Hence, grad u, grad v and grad w are coplanar vectors.   Find the directional derivative of φ = 5x2 y – 5 y 2 z +

5 2  z x  at the point 2

x −1 y − 3 z = = · (U.P.T.U., 2003) −2 2 1 5 2  2  ∇φ = 10 xy + z  i + ( 5x − 10 yz)  j + ( −5 y 2 + 5zx)k  2   25 i − 5j ∇ φ at P(1, 1, 1) = 2 y−3 z  x − 1 Direction Ratio of the line = =   are 2, – 2, 1 2 1 −2 −2 1 2 , Direction cosines of the line are , 2 2 2 2 + −2 + 1 2 2 + −2 + 1 (2) 2 + ( −2)2 + (1) 2 2 −2 1 , i.e., , 3 3 3 Directional derivative in the direction of the line

P(1, 1, 1) in the direction of the line

=

 25 i − 5 j ⋅  2 i − 2 j + 1 k      2    3 3 3  

25 10 + 3 3 35 = · 3 =

 f (r)r  1 d 2    = 2 ( r f ).  r  r dr  xi + yj + zk    f (r)r      ∇⋅  r  = ∇ ⋅  f (r) r  ∂  f (r )x  ∂  f (r )y  ∂  f (r )z  +  +   = ∂  x  r  ∂y  r  ∂z  r  ∂  f (r )x  d  f (r)  ∂r  f (r )   = x   + r ∂x  r  dr  r  ∂x

  Prove that ∇·



Now,

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A TEXTBOOK OF ENGINEERING MATHEMATICS—I

1 df  −  f (r)  x +  f (r)  r dr r 2  r r

= x = Similarly,

    ∂  f (r )z    ∂ z  r 

∂  f ( r ) y r ∂ y

as

∂r x = ∂x r

f (r ) x 2 df (r ) x 2 − 3  f ( r ) + 2 dr r r r

 y 2 df (r )  y 2 f (r ) − 3  f (r ) + = 2 r r dr r

f (r)  z 2 df (r)  z 2 − 3  f (r ) + and = 2 r r dr r Now using these results, we get ∇·

 f (r)r   r 

df (r ) 2 +  f (r ) dr r 1 d = 2 (r 2 f ) ⋅   r dr =

  Find  f (r) such that ∇  f  =

r   and  f (1) = 0. r5

 It is given that ∂ f   ∂ f   ∂f  xi + yj + zk   r i +  j + k  = ∇  f  =  = 5 ∂x ∂ y ∂ z r5 r x ∂ f  ∂ f  ∂ f   y  z So, = 5,  = 5  and  = 5 ∂ y r ∂x ∂ z r r ∂ f   ∂ f   ∂ f  y x z df  = ∂x dx + ∂ y dy + ∂ z dz  = 5 dx + 5 dy + 5 dz We know that r r r xdx + ydy + zdz rdr df  =  =  = r–4dr r5 r5 r −3 +c Integrating f (r) = −3 1 Since 0 = f (1) = − + c 3 1 c = So, 3 1 1 1 Thus, f (r) =  – · 3 r3 3

EXERCISE 5.2  Find grad  f   where  f  = 2xz4 – x2 y  at (2, –2, –1).  Find ∇  f   when  f  = (x2 +  y2 +  z2) e −

x2 + y2 +z2

·

 10i − 4 j − 16 k 

2 − re −r r

351

VECTOR CALCULUS

 Find the unit normal to the surface x2 y + 2xz = 4 at the point (2, –2, 3).

 

1  i − 2 j − 2k   3  2 2  Find the directional derivative of  f  = x  yz + 4xz   at (1, –2, –1) in the direction 37   2i –  j – 2k.  3   Find the angle between the surfaces x2 +  y2 +  z2  = 9 and  z = x2 +  y2  – 3 at the point

  

(2, –1, 2) .

±

 8 21      63  

 θ =cos −1 

 Find the directional derivative of  f  = xy +  yz +  zx in the direction of vector i + 2 j + 2k at 10 the point (1, 2, 0). 3

 

 

 f = x 2 yz 3 + 20

 If ∇ f  = 2xyz3i + x2 z3 j + 3x2 yz2k , find  f  (x,  y,  z) if  f   (1, – 2, 2) = 4.

 f = x 2 + 2y 2 + 4z 2

 Find  f   given ∇ f  = 2xi + 4 yj + 8 zk .

 Find the directional derivative of φ = (x2 + y2 + z2)–1/2 at the point P(3, 1, 2) in the direction

 

of the vector,  yzi + xzj + xyk.  Prove that ∇2 f (r) =  f ′′(r) +  Show that ∇2

−

9 49

  14 

2  f ′ (r) · r

  x     = 0, where  r 3  

r  is the magnitude of position vector r = xi + yj + zk. 

[U.P.T.U. (C.O.), 2002]  Find the direction in which the directional derivative of  f (x,  y) = (x2 –  y 2)/xy at

 

(1, 1) is zero.  Find the directional derivative of

1+ i  2

1   in the direction of r , where r  = xi +  yj +  zk. r 1  − 2  (U.P.T.U., 2003)  r 

 Show that ∇r–3 = – 3r–5 r .  If φ = log | r |, show that ∇φ =

5.10

r · r2

[U.P.T.U., 2008]

DIVERGENCE OF A VECTOR POINT FUNCTION

If  f   (x, y, z) is any given continuously differentiable vector point function then the divergence of 

 f    scalar function defined as

(U.P.T.U., 2006)

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A TEXTBOOK OF ENGINEERING MATHEMATICS—I

∇ ⋅ f  =

5.11

 i ∂ +  j ∂ + k  ∂  ⋅ f = i ⋅ ∂ f  + j ⋅ ∂ f  + k ⋅ ∂ f  = div f    ∂x ∂ y ∂ z    ∂x ∂ y ∂ z

.

PHYSICAL INTERPRETATION OF DIVERGENCE

Let v  = vxi + v y j + v z k   be the velocity of the fluid at P(x,  y,  z). Here we consider the case of fluid flow along a rectangular parallelopiped of dimensions δx, δy, δ z Mass in = v·x δ yδ z (along x-axis) Z Mass out = vx(x + δ x) δ yδ z =

  v + ∂vx δx  δyδz   x ∂x  

S

|By Taylor's theorem Net amount of mass along x-axis

 

= vx δ  yδ   z –  vx +

Vx R z

    

 ∂vx δx δyδz ∂x

∂v x δxδyδz ∂x |Minus sign shows decrease. Similar net amount of mass along  y-axis

Q

= –

∂ y

C

x

B X

Y

 

δxδyδz

and net amount of mass along z-axis = −

 ∂ v z δxδyδz ∂ z

∴  Total amount of fluid across parallelopiped per unit time = −

 ∂vx + ∂vy + ∂vz  δxδyδz   ∂x ∂ y ∂ z   

Negative sign shows decrease of amount ⇒

A

P

O

= –

∂v y

D vx (x + x)

y

Decrease of amount of fluid per unit time =

 ∂vx + ∂v y + ∂vz  δxδyδz   ∂x ∂ y ∂ z   

Hence the rate of loss of fluid per unit volume =

 ∂vx + ∂v y + ∂vz     ∂x ∂ y ∂ z   

= ∇ ⋅ v  = div v. Therefore, div v  represents the rate of loss of fluid per unit volume.

353

VECTOR CALCULUS

 For compressible fluid there is no gain no loss in the volume element ∴ div v = 0

then v  is called Solenoidal vector function.

5.12

CURL OF A VECTOR

If  f    is any given continuously differentiable vector point function then the curl of  f    (vector function) is defined as Curl f   = ∇ ×  f  = i ×

 ∂ f   ∂ f   ∂ f  + j × + k × ∂x ∂ y ∂ z

(U.P.T.U., 2006)

f  = f x i +  f  y j +  f  z k , then

Let

i j k  ∇ ×  f  = ∂ / ∂x ∂ / ∂y ∂ / ∂z ·  fx fy f z

5.13

PHYSICAL MEANING OF CURL

Here we consider the relation v  = w × r , w  is the angular velocity r  is position vector of a point on the rotating body (U.P.T.U., 2006) curl v = ∇ × v = ∇ × (w × r ) = ∇ × [(w1 i + w2 j + w3 k ) × (xi + yj + zk )]  

i = ∇ × w1 x

j w2 y

w = w1i + w2 j + w3k   r = xi + yj + zk   

k  w3 z

= ∇ × [( w2 z − w 3 y)i − (w1 z − w 3 x) j + (w1 y − w2 x )k ]  =

 i ∂ +  j ∂ + k  ∂  × [(w z − w y)i − (w z − w x) j + (w1 y − w2 x)k]  1 3   ∂x ∂ y ∂ z    2 3

i ∂ = ∂x w2 z − w 3 y

j k  ∂ ∂ ∂y ∂z w3 x − w1 z w1 y − w2 x

= (w1 + w1) i – (–w2 – w2) j + (w3 + w3) k  = 2 (w1i + w2 j + w3k ) = 2w Curl v  = 2w  which shows that curl of a vector field is connected with rotational properties of the vector field and justifies the name rotation used for curl.

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A TEXTBOOK OF ENGINEERING MATHEMATICS—I

   If curl  f   = 0, then the vector  f   is said to be irrotational. Vice-versa, if 

 f    is irrotational then, curl  f  = 0.

5.14

VECTOR IDENTITIES grad uv = u grad v + v  grad u

 

grad (uv) = ∇ (uv)



=

 i ∂ +  j ∂ + k  ∂  (uv)   ∂x ∂ y ∂ z   

= i

∂ ∂  ∂ (uv) (uv) + j (uv) + k  ∂x ∂ y ∂ z

  u ∂v + v ∂u    +  j  u ∂v + v ∂u   + k   u ∂v + v ∂u    i =   ∂x ∂x     ∂ y ∂ y       ∂ z ∂ z     ∂v ∂v ∂v   + v i ∂u +  j ∂u + k ∂u   = u i ∂x +  j ∂ y + k  ∂ z       ∂x ∂ y ∂ z    grad uv = u  grad v + v  grad u .

or











 

grad ( a · b ) = a   × curl b  + b   × curl a +  a ⋅ ∇ b + (b ⋅ ∇ ) a



    grad ( a · b ) = Σi ∂ a ⋅ b = Σi ∂a ⋅ b + a ⋅ ∂b ∂x ∂x ∂x

= Now, ⇒ ⇒ ⇒ ⇒

   ∂ b   a ×  i ×     ∂x     ∂b    a ⋅ ∂x   i     ∂b   Σ  a ⋅  i   ∂x       ∂b   Σ  a ⋅  i   ∂x       ∂b   Σ  a ⋅  i   ∂x   

= = = =

                ∂a      ∂b  .   Σi  b ⋅   + Σi  a ⋅     ∂x     ∂x      ∂b    ∂b  a ⋅ ∂x   i − ( a ⋅ i) dx    ∂b    ∂b  a ×  i ×   +  a ⋅ i   ∂x   ∂x      ∂b     ∂b ∑ a ×  i ×   + ∑  a ⋅ i   ∂x   ∂x     ∂b     ∂     a × ∑  i ×   + ∑  a ⋅ i  b   ∂x     ∂x   











...(i)

= a × curl b +  a ⋅ ∇ b ·

...(ii)



Interchanging a  and a , we get    ∂a     Σ  b ⋅  i   ∂x  





= b × curl a + b ⋅ ∇ a

...(iii)

From equations (i), (ii) and (iii), we get  











 







grad ( a  · b ) = a   × curl b  + b   × curl a + ( a  · ∇) b + ( b  · ∇) a

.

355

VECTOR CALCULUS

 

div (u a ) = u div a  + a   · grad u



div (u a ) = ∇·(u a ) =

 i ∂ +  j ∂ + k  ∂  ⋅ (ua )   ∂x ∂ y ∂ z   

= i⋅ = =

(U.P.T.U., 2004)

∂  ∂   ∂  (ua) + j ⋅ (ua) + k ⋅ (ua) ∂x ∂ y ∂ z

   ∂u  ∂a  ∂u  ∂a  ∂u  ∂a     i ⋅  a + u  +  j ⋅  a + u  + k ⋅  a + u  ∂x  ∂ y  ∂ z   ∂x  ∂ z  ∂ y     ∂a  ∂a  ∂a    ∂u ∂u ∂u   u i ⋅ + j ⋅ + k ⋅  + a ⋅ i +  j + k    ∂x ∂ y ∂ z   ∂x ∂ y ∂ z 

div (u a ) = u div a  + a  · grad u .

or  

div ( a  × b ) = b · curl b  – a · curl b



div ( a  × b ) = ∇ · ( a  × b )  ∂  = Σi ⋅ ( a × b ) ∂x = = =

[U.P.T.U. (C.O.), 2003]

 ∂a    ∂b   Σi ⋅   ∂x × b + a × ∂x        ∂b   ∂a      Σi ⋅   ∂x × b   + Σi ⋅  a × ∂x        ∂b     ∂ a      Σ  i ×   ⋅ b − Σ  i ×  ⋅ a   ∂x     ∂x   





= (curl a ) · b  – (curl b) · a 







div ( a  × b ) = b  · curl a  – a  · curl b

or

 



 

 

.



 

curl (u a ) = u curl a  + (grad u) × a



curl (u a ) = ∇ × (u a ) ∂  = Σi × (ua ) ∂x

[U.P.T.U. (C.O.), 2003]

 ∂u a + u ∂a     ∂x ∂x    ∂u     ∂a       Σ  i   × a + uΣ i ×     ∂x     ∂x  

= Σi ×  =





= (grad u) × a  + u curl a or

 





curl (u a ) = u curl a  + (grad u) × a

.

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A TEXTBOOK OF ENGINEERING MATHEMATICS—I

  

 













 



 

curl( a  × b ) = a  div b – b div a + ( b ·∇) a – ( a · ∇) b       curl ( a  × b ) = ∇ × ( a  × b ) ∂   = Σi × ( a × b ) ∂x = =

 ∂a    ∂ b   Σi ×   ∂x × b + a × ∂x        ∂b   ∂a      Σi ×   ∂x × b   + Σi ×  a × ∂x       

  ∂ x  

= Σ ( i. b) ∂a − Σ    i ⋅ ∂ a   b + Σ  i ⋅ ∂b   a − Σ ( i. a ) ∂b ∂x



= Σ (i. b ) 

 

∂x

   

 



   

    



∂x

    

   

    

   ∂ ∂  ∂a   ∂b  a − Σ i⋅ b + Σ i⋅ a − Σ a⋅i b ∂x ∂x ∂x ∂x













 

= ( b · ∇) a – (div a ) b   + (div b ) a – ( a .∇) b          = ( b · ∇) a – ( a  · ∇) b  + a  div b  – b  div a .  











 

∂ 2 f   ∂ 2 f   ∂ 2 f  + 2 + 2 = ∇ 2 f  div grad  f  = ∇ · (∇ f ) = 2 ∂x ∂ y ∂ z



div grad  f  = ∇ · (∇ f ) = = =

 



  

.

 i ∂ + j ∂ + k  ∂   ⋅  i ∂f  + j ∂f  + k ∂f     ∂x ∂ y ∂ z      ∂x ∂ y ∂ z    ∂  ∂ f   ∂  ∂ f   ∂  ∂ f      +    +    ∂x  ∂x   ∂y  ∂ y   ∂z  ∂ z   ∂ 2 f   ∂ 2 f   ∂ 2 f  + + ∂x 2 ∂ y 2 ∂ z 2

div grad  f  = ∇2  f

or

.

curl grad  f  = 0 curl grad  f  = ∇ × (∇ f ) = Σi = = =

or





curl ( a  × b ) = a  div b  – b  div a + ( b  · ∇) a – ( a  · ∇) b

or

   

    

∂f  ∂f  ∂f  ∂ × i + j + k  ∂x ∂x ∂ y ∂ z

  ∂ 2 f  ∂ 2 f  ∂ 2 f    Σi ×  i 2 +  j + k    ∂x ∂x∂y ∂x∂z      ∂ 2 f  ∂ 2 f    Σ  k  −  j   ∂x∂y ∂x∂z     k  ∂ 2 f  − j ∂ 2 f    +  i ∂ 2 f  − k  ∂ 2 f    +   j ∂ 2 f  − i ∂2 f      ∂x∂y ∂x∂z      ∂ y∂z ∂ y∂x      ∂ z∂x ∂ z∂y   

curl grad  f  = 0 .

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

 

div curl  f  = 0



div curl  f  = ∇ · (∇ ×  f  )





  







 ∂ f   ∂ f   ∂ f  ∂ = Σi ⋅ i× +  j × + k × ∂x ∂x ∂ y ∂ z

=

=

=

= 

  

    ∂ 2 f  ∂ 2 f  ∂ 2 f    Σi ⋅ i × 2 +  j × + k ×  ∂ x∂y ∂x∂z    ∂x      ∂ 2 f  ∂ 2 f  ∂ 2 f    Σ (i × i) ⋅ 2 + (i × j ) ⋅ + (i × k ) ⋅  ∂x∂y ∂x∂z   ∂x   ∂ 2 f   ∂ 2 f   Σ k ⋅ −  j ⋅   ∂x∂y ∂x∂z   ∂ 2 f  ∂ 2 f    ∂ 2 f  ∂ 2 f   k ⋅ ∂x∂y −  j ⋅ ∂x∂z  + i ⋅ ∂ y∂z − k ⋅ ∂ y∂x  +    

 ∂ 2 f  ∂2 f    j ⋅ ∂ z∂x − i ⋅ ∂ z∂y   

div curl  f  = 0 . 







 ∂ 2 f   ∂ 2 f   ∂ 2 f  grad div  f  = curl curl  f  + 2 + 2 + 2 ∂x ∂ y ∂ z

 







 Curl curl  f  = ∇ × (∇ ×  f  )

  







 ∂ f   ∂ f   ∂ f  ∂ = Σi × i × +  j × + k × ∂x ∂x ∂ y ∂ z



⇒  Curl curl  f   +

Again,

  

    ∂ 2 f  ∂ 2 f  ∂ 2 f    = Σi × i × 2 +  j × + k ×  ∂x∂y ∂x∂z    ∂x     ∂ 2 f     ∂ 2 f     ∂ 2 f   ∂ 2 f    = Σ  i ⋅ 2  i − (i ⋅ i ) ⋅ 2  +  i ⋅  j − (i ⋅ j)       ∂x∂y   ∂x∂y  ∂x    ∂x        ∂ 2 f    ∂ 2 f   +  i ⋅   k − (i ⋅ k ) ∂x∂z    ∂ ∂ x z          ∂2 f      ∂ 2 f       ∂ 2 f     ∂ 2 f  = Σ  i ⋅ 2  i +  i ⋅  j +  i ⋅  k  − Σ   ∂x     ∂x∂y      ∂x∂z    ∂x 2      ∂ 2 f      ∂ 2 f       ∂ 2 f     ∂ 2 f   ∂ 2 f   ∂ 2 f   j +  i ⋅  = Σ  i ⋅ 2  i +  i ⋅ ...( i) + +  k  · ∂x 2 ∂ y 2 ∂ z 2   ∂x     ∂x∂y      ∂x∂z         ∂f   ∂ f  ∂f     ∂  grad div  f  = Σi i ⋅ +  j ⋅ + k ⋅   ∂x   ∂x ∂ y ∂ z     ∂ 2 f  ∂ 2 f  ∂ 2 f    = Σi i ⋅ 2 +  j ⋅ + k ⋅  ∂ x∂y ∂x∂z    ∂x

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=

=

  ∂ 2 f      ∂ 2 f        ∂ 2 f     Σ  i ⋅ 2  i +  j ⋅ i +  k ⋅  i    ∂x     ∂x∂y      ∂x∂z      ∂ 2 f       ∂ 2 f       ∂ 2 f     k  Σ  i ⋅ 2  i +  i ⋅   j + i ⋅   ∂x     ∂ z∂x     ∂ y∂z    

From eqns. (i (i) and (ii (ii), ), we prove 



ii)) ...(ii ...(



grad div  f  = curl cu curl rl  f   + ∇2  f 







  If  f   = xy2 i + 2x2 yz  yz j  j –  – 3 yz2 k  then   then find div  f   and curl  f   at the point point (1, – 1, 1). 

 We have  f  = xy2 i +2 +2xx2 yz  j – 3 yz2 k   ∂ ∂ ∂ (xy 2 ) + ( 2x 2 yz) + (–3 yz 2 ) div  f  = ∂x ∂ y ∂ z 2 2 = y + 2x  z – 6 yz = (– 1)2  + 2 (1) 2  (1) – 6 (– 1)(1) 1)(1) at (1, – 1, 1) = 1 + 2 + 6 = 9. 

Again,

curl  f  = curl [xy2 i + 2x2 yz  j – 3 yz2 k ]

i ∂ = ∂x xy 2

j k  ∂ ∂ ∂y ∂z 2 2x yz −3yz 2

 ∂ (−3 yz 2 ) − ∂ (2 x 2 yz) + j  ∂ (xy 2 ) − ∂ (−3 yz 2 )   ∂ z ∂ z ∂x   ∂ y  ∂ ∂ + k   ( 2 x 2 yz ) − ( xy 2 ) ∂ y  ∂x 

= i

 y]] +  j  j   [0 – 0] + k  [4 xyz – 2xy xy]] = i [–3 z2 – 2x2 y [4xyz  y))i + (4 xyz – 2xy xy)) k  = (–3 z2 – 2x2 y (4xyz 2 2 = {–3 (1)   – 2(1) (–1)} i  + {4(1)(–1)(1) – 2(1)(–1)} k   at (1, –1, 1) = – i – 2k .   Prove that   (i) div r   = 3. (ii ii)) curl r = 0.    (i) div r = ∇ · r =

 i ∂ +  j ∂ + k  ∂  ⋅ ( xi + yj + zk )    ∂x ∂ y ∂ z   

∂ ∂ ∂ ( x) + ( y) + ( z) ∂x ∂ y ∂ z = 1 + 1 + 1 = 3.   curl r = ∇ × r ∂ ∂ ∂  × ( xi + yj + zk )  = i ∂x +  j ∂ y + k ∂ z  ×

=

(ii ii))

   

=

i ∂ ∂x x

    

j ∂ ∂y y

k  ∂ ∂z z

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VECTOR CALCULUS

 ∂ ( z) − ∂ ( y) + j  ∂ (x ) − ∂ ( z) + k  ∂ ( y ) − ∂ (x )   ∂ z  ∂x  ∂ z ∂x  ∂ y  ∂ y

= i

= i (0) +  j  j (0)  (0) + k  k (0) (0) = 0 + 0 + 0 = 0.   Find the divergence and curl of the vector

(x2 –  y2) i + 2xy  j + ( y  y2 – xy xy)) k . 

 Let

 f  = (x2 –  y2) i + 2xy  j + ( y  y2 – xy xy)) k . 

Then

∂ 2 ∂ ∂ ( x − y 2 ) + (2 xy) + ( y 2 − xy) ∂x ∂ y ∂ z = 2x + 2x  + 0 = 4x 4x

div  f  =

i ∂ curl  f  = ∂x 2 x − y2 

and

j k  ∂ ∂ ∂y ∂z 2 2xy y − xy

 ∂ ( y 2 − xy) − ∂ (2 xy ) + j  ∂ (x 2 − y 2 ) − ∂ ( y 2 − xy )   ∂ z  ∂ z ∂x  ∂ y ∂ ∂  + k   (2 xy ) − ( x 2 − y 2 ) ∂ y  ∂x 

= i

= i [2 y  y –  – x) – 0] +  j  j   [0 – (– y  y)] )] + k  [(2 y  y)) – (– 2 y  y)] )] = (2 y –  y – x) i +  y j + 4 y k .  



  find divergence and curl of  f  (x,  y   If  f  (x,  y  y,,  z  z)) = xz3 i – 2x 2 x2 yz  j + 2 yz4 k   y,,  z  z)) 2006 ) (U.P.T.U., 2006)





div  f  =

=

 i ∂ +  j ∂ + k  ∂  ⋅ xz 3 i − 2 x 2 yz j + 2 yz 4 k     ∂x ∂ y ∂ z    ∂ ∂ ∂ ( xz 3 ) − (2x 2 yz) + ( 2 yz 4 ) ∂x ∂ y ∂ z

= z3 – 2x2 z + 8 yz3

i ∂ curl  f  = ∂x xz 3 

j k  ∂ ∂ ∂y ∂z 2 −2 x yz 2 yz 4

 ∂ (2 yz 4 ) + ∂ (2x 2 yz) − j  ∂ (2 yz 4 ) − ∂ (xz 3 )   ∂x ∂ z ∂ z   ∂ y  ∂ ∂ 2 3  + k  ( −2x yz) − ( xz ) ∂ y  ∂x 

= i

 y)) –  j  j   (0 – 3 z2x) + k  (– 4xyz – 0) = i (2 z4 + 2x2 y = 2 (x2 y  y +  +  z4) i + 3 z2xj – 4xyz xyz··k.

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A TEXTBOOK OF ENGINEERING ENGINEERING MATHEMA THEMATICS—I TICS—I



   Find the directional derivative of ∇. u  at the point (4, 4, 2) in the direction of  

the corresponding outer normal of the sphere x2 +  y2 +  z2  = 36 where u  = x4 i +  y4  j + z4 k . 

∇. u



(∇ f )(4, 4, 2)

∴

= ∇. (x4i +  y4 j  j +  +  z4k ) = 4(x 4( x3 +  y3 +  z3) =  f   (say) = 12 (x2i +  y2 j + z2k )(4, 4, 2) = 48(4i 48(4i + 4 j  j +  + k )

Normal to the sphere  g ≡ x2 +  y2 +  z2  = 36 is (∇ g)(4, 4, 2) = 2 (xi xi +  +  yj + zk )(4, 4, 2) = 4(2i 4(2i + 2 j  j +  + k )

a

= un unit it nor norma mall =





4 2i + 2 j + k  ∇ g = ∇ g 64 + 64 + 16

2i + 2 j + k  3 The required directional derivative is

=

∇ f.  a

2i + 2 j + k  3 = 16(8 16(8 + 8 + 1) = 272. 272.

= 48 (4i  j + (4i + 4 j  + k ). ).

  A fluid motion is given by v  = ( y  y +  + z  z))i + ( z ( z +  + x) j +  j + (x ( x +  + y  y))k  show  show that the motion is irrotational and hence find the scalar potential. ( U.P.T.U., 2003) 2003)

Curl v



= ∆ × v =

 i ∂ + y ∂ + k  ∂  ×  y + zi + z + x j + x + y k    ∂x ∂ y ∂ z   

=

i j k  ∂ ∂ ∂  = i (1 – 1) –  j (1 – 1) + k  k (1 (1 – 1) = 0 ∂x ∂y ∂z  y + z z + x x + y

Hence v  is irrotational. dφ =

Now

= =

∂φ  ∂φ  ∂φ dx + dy + dz ∂x ∂ y ∂ z

 i ∂φ + j ∂φ + k ∂φ   ⋅ ⋅ idx + jdy jdy + kdz    ∂x ∂ y ∂ z     i ∂ +  j ∂ + k  ∂  φ  ⋅ dr = ∇φ ⋅ dr = v⋅ dr   ∂x ∂ y ∂ z   

= [( y  y + z) i + ( z +  z + x) j + (x + y) k ]. ]. (idx idx +  +  jdy  jdy +  + kdz kdz)) = ( y  y +  z  z)) dx + ( z  z + x) dy + (x +  y  y)) dz  zdxx +  zd  zdyy + xdy + xdz +  ydz =   ydx +  zd On integrating

φ =

=

            ydx + xd xdy +

d xy + d yz + d zx

φ = xy +  yz +  zx + c

Thus,

velocity potential = xy +  yz +  zx + c.

  zdx + xdz

zdy + yd ydz +



v = ∇φ

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VECTOR CALCULUS









      



   Prove that a × ∇ × r = ∇ a ⋅ r − a ⋅ ∇ r   where a   is a constant vector and  (U.P.T.U., 2007) r  = xi +  yj +  zk.   Let a = a1i + a2 j + a3k   r = r1i + r2 j + r3k 

∴

∇ × r =

i ∂ ∂x r1

j ∂ ∂y r2

k  ∂r3  ∂r2 ∂r3  ∂r1 ∂r2  ∂r1 ∂  = i ∂ y − ∂ z – ∂x − ∂ z  j + ∂x − ∂ y k  ∂z r3

   

i Now a × ∇ × r = a1 ∂r3  ∂r2 − ∂ y ∂x



j a2 ∂r1  ∂r3 − ∂ z ∂x



        

       

    

k  a3 ∂r2  ∂r1 − ∂x ∂ y

  ∂r2 − ∂r1     ∂r1 − ∂r3    a 2 ∂x a2 ∂ y   –  a3 ∂ z a3 ∂x   i     ∂r2 − ∂r1   −   ∂r3 − ∂r2   –  a1 ∂x a1 ∂ y    a 3 ∂ y a3 ∂ z   j           ∂r3 ∂r1   −  a ∂r3 − a ∂r2  k  +  − a1 ∂x + a1 ∂ z       2 ∂ y 2 ∂ z       ∂ ∂ ∂    ∂ +a ∂ +a ∂  r i+r j +r k  =  i +  j + k    a1 r1 + a2 r2 + a3 r3  – a1      ∂x ∂ y ∂ z    ∂x 2 ∂ y 3 ∂ z  1 2 3 = ∇  a1 j + a2 j + a3 k  ⋅ r1 j + r2 j + r3 k      ∂ ∂ ∂   –  a1 i + a 2 j + a3 k ⋅  i +  j + k    r1 i + r2 j + r3 k    ∂x ∂ y ∂ z    = ∇ a ⋅ r −  a ⋅ ∇r ·   =

  Find the directional derivative of .( f ) at the point (1, –2, 1) in the direction of  the normal to the surface xy2 z = 3x +  z2 where  f  = 2x3 y2 z4. (U.P.T.U., 2008) 

∇ f  =

∇.(∇ f ) =

 i ∂ +  j ∂ + k  ∂   (2x y z ) = (6x y z )i + (4x yz ) j + 8x y z )k    ∂x ∂ y ∂ z     i ∂ +  j ∂ + k  ∂   . {(6x  y z )i + (4x  yz ) j + (8x  y z )k }   ∂x ∂ y ∂ z    3 2 4

2

2 2 4

2 4

3

3

4

4

3 2 3

3

2 3

∇.(∇ f ) = 12x  y2 z4 + 4x3  z4 + 24x3  y2 z2 = F(x, y, z) (say)

or Now

∇F =

 i ∂ +  j ∂ + k  ∂   (12x  y z   ∂x ∂ y ∂ z   

2 4

+ 4x3  z4 + 24x3  y2 z2)

= (12 y2  z4 + 12x2  z4 + 72x2  y2 z2)i  + (24xyz4 + 48x3  yz2)  j + (48x  y2 z3 + 16x3 z3 + 48x3 y2 z)k 

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A TEXTBOOK OF ENGINEERING MATHEMATICS—I

∴

(∇F)(1, –2, 1) = 348i  – 144 j  + 400k  g(x, y, z) = xy2 z – 3x –  z2 = 0

Let

∇ g =

(∇ g)(1, –2, 1)

 i ∂ +  j ∂ + k  ∂   (xy z – 3x –  z )   ∂x ∂ y ∂ z    2

2

= ( y2 z – 3) i + (2xyz) j + (xy2 – 2z) k  = i – 4 j + 2k 

a = unit normal =

∇ g ∇ g

i − 4 j + 2k

=

1 + 16 + 4

=

i − 4 j + 2k   21

Hence, the required directional derivative is ∇F . a = (348i  – 144 j  + 400k ) .

=

348 + 576 + 800 21

=

i − 4 j + 2k  21

1724 . 21

  Determine the values of a and b so that the surface ax2 – byz = (a + 2)x  will be orthogonal to the surface 4x2 y +  z3 = 4 at the point (1, –1, 2).

f  ≡ ax2 – byz – (a + 2)x  = 0  g ≡ 4x2 y +  z3  – 4 = 0

 Let

grad  f  = ∇ f  =

(∇ f )(1, –1, 2)

 i ∂ +  j ∂ + k  ∂   {ax  –   ∂x ∂ y ∂ z    2

byz – (a + 2)x}

= (2ax – a – 2)i + (–bz) j + (–by)k  = (a – 2)i – 2bj + bk 

grad  g = ∇ g =

 i ∂ +  j ∂ + k  ∂   (4x y +  z   ∂x ∂ y ∂ z    2

= (8xy)i + (4x2) j + (3 z2)k  (∇ g)(1, –1, 2) = – 8i + 4 j + 12k  Since the surfaces are orthogonal so

∇ f  . ∇g

...(i) ...(ii)

3

...(iii)

– 4)

...(iv)

= 0

⇒ [(a – 2)i – 2bj + bk ] . [– 8i + 4 j + 12k ] = 0

– 8(a  – 2) – 8b + 12b = 0 ⇒ – 2a + b  + 4 = 0 But the point (1, –1, 2) lies on the surface ( i), so a + 2b – (a  + 2) = 0 ⇒ 2b − 2 = 0 ⇒ b = 1 Putting the value of b in (v), we get − 2a + 1 + 4 = 0 ⇒ a =

Hence, a =

5 , b = 1. 2

5 2

...(v)

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VECTOR CALCULUS

  Prove that  A = (x2 –  yz)i + ( y2 –  zx) j + ( z2 – xy)k   is irrotational and find the

scalar potential  f   such that  A  = ∇ f .

i j k  ∂ ∂ ∂  ∇ ×  A  = = (–x + x)i – (– y +  y) j + (– z +  z) = 0 ∂x ∂y ∂z x 2 − yz y 2 − zx z 2 − xy Hence,  A  is irrotational.

A = ∇ f  = i

Now

∂f  ∂f  ∂f  +  j + k  = (x2 –  yz)i + ( y2 –  zx) j + ( z2 – xy)k  ∂x ∂ y ∂ z

Comparing on both sides, we get ∂ f  ∂x

∴

= (x2 –  yz),

df  =

∂ f  ∂ f  = ( y2 –  zx) and = ( z2 – xy) ∂ z ∂ y

∂ f   ∂ f   ∂ f  dx + dy + dz = (x2 –  yz)dx + ( y2 –  zx)dy + ( z2 – xy)dz ∂x ∂ y ∂ z

= (x2 dx +  y2 dy +  z2 dz) – ( yzdx +  zxdy + xydz) or

df  =

1 d(x3 +  y3 +  z3) – d(xyz) 3

 f  =

1 (x3 +  y3 +  z3) – xyz + c. 3

On integrating, we get

EXERCISE 5.3  Find div  A , when  A  = x2 zi – 2 y3 z2 j + xy2 zk .  If V  =

xi + yj + zk   x2 + y2 + z2

.

2xz – 6 y 2 z 2 + xy 2

(U.P.T.U., 2000)

, find the value of div V .

 

.

x 2 + y 2 + z 2  

2 division

 Find the directional derivative of the divergence of  f  (x, y, z) = xyi + xy2 j + z2k  at the point 13 (2, 1, 2) in the direction of the outer normal to the sphere, x2 +  y2 +  z2 = 9. 3

 



 3

 Show that the vector field  f   = r / r

potential.

.

 

 is irrotational as well as solenoidal. Find the scalar (U.P.T.U., 2001, 2005)

 Ans. 

–

  2 2 2  x + y +z  1

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A TEXTBOOK OF ENGINEERING MATHEMATICS—I

  k × grad 1    + grad   r  

 If r is the distance of a point ( x, y, z) from the origin, prove that curl

  k. grad 1   = 0, where k   is the unit vector in the direction   r  

OZ.

(U.P.T.U., 2000)

 Prove that  A  = (6xy +  z3)i + (3x2 – z)j + (3xz2 –  y) k  is irrotational. Find a scalar function

= 3 x 2 y + xz 3 – zy + c 2

Ans.  f

 

 f (x,  y,  z) such that  A  = ∇ f .

 Find the curl of  yzi + 3 zxj +  zk  at (2, 3, 4).

Ans.  

 

– 6i + 3 j + 8 k 

 If  f  = x2 yz,  g = xy – 3 z2, calculate ∇.(∇ f  × ∇ g).   Ans.  zero  Determine the constants a and b such that curl of (2xy + 3 yz)i + (x2 + axz – 4 z2) j + (3xy

+ 2byz) k  = 0.

Ans.  a

 

= 3, b = −4

 Find the value of constant b such that 3 2 2  A = (bxy –  z ) i + (b – 2)x  j + (1 – b) xz k   has its curl identically equal to zero. Ans.

  1    = a ⋅ r ·  r   r 3 Prove that ∇  a ⋅ u  =  a ⋅ ∇u + a   × curl   x   = 0. Prove that ∇2   3   r  

b =4



 Prove that a . ∇  

u a  is a constant vector.

2 f ′ (r). r

 Prove that ∇2 f (r) =  f ″(r) +

 If u = x2  +  y2 +  z2 and v  = xi + yj + zk  , show that div  Prove the curl

 a × r     r 3    =

uv

= 5u

  

a 3r a ⋅ r − 3 + . r r5





    Find the curl of v  = exyz i + j + k   at the point (1, 2, 3).  Prove that ∇ × ∇ f   = 0 for any  f (x,  y,  z).

 

Ans.

 Find curl of  A  = x 2 yi – 2xzj + 2yzk   at the point (1, 0, 2).





e 6 i − 21 j + 3k  

 

Ans.

4 j

 

 Determine curl of xyz2i +  yzx2 j +  zxy2 k  at the point (1, 2, 3).

Ans.

 Find  f (r) such that  f (r) r  is solenoidal.



   







xy 2z – x i + yz 2x − y j + zx 2y − z k ;10i + 3k  

 

  3 r  c

.

 Find a, b, c when  f  = (x + 2 y + az)i + (bx – 3 y –  z) j + (4x + cy + 2 z)k   is irrotational.

[ a  = 4, b  = 2, c  = –1]  Prove that ( y  –  z + 3 yz – 2x)i + (3xz + 2xy)i + (3xy – 2xz + 2 z)k   are both solenoidal and irrotational. [U.P.T.U., 2008] 2

2

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VECTOR CALCULUS

5.15 VECTOR INTEGRATION Vector integral calculus extends the concepts of (ordinary) integral calculus to vector functions. It has applications in fluid flow design of under water transmission cables, heat flow in stars, study of satellites. Line integrals are useful in the calculation of work done by variable forces along paths in space and the rates at which fluids flow along curves (circulation) and across boundaries (flux).

5.16

LINE INTEGRAL



Let F r  be a continuous vector point function. Then

  C

F ⋅ dr , is known as the line integral of 



F r   along the curve C. Let F  = F1 i + F2 j + F3 k   where F1, F 2, F3 are the components of F  along the coordinate axes and are the functions of x,  y,  z each.    r =   xi + yj + zk  

Now,

   dr =   dxi + dyj + dzk 

∴ ∴

  C

F ⋅ dr =

=

    C C





F1i + F2 j + F3 k ⋅ dxi + dyj + dzk  



F1 dx + F2 dy + F3dz .

Again, let the parameteric equations of the curve C be x = x (t)  y = y (t)  z = z (t) then we can write

 

F ⋅ dr = C

   t2

t1

dy dz  + F2 t + F3 t dt   dx dt dt dt 

F1 t

were t1 and t2  are the suitable limits so as to cover the arc of the curve C. 

work done =

  The line integral

   



C

C

F ⋅ dr 

F ⋅ dr   of a continuous vector point functional F   along a

closed curve C  is called the circulation of F   round the closed curve C. This fact can also be represented by the symbol .    A single valued vector point Function F  (Vector Field F ) is called irrotational in the region R, if its circulation round every closed curve C  in that region is zero that is

    C

or

F ⋅ dr = 0

F ⋅ dr = 0.

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A TEXTBOOK OF ENGINEERING MATHEMATICS—I

5.17

SURFACE INTEGRAL

Any integral which is to be evaluated over a surface is called a surface integral.



Let F r   be a continuous vector point function. Let r  = F (u, v) be a smooth surface such that F (u, v) possesses continuous first order partial derivatives. Then the normal surface integral



of F r  over S is denoted by

   

F r ⋅ da =

S



   

 F r ⋅ ndS

S

where da   is the vector area of an element dS and n   is a unit vector normal to the surface dS. Let F1, F2, F3 which are the functions of x, y, z be the components of F along the coordinate axes, then Surface Integral = = = =

          S S

 F ⋅ ndS

F ⋅ da S

S





 F1 i + F2 j + F3 k ⋅ dydzi + dzdxj + dxdyk 



F1 dy dz + F2 dz dx + F3 dx dy .

5.17.1 Important Form of Surface Integral





 Let dS = dS cos αi + cos βj + cos γ k 

...(i)

where α, β  and γ   are direction angles of dS. It shows that dS  cos α, dS  cos β, dS  cos γ  are orthogonal projections of the elementary area dS on yz. plane, zx-plane and xy-plane respectively. As the mode of sub-division of the surface S is arbitrary we have chosen a sub-division formed  by planes parallel to coordinate planes that is  yz-plane,  zx-plans and xy  plane. Clearly, projection on the coordinate planes will be rectangles with sides dy and dz on  yz plane, dz and dx on xz plane and dx and dy on xy  plane.





 i i ⋅ dS = dS cos αi + cos βj + cos γ kj ⋅

Hence

i ⋅ n dS

= dS cos α = dy dz dy dz dS = . i.  n

Hence

   respectively, we have Similarly, multiplying both sides of (i) scalarly by  j  and k  dz dx dS =  j ⋅ n dx dy and dS = k ⋅ n  dy dz F ⋅ n  F ⋅ ndS Hence = S1 i ⋅ n  dz dx F ⋅ n = S2  j ⋅ n

 

=

     

F ⋅ n

 dx dy

k ⋅ n where S1, S2, S3  are projections of S on  yz,  zx and xy  plane respectively. S3

...(ii)

...(iii) ...(iv) ...(v)

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VECTOR CALCULUS

5.18 VOLUME INTEGRAL



Let F r is a continuous vector point function. Let volume V   be enclosed by a surface S given by

 

r = f u, v sub-dividing the region V  into n elements say of cubes having volumes ∆V 1, ∆V 2, .... ∆V n

...(i)

∆V k  = ∆xk  ∆ yk  ∆ zk 

Hence

k  = 1, 2, 3, ... n where (xk , yk , zk ) is a point say P on the cube. Considering the sum

P

n

∑ Fx k , y k , z k ∆Vk    

 

k =1

taken over all possible cubes in the region. The limits of sum when n → ∞ in such a manner that the dimensions ∆V k  tends to zero, if it exists is denoted by the symbol



V



F r dV ⋅ or

 

V 

FdV or

  

F dx dy dz

V 

is called volume integral or space integral. If 

F

   

=

F r dV  =

V 

F1i + F2 j + F3 k  , then i



V

F1 dx dy dz + j



V

F2 dx dy dz + k

   V 

F3 dx dy dz

where F1, F2, F3  which are function of x,  y,  z  are the components of F   along X , Y , Z  axes respectively.    If in a conservative field F

C F ⋅ dr = 0 along any closed curve C. Which is the condition of the independence of path.

 

F ⋅ dr   where F = x 2 y 2 i + yj   and the curve C, is  y2 = 4x  in the

   Evaluate

C

xy-plane from (0, 0) to (4, 4).  We know that r ∴

dr

∴

F ⋅ dr

∴

 

F ⋅ dr

C

=   xi + yj =   dxi + dyj =

x 2 y 2i + yj ⋅ dxi + dyj

= x2 y2dx +  ydy =

   C

x2 y2 dx + ydy



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A TEXTBOOK OF ENGINEERING MATHEMATICS—I

=



C

x 2 y 2 dx +

  C

y dy ⋅

But for the curve C, x and  y both vary from 0 to 4.

 

∴

C





F ⋅ dr

=

        4xdx +

4 2 x 0 4

4

0

0

4

0

[  y2 = 4x]

y dy

3 = 4 x dx + y dy 4 4   x 4    y 2   4    +      4  0   2  0

=

= 256 + 8 = 264.

 



x dy – y dx   around the circle x2 +  y2 = 1.

   Evaluate

 Let C denote the circle x2 +  y2 = 1, i.e., x = cos t, y = sin t. In order to integrate around C, t varies from 0 to 2 π.

 C x dy − y dy

∴

= = =

 

  Evaluate

C

     2π

0

  x dy − y dx  dt   dt dt  

2π

0

 

2π

0



cos 2 t + sin 2 t dt

dt

= (t)20π = 2π. 





F ⋅ dr , where F = (x2 +  y2) i – 2xy j, the curve C is the rectangle in Y

the xy-plane bounded by  y = 0, x = a,  y = b, x = 0. 

  C





F ⋅ dr =

    C

  



x 2 + y 2 i – 2xyj ⋅ dxi + dyj 2

2





x + y dx − 2xy dy   = OACB. Now, C is the rectangle C  On OA,  y = 0 ⇒ dy = 0

 y=b B (0, b)

...(i)

On CB,  y = b ⇒ dy = 0  On BO, x = 0 ⇒ dx = 0 ∴  From (i),

  C



F ⋅ dr =

   OA

a

2

O (0, 0)



x + 0 dx +



AC

–2aydy +  CB x

    b

2 =   x dx − 2 a y dy + 0

0

0

a x 2 + b2 dx +

b a 0     y 2    x 3 x 3   2   =    − 2a   +  + b x  + 0   3  0   2  0   3  a

x=a

x=0

 On  AC, x = a ⇒ dx = 0



C (a, b)

2

2



+ b dx +

  0

b

0 dy

y=0

 

BO

 

0dy

A (a, 0)

X

369

VECTOR CALCULUS

  Evaluate

  C

a3 a3 2 − − − ab 2 ab = 3 3 2 = – 2ab . 





F ⋅ dr , where F  = yz i + zx j + xy k  and C is the portion of the curve



r = (a cos t) i + (b  sin t)  j + (ct) k   from t = 0 to 

r = (a cos t) i + (b sin t) j + (ct) k .

 We have

Hence, the parametric x =  y =  z = 

Also, Now,

π . 2

  C



dr dt

equations of the given curve are a cos t b sin t ct

= (– a sin t) i + (b cos t)  j + ck 



F ⋅ dr =

= = = = = =

 





dr F ⋅ dt C dt

   yzi + zxj + xyk ⋅  −a sin ti + b cos t j + ckdt   bct sin t i + act cos t j + ab sin t cos t k ⋅ −a sin ti + b cos t j + ckdt   −abc t sin t + abc t cos t + abc sin t cos t dt abc   t cos t sin t sin t cos t dt sin 2 t     abc    t cos 2 t +  dt   2   sin 2 t     abc    t cos 2 t +  dt   2   C C

2

2

C

2

C

−

2

+

C

π 2 0

π

 sin 2t + cos 2t − cos 2t  2 =   abc t 4 4  0  2 abc = t sin 2t 2



  Evaluate

π

02 = 0.

Y

C

 

(4,12)

F. dr  where F  = xyi + (x2 +  y2) j and C

c

is the x-axis from x = 2 to x = 4 and the line x = 4 from  y  = 0 to  y  = 12.  Here the curve C  consist the line  AB  and BC.

r = xi +  yj (as  z = 0)

Since

so

 

F . dr =

C

A

dr = dxi + dyj

  xyi + x

 ABC

O

2

 

+ y 2 j . dxi + dyj



(2, 0)

 

B (4, 0)

X

370

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

=

 



 

xy dx + x 2 + y 2 dy +

 AB

=

 



 

xy dx + x 2 + y 2 dy

BC

 

4

x=2

0 . dx +

 

12 0  y = 0





+ 16 + y 2 dy =

  16 + y dy 12

2

0

12  y3  = 16 y +    = [192 + 576] = 768. 3   0

  If F  = (–2x +  y)i + (3x + 2 y) j, compute the circulation of F  about a circle C in the xy  plane with centre at the origin and radius 1, if C is transversed in the positive direction.  Here the equation of circle is x2 +  y2 = 1 x = cos θ, y  = sin θ |As r = 1 Let

2

2

x + y =1 C

F = (–2cos θ + sin θ)i + (3 cos θ + 2sin θ) j

r = xi +  yj  = (cos θ)i + (sin θ)i So

dr = {(– sin θ)i + (cos θ) j}dθ

Thus, the circulation along circle C =

O

 F . dr

C

=

         2π

 

θ=0

2π

= =

0

2π

0

 

 

  − sin θi + cos θ jdθ

− 2 cos θ + sin θ i + 3 cos θ + 2 sin θ j .



2 sin θ cos θ − sin 2 θ + 3 cos 2 θ + 6 sin θ cos θ dθ



8 sin θ cos θ + 4 cos 2 θ − 1 dθ = 2π

2π

2π

= −2 cos 2θ 0 + sin 2θ 0 + θ 0 = 2π .

 

2π

0

 4 sin 2θ + 2 cos 2θ + 1dθ

= − 2 cos 4π − cos 0 + sin 4 π − sin 0 + 2 π

  Compute the work done in moving a particle in the force field (2xz –  y) j +  zk   along.

is

(i) A straight line from P(0, 0, 0) to Q(2, 1, 3). (ii) Curve C : defined by x2 = 4 y, 3x3 = 8 z from x = 0 to x = 2.  (i) We know that the equation of straight line passing through ( x1, y1, z1) and (x2, y2, z2)

x − x1 y − y1  z − z1 = = x 2 − x1  y 2 − y1  z 2 − z1 ⇒

or

F = 3x2 i +

x − 0 y−0 z−0 = = 2−0 1−0 3−0

⇒

 y z x =  = 1 3 2

 y z x =  = t  (say), so x = 2t, y = t, z = 3t = 1 3 2

371

VECTOR CALCULUS

∴

r = xi +  yj +  zk  = 2ti + tj + 3tk 

⇒

dr = (2i +  j + 3k )dt F = (12t2)i  + (12t2 – t) j + (3t)k 

and The work done =

 

Q

P

F . dr =

=

(ii) F  =

    1

12t 2 i + 12 t2 − t j + 3 t k . 2 i + j + 3 k dt  0 1

0



24t 2 + 12t 2 − t + 9t dt =

  1

0

As t varies from 0 to 1



36t 2 + 8t dt

1  36t 3 8t 2  =  +    = 12 + 4 = 16.  3 2 0   3x 3 − x 2   j + 3x 3 k = 3x 2i +  3x 4 − x 2   j + 3x 3 k  3x 2i +  2x.   4 4    8   8 4    8

x2 3x 3 xi +  yj +  zk  =   = r xi +  j + k  4 8

dr

Work done =



2

x=0

F . dr =

  x 9x 2   k dx =  i +  j +   2 8    2  3x 4 − x 2   j + 3x 3 k  . i + x  j + 9x 2 kdx 3x 2i +      8   2 8  0 4 4     2  3x 5 x 3 27x 5   2 = + − + 3 x dx  0   8 8 64     2  27 x 6  x6 x 4 3 = x + − +   16 32 64 × 6  0

 

 

= 8+

64 16 27 × 64 1 9 − + = 8+ 4− + 16 32 64 × 6 2 2

= 16.   If V   is the region in the first octant bounded by  y2 +  z2  = 9 and the plane

x  = 2 and F = 2x2 yi –  y2 j + 4xz2k . Then evaluate

  V 



∇ ⋅ F dV .

∇ ⋅ F = 4 xy − 2y + 8xz



The volume V  of the solid region is covered by covering the plane region OAB while x varies from 0 to 2. Thus,

         V 

=

2

∇ ⋅ F dV  3

9− y 2

x =0  y =0  z=0

4xy − 2y + 8xzdz dy dx

372

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

= =

=

= =

             2 3

0 0

0 0

0

2

0

9 − y 2 0

dy dx

 dy dx    y   1     4x − 2 −  9 − y   + 4x 9 y −       3     3  9 4 x − 2  + 72x dx

2 3

2

4xyz − 2 yz + 4xz 2





4xy − 2y 9 − y2 + 4 x 9 − y 2

2

18x 2 − 18x + 36x 2

3 2

3

3

dx 0

2 0

= 180.

 

  

  Evaluate

S



 yzi + zxj + xyk ⋅ dS , where S  is the surface of the sphere x2 +  y2 +

 z2 = a2  in the first octant. 

 



  yzi + zxj + xyk ⋅ dS S

= = =

        S S

a

(U.P.T.U., 2005)



yz dy dz + zx dz dx + xy dx dy a2 − z2

0 0

yz dy dz +

a2 − x 2

0 0

zx dz dx +

0 0

=

1 a 2 1  z a − z 2 dz + 2 0 2

−

2

2

  

−

2

2

−

a2 −y 2

xy dx dy

2

 0 xa2 − x2 dx + 12  0 ya 2 − y 2dy a a a 1  a2 z 2 z 4   1  a 2 x 2 x 4   1  a 2 y 2 y 4   −   +  −   +  − 2   2 4  0 2   2 4  0 2   2 4    0 a

a

1 a4 1 a 4 1 a4 3a4 . + + = = 8 2 4 2 4 2 4

F . n ds , where F =  zi + xj – 3 y2  zk   and S  is the surface of the

S

cylinder x2 +  y2  = 16 included in the first octant between  z  = 0 and  z = 5.  Since surface S : x2 +  y2 = 16 Let f  ≡ x2 +  y2 – 16 ∇ f  =

   a

=

=

  

   a



a z a x  x 2   a y  y 2    z 2   a   a   a  0 z  2   0 dz +  0 x  2   0dx +  0 y   2   0dy 2

  Evaluate



 yzi + zxj + xyk ⋅ dy dz i + ⋅ dz dx j + dx dy k 

 i ∂ +  j ∂ + k  ∂   (x  +  y   – 16)   ∂x ∂ y ∂ z   

= 2xi + 2 yi

2

2

37 3

VECTOR CALCULUS

unit normal

n = 

n = 

Now

∇ f  ∇ f 

Z

2xi + 2 yj

=

E

4x 2 + 4y 2

2bxi + yj g

xi + yj

=

2 x2 + y2

C

16

F . n = (zi  + xj  – 3 y 2 zk ). 

xi + yi / 4

=

F G xi + yj I J  = H  4 K 

1 (zx  + xy ) 4

Here the surface S   is perpendicular to xy -plane so we will take the projection of S  on zx -plane. Let R   be that projection.

A

O

X

dx dz dx dz 4dx dz = =  y  y n. j 4 xi + yj 4 F . n ds = zi + xj − 3 y 2 zk   . . dx dz 4  y S

ds  =

∴

D

B

2

2

Y

x +y =16

 



z z 

z z  z z F GH  e



R

=

z z 

=

=

z   z  L z MN 5

4

z=0 x=

5

0

g

16 − x 2 on S . x   is also varies from 0 to 4.

Since z   varies from 0 to 5 and  y  = ∴

b

zx + xy I  dx dz  y J  K 

R

F zx + xy I dx dz GH   y J K  R

j

F  G 0G H 

xz

16 − x 2

I  + xJ dx dz J K 

x2 O − z 16 − x + P dz = 2 Q0 4

2

z b 5

0

4z + 8g dz

= (2z 2 + 8z )05  = 50 + 40 = 90.

z z z 

φdV ,

  Evaluate

where φ = 45x 2  y   and V   is the closed region bounded by

V 

the planes 4 x  + 2 y  + z   = 8, x   = 0,  y   = 0, z  = 0.  Putting  y   = 0, z   = 0, we get 4 x   = 8 or x  = 2 Here x   varies from 0 to 2 y   varies from 0 to 4 – 2 x  and z   varies from 0 to 8 – 4 x  – 2 y  Thus

z z z 

φdV 

=

z z z  z  z  z  z z  z z  b

45x 2 y dx dy dz

V 

V 

=

2

4 − 2x

x = 0  y = 0

= 45

= 45

2 4− 2 x

0 0

2 4 − 2x

0 0

8 − 4x − 2 y

z=0

x2y z

45x 2 y dx dy dz

8 − 4x − 2 y dx 0

dy

x 2 y 8 − 4x − 2 yg dx dy

374

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

         

4− 2x 2 3  2 2 dx 45 x  4 y − 2xy − y  0 3 0  2 2 2 2 3  45 x 2  4 4 − 2x − 2x4 − 2x −  4 − 2x  dx 0 3   2

= =

2

45 3

=

2

0

3

x 2 4 − 2x dx

2



= 15 x 2 64 − 8 x 3 − 96 x + 48x 2 dx 0

2  64x 3 8 x 6 96x 4 48x 5  − − + 15  3 6 4 5 0   512 − 256 − 384 + 1536  15 5  3 3

=

=

= 128.

EXERCISE 5.4  Find the work done by a Force F  =  zi + xj +  yk  from t = 0 to 2 π, where r  = cos t i + sin tj + tk .

  

 



Work done =

 Show that

C

  C



  

F ⋅ dr .

Ans.

F ⋅ dr  = –1, where F  = (cos y) i – xj – (sin y) k  and C is the curve y =

3π

1 − x2

in xy-plane from (1, 0) to (0, 1).  Find the work done when a force F = (x2 –  y2 + x) i – (2xy +  y) j moves a particle from 2   Ans. origin to (1, 1) along a parabola  y2 = x. 3 

 

  C

 

xy3 dS   where C is the segment of the line  y = 2x in the xy plane from  A(– 1, – 2, 0) to

Ans. 

 

B(1, 2, 0).



 

16 5

 



 F  = 2xzi + x 2 − y j + 2z − x 2 k  is   conservative or not. Ans.





2  If F  = 2x − 3z i − 2xyj − 4xk ,  then evaluate

x = 0,  y = 0,  z  = 0 and 2x + 2 y +  z = 4.

 

∇ × F ≠ 0, so non - conservative



∇FdV , where V  is bounded by the plane

V 

 

Ans. 8   3 

375

VECTOR CALCULUS

 Show that

 

  F ⋅ ndS  =

S

 bounded by the planes;

3   where F = 4xzi –  y2 j +  yzk   and S is the surface of the cube 2

Ans. 3   2 

 

x = 0, x = 1,  y = 0,  y = 1,  z = 0,  z = 1. 

2  If F  = 2 yi − 3 j + x k   and S  is the surface of the parobolic cylinder  y2 = 8x  in the first

octant bounded by the planes  y = 4, and  z  = 6 then evaluate

x − yi + x + y j evaluate

 If  A =

  S

 F ⋅ ndS .

 

Ans.  132

  A ⋅ dr around the curve C  consisting of  y = x2 and

Ans. 2   3 

 

 y2 = x.

 Find the total work done in moving a particle in a force field  A  = 3xyi − 5zj + 10xk along  

the curve x = t2  + t,  y =2t2, z = t3 from t = 1 to t = 2.   Ans. 303  Find the surface integral over the parallelopiped x = 0,  y = 0,  z = 0, x = 1,  y = 2,  z = 3 

when  A = 2xyi +  yz2 j + xzk .  Evaluate

 

Ans.

 

  and V   is the closed region in the ∇ × AdV , where  A  = (x + 2 y) i – 3 zj + x k 

V 

Ans. 8 3i − j + 2k    3

 

first octant bounded by the plane 2 x + 2 y +  z = 4.  Evaluate

33

 

 fdV    where  f  = 45x2 y  and V   denotes the closed region bounded by the

V 

planes 4x + 2 y +  z = 8, x = 0,  y = 0,  z = 0.

 

Ans.  128

  V   is the closed region bounded by the planes x = 0, 2x 2 − 3zi − 2xyj − 4xk and Ans. 8  j − k     y = 0,  z  = 0 and 2x + 2 y +  z = 4, evaluate  ∇ × AdV .   V  3 3 3    If  A  = x − yzi − 2x yj + 2 k evaluate ∇ ⋅ AdV  over the volume of a cube of side b. V  Ans. 1 b3   3   If  A  =

 

 Show that the integral

    3, 4

1, 2







xy 2 + y 3 dx + x 2 y + 3xy 2 dy  is independent of the path joining the points (1, 2) and

(3, 4). Hence, evaluate the integral.

 

Ans.

254

 If F = ∇ φ show that the work done in moving a particle in the force field F from (x1, y1,  z1) to B (x2,  y2,  z2) is independent of the path joining the two points.  If F  = ( y – 2x)i + (3x + 2 y) j, find the circulation of F  about a circle C in the xy-plane with centre at the origin and radius 2, if C is transversed in the positive direction.[  8π] 3

 If F (2) = 2i –  j + 2k , F (3) = 4i – 2 j + 3k   then evaluate

 2 F . dFdt dt .

[ 10]

37 6

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

   Prove that F = (4xy  – 3x 2z 2)i  + 2x 2  j  – 2x 3 zk   is a conservative field.    Evaluate z  z F . n ds   where F = xyi  – x 2  j  + (x  + z )k , S   is the portion of plane 2 x  + 2 y  + z  

S

LAns. MN

 

= 6 included in the first octant.

27 O 4

PQ

   Find the volume enclosed between the two surfaces S 1 : z   = 8 – x 2 –  y 2  and S 2 : z 

= x 2 + 3 y 2.

5.19

[ Ans. 8π 2 ]

GREEN’S* THEOREM

If C   be a regular closed curve in the xy -plane bounding a region S   and P (x ,  y ) and Q  (x ,  y ) be continuously differentiable functions inside and on C   then (U.P.T.U., 2007 )

zz  bPdx C

+ Qdy g

=

F ∂Q

zz GH  x ∂

S

−

 ∂ P I 

Y

J dx dy ∂ y K 

y = f2(x)

 Let the equation of the curves  AEB   and AFB  are  y  =  f  1(x ) and  y =  f   2 (x ) respectively.

Consider

 f 2 a xf

b

∂

∂P

P zz   y dx dy = z   z 

x = a  y = f1 a xf ∂ y

S∂

= = = =

a

 y = f1 a xf

−

2 −

−

2

E

−

a

O

b

1

a

X

b

 

−

BFA

=

y = f1(x)

1

a

b

B

dx

z  Pbx, f g Pbx, f g dx z  Pbx, f gdx z Pbx, f gdx z Pbx, yg dx z Pbx, yg dx a

C

A c

 y = f 2 a xf

b

S

dy dx

b

z  Pbx, yg

F

d

AEB

z Pb x, yg dx

−

BFAEB

⇒

∂P

zz   y dx dy = z   Pbx, yg dx S∂

...( i )

−

C

Similarly, let the equations of the curve EAF  and EBF  be x  = f  1 ( y ) and x  =  f  2 ( y ) respectively,

zz 

then

∂Q

S ∂x

dx dy =

= ⇒

zz 

∂Q

S ∂x

dx dy =

* George Green (1793–1841), English Mathematician.

d

 f2 b yg

z  z 

∂Q

 y =c x = f1 b y g ∂x d

dx dy =

C

c

2

− Qb f1 , yg

dy

c

z  z  z  Q bx, yg dy c

d

z  Qb f , yg

Qb f 2 , ygdy + Qb f1 , ygdy d

...(ii )

377

VECTOR CALCULUS

Adding eqns. (i) and (ii), we get

 

Pdx + Qdy



=

C

    

 

F ⋅ dr =

C

Let

F =

Curl F =

   

    

∂Q  ∂P − dx dy ∂x ∂ y

S

  S



∇ × F ⋅ kdS

Pi + Qj, then we have

i ∂ ∂x P

j ∂ ∂y Q

k  ∂  = ∂z 0

 ∂Q − ∂P  k   ∂x ∂ y   

∇ × F ⋅ k  =   ∂∂Qx − ∂∂ yP    

⇒

 

F ⋅ dr =

Thus,

C

  S



∇ × F ⋅ k dS

 Area of the plane region S  bounded by closed curve C. Let Q = x, P = –  y, then

 

xdy − ydx



=

C

       S



1 + 1 dxdy

= 2 S dx dy = 2A area  A =

Thus,

1 ( xdy − ydx ) 2C

  Verify the Green’s theorem by evaluating

 







x 3 − xy 3 dx + y 3 − 2xy dy   where

C

C is the square having the vertices at the points (0, 0), (2, 0) (2, 2) and (0, 2). ( U.P.T.U., 2007)

 

 We have





x 3 − xy 3 dx + y 3 − 2xy dy



C

By Green’s theorem, we have

 

Pdx + Qdy



C

Here ∴

=

 ∂Q − ∂P  dx dy  ∂ y    S  ∂x

 

P = (x3 – xy3), Q = ( y3 – 2xy) ∂P ∂ y

= – 3xy2,

∂Q = – 2 y ∂x

...( A)

378

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

So

 S   ∂∂Qx − ∂∂ yP    dx dy    

= –2

2

x =0

dx

2

 y = 0

         

ydy + 3

2

=

2

x =0  y =0

2

x= 0

2

xdx

 y = 0

Y 2



–2 y + 3xy dx dy

y 2dy C (0, 2)

 y   x2 2  y  = –2 x   + 3    = − 8 + 16  2  0  2 0  3 0  ∂Q − ∂P  dx dy = 8 ⇒ S  ∂x ∂ y    2 2

2 0

3 2

 

O

...(i)



Pdx + Qdy =

    







x 3 − xy 3 dx + y 3 − 2xy dy

C

C

=







x 3 − xy 3 dx + y 3 − 2xy dy +

OA

 







x 3 − xy 3 dx + y 3 − 2xy dy +

BC

∴

along OA,  y along  AB, x along BC,  y along CO, x

 



=0 =2 =2 =0

⇒ ⇒ ⇒ ⇒

dy = dx = dy = dx =





 





x 3 − xy 3 dx + y 3 − 2xy dy



CO

and x  = and  y  = and x  = and  y  =

0 0 0 0

    2

2

0

0

Pdx + Qdy =   x 3dx +

C



x 3 − xy 3 dx + y 3 − 2xy dy

 AB

  

+ But

X

A (2, 0)

(0, 0)

 

Now, the line integral

 

B (2, 2)

0 0 2 2



to to to to

y 3 − 4y dy +

2 2 0 0

  0



  0

x 3 − 8x dx + y 3dy

2

2

 x 4  2 +  y 4 − 2 y 2  2 +  x 4 − 4x 2  0 +  y 4  0   4   4  =  4  4  0  2  2 0  ⇒

= 4 – 4 + 12 – 4

 



Pdx + Qdy = 8

...(ii)

C

Thus from eqns. (i) and (ii) relation ( A) satisfies. Hence, the Green’s theorem is verified.     Verify Green’s theorem in plane for

  C

 boundary of the region defined by  y2 = 8x and x = 2.  By Green’s theorem

 

 Pdx + Qdy 

C

=







x 2 − 2xy dx + x 2 y + 3 dy , where C is the

 ∂Q − ∂P  dx dy  S  ∂x ∂ y   

379

VECTOR CALCULUS

i.e.,

Line Integral (LI ) = Double Integral (DI ) P = x2 – 2xy, Q = x2 y + 3

Here,

∂P ∂ y

= −2x,

∂Q = 2 xy ∂x

So the R.H.S. of the Green’s theorem is the double integral given by DI  = =

      

    

∂Q  ∂P − dx dy ∂x ∂ y

S

 

2 xy − −2 x dx dy

S

The region S is covered with  y varying from –2 2 x  of the lower branch of the parabola to its upper branch 2 2 x  while x varies from 0 to 2. Thus DI  = =

        

8x

2

x = 0  y =− 8 x

2xy + 2xdy dx

2

0

xy 2 + 2xy 3 x 2 dx

8x dx − 8x

128 0 5 The L.H.S. of the Green’s theorem result is the line integral 2

= 8 2

LI  =

 

=







x 2 − 2 xy dx + x 2 y + 3 dy ⋅

C

Here C  consists of the curves OA, ADB. BO. so LI  =

      =

C

=

OA+ ADB + BO

OA

+

ADB

+

 

= LI1 + LI 2 + LI 3 

BO

Y

2   y = –2 2 x , so dy = – dx x LI 1 =

     OA 2





=

 8 x  y  =

B (2, 4)

 2



x 2 − 2xy dx + x 2 y + 3 dy



   2  dx + x 2 −2 2 x  + 3  −   x    1 2  2 −   32  0  5x + 4 2 ⋅ x − 3 2x 2   dx

=   x 2 − 2x –2 2 x dx 0

x=2

D (2, 0)

O (0, 0)

A (2, –4)

 

X

380

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

=

 5x 3  3 +4 

5

2 2 x2 − 3 2⋅ 2 x 5

2   0

40 64 + − 12 3 5 x = 2, dx = 0 =







LI 2 = = 



       ADB 4







x 2 − 2xy dx + x 2 y + 3 dy



4 y + 3 d y = 2 4

–4

y = 2 2 x , with x  : 2 to 0.



2 dx x

dy = LI 3 =

        BO



0

=

= −

2

5x − 4

2





x 2 − 2xy dx + x 2 y + 3 dy 3 2x 2

40 64 + − 12 3 5

LI  = LI 1 + LI 2 + LI 3 =

+ 3 2x

−

1 2

  dx

  40 + 64 − 12   + 24 +  − 40 + 64 − 12   = 128   3 5     3 5   5

⇒ Hence the Green’s theorem is verified.

  Apply Green’s theorem to evaluate

   C







2x 2 − y 2 dx + x 2 + y 2 dy  where C is the

 boundary of the area enclosed by the x-axis and the upper half of the circle x2 + y2 = a2. (U.P.T.U., 2005)  By Green’s theorem 2 2

  C



Pdx + Qdy =

=

      

     a

=

−a 0

a

=

y= a –x

 ∂Q − ∂P  dx dy S  ∂x ∂ y    a a −x  ∂  ∂ x 2 + y 2  − 2x 2 − y 2 dx dy   x =− a  y = 0 ∂ y  ∂x 

−a

2

a 2 −x2

S

2

 2x + 2ydx dy = 

2xy + 2y 2  a  a 2 0 

  a

−



2x a 2 − x 2 + a 2 − x 2 dx = 0 + 2

  a

0

2

 – a

O

a

 

−x 2

dx.



a 2 − x 2 dx

[First integral vanishes as function is odd] a  x3  2 = 2 a x −  3 0 

  

= 2 a3 −

a3 3

  

=

4 3 a . 3

382

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

Let

y = 3 sin θ  in first and  y = sin φ in second integral. L.H.S. = 4 × 81

π 2 0



sin 4 θ cos θ dθ − 4 cos θ

π 2 0

 

sin 4 φ cos φ dφ cos φ

5 1 5 1 = 4 × 81 2 2 − 4 2 2 = 60 π 2 3 2 3 ⇒ L.H.S. = R.H.S. Hence Green’s theorem is verified.   Find the area of the loop of the folium of Descartes

 Let ∴

giving Hence,

x3 +  y3 = 3axy, a > 0. y = tx. x3 + t3x3 = 3ax.tx

...(i)

3at x = 1 + t3

...(ii)

1 required area = 2

          

  y

t=1

x dy − y dx

C

x    O +    y    +   a    =   0   



1 2

=

1 2

=

1 2 x dt , as  y = tx 2 C

=

1 2

C

C

x 2d

C

= 3a 2

= 3a

2

x2

  y     x  

9a 2 t 2

1+t

 

t=0

x dy − y dx

=

x2 ⋅

  x  =

3 2



1

3t 2

0

1 + t3



 

dt  using (ii)

dt,  by summetry

2

− 1 1 = 3 a 2 .  1 + t 3  0 2

  Using Green’s theorem, find the area of the region in the first quadrant bounded x 1  by the curves  y = x,  y = ,  y = . (U.P.T.U., 2008) x 4  By Green’s theorem the area of the region is given by

 A =

  

1 xdy − ydx 2 C

1 = 2



   xdy − ydx  + xdy − ydx  + xdy − ydx    C  C C

 

 

 

1

2

3

...(i)

383

VECTOR CALCULUS

Y

  )  1 ,1   B (  y=1 x

C3   x  =   y

C2 A (2,1) 2

C1

 y = x  4

(0,0) O

X

 

Now along the curve C1 :  y =

 

xdy − ydx



    2

=

C1

    

x x dx − dx = 0 4 4

0

...(ii)

1 1 , dy = − 2 dx and x varies from 2 to 1. x x

along, the curve C2 :  y =

   

xdy − ydx



=

C2

xdy − ydx

or

x dx or dy =  and x varies from 0 to 2. 4 4



C2

x.   1   dx − 1 dx = 2   − x 2   x 

 

 

1

1

−

2

2 dx x

1

= − 2 log x 2  = – 2[log 1 – log 2] = 2 log 3

...(iii)

along, the curve C3 :  y = x, dy = dx and x  varies from 1 to 0.

 

xdy − ydx

  0

=

C3

1



xdx − xdx = 0

...(iv)

Using (ii), (iii) and (iv) in (i), we get the required area  A =

1 [0 + 2 log 2 + 0] = log 2. 2

  Verify Green’s theorem in the xy-plane for

  C

C  is the boundary of the region enclosed by  y = x2 and  y2 = x.  Here P(x, y) = 2xy – x2 Q(x, y) = x2 +  y2 ∂Q = 2x, ∂x

Pdx + Qdy

C



=

 ∂Q − ∂P  dx dy  ∂x ∂ y    S

  





Y

2

x =y 2

y=x

A

∂P = 2x ∂ y

C2

(1, 1) C1

By Green’s theorem, we have

 



2xy − x 2 dx + x 2 + y 2 dy , where

O

...(i)

 

X

384

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

∴

    



2x − 2x dx dy = 0

R.H.S. =

S

and =

 



C







2xy − x 2 dx + x 2 + y 2 dy

L.H.S. =





2xy − x 2 dx + x 2 + y 2 dy +

C1

 







2xy − x 2 dx + x 2 + y 2 dy

...(ii)

C2

Along C1 :  y = x2 i.e., dy = 2xdx and x varies from 0 to 1

 







2xy − x 2 dx + x 2 + y 2 dy

    1

=

0

C1

1

=

0







2xy − x 2 dx + x 2 + y 2 dy

2 x1 2

   1

C2

2

5



4x − x + 2x dx =

0

=

3

dx

Along C2 :  y2 = x, 2 y dy = dx or dy =

 



2x 3 − x 2 + 2x 3 + 2x 5 dx

x 4 − x 3 + x 6 1  3 3  0

= 1

 and x varies from 1 to 0.

dx 2x ⋅ x 1 2 − x 2 dx + x 2 + x . 1 2 2x







 

   

  2x 3 2 − x 2 + 1 x 3 2 + 1 x1 2   dx   1   2 2 0 5   x 3 2 − x 2 + 1 x1 2   dx   1  2 2 0 0 0  5  x5 2  x3  1 x3 2    −  +   2  5 2  1  3  1 2  3 2 1 0

= =

=

= −1+

1 1 − = −1 3 3

Using the above values in (ii), we get L.H.S. = 1 – 1 = 0 Thus L.H.S. = R.H.S.; Hence, the Green’s theorem is verified.

EXERCISE 5.5  Using Green’s theorem evaluate

  C



x 2 y dx + x 2 dy ,  where C  is the boundary described

counter clockwise of the triangle with vertices (0, 0), (1, 0), (1, 1).

 





( U.P.T.U., 2003) 5 Ans. 12

 

 

xy + y 2 dx + x 2dy , where C is the closed curve of  1   Ans. – the region bounded by  y = x2 and  y = x. 20

 Verify Green’s theorem in plane for

C

 

 

385

VECTOR CALCULUS

 Evaluate

 

(cos x sin y − xy)dx + sin x ⋅ cos y dy   by Green's theorem where C  is the circle

C

x2 +  y2  = 1.  Evaluate by Green's theorem

 

[ 0]



e − x sin y dx + e − x cos y dy , where C is the rectangle with

C

vertices (0, 0) ( π, 0),

 π, 1 π   0, 1 π .   2     2  

[ 2(e–π–1)]

 Find the area of the ellipse by applying the Green's theorem that for a closed curve C in the xy-plane. [ Parametric eqn. of ellipse x = a cos φ, y = a sin φ and φ vary from φ1 = 0 to φ2 = 2π] [ π ab]  Verify the Green's theorem to evaluate the line integral

 

(2 y 2 dx + 3x dy) , where C is the

Ans. 27   4 

C

 

 boundary of the closed region bounded by  y = x and  y = x2 .

 Find the area bounded by the hypocycloid x2/3 +  y2/3 = a2/3 with a > 0. O

[ x = a cos3 φ,  y = a sin3 φ, φ varies from φ1 = 0 to φ1 = π/2

 Verify Green's theorem

A]

Ans. 3πa2    8 

 

( 3x + 4 y ) dx + (2x − 3y ) dy  with C; x2 +  y2 = 4.

C

[   Common value: – 8 π] 3

3

 Find the area of the loop of the folium of descartes x  +  y = 3axy, a > 0.

 Evaluate the integral

Ans. 3a2    2 

[Hint: Put  y = tx, t : 0 to ∞]

 

(x 2 − cos hy ) dx + ( y + sin x ) dy ,   where C: 0 ≤ x ≤ π, 0 ≤  y ≤ l.

C

11.

 

[ π  (cos h  1 – 1)]

Verify Green’s theorem in the xy-plane for

  







3x 2 − 8 y 2 dx + 4y − 6xy dy , where C is the

C

region bounded by parabolas  y =

2 x  and  y = x .

12 .

Find the area of a loop of the four - leafed rose r = 3 sin2 θ.

13 .

Verify the Green’s theorem for 1 ≤ x ≤ 1 and –1 ≤  y ≤ 1.

 



 

 

  A 

=

1 2

 0 2 r 2 dθ π

 2  9π  8 

Ans.

=

3

 y 2dx + x 2dy , where C is the boundary of the square –

C

38 6

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

Using Green’s theorem in the plane evaluate

14 .

 

 





2 tan −1 y x dx + log x 2 + y 2 dy , where C

C

is the boundary of the circle ( x – 1)2 + ( y + 1)2 = 4.

5.20

STOKE’S THEOREM



If F   is any continuously differentiable vector function and S is a surface enclosed by a curve C, then

 

F ⋅ dr =

C

 

 . (∇ × F ) ⋅ ndS

S

where n   is the unit normal vector at any point of S. (U.P.T.U., 2006)  Let S  is surface such that its projection on the xy,  yz  and xz  planes are regions  bounded by simple closed curves. Let equation of surface  f (x,  y, z) = 0, can be written as  z = f 1 (x,  y)  y = f 2 (x,  z) x = f 3 ( y, z) 

Let F Then we have to show that

  S

= F1 i + F2  j + F3 k 

 



 ∇ × { F1 i + F2 j + F3 k } ⋅ ndS =

Considering integral

 



F ⋅ dr

C

 ,   we have ∇ × ( F1i ) ⋅ ndS

S

[∇ × (F1 i)]· n dS = =

= 

i ∂ + j ∂ + k  ∂  × F i ⋅ ndS  ∂x ∂ y ∂ z  1     ...(i)    ∂F1  j − ∂F1 k ⋅ ndS  ∂ z ∂ y    ∂F1 n ⋅ j − ∂F1 n ⋅ k dS  ∂ z  ∂ y

 

...(ii)

r = xi +  yj +  zk  = xi +  yj +  f 1(x,  y)k 

Also,



∂r ∂ f  = j + 1 k  ∂ y ∂ y

So,

[As  z =  f (x,  y)]



∂r But   is tangent to the surface S. Hence, it is perpendicular to n . ∂ y

So,

n ⋅

 ∂ r ∂ y

= n ⋅ j +

 ∂ f 1 n ⋅ k  = 0 ∂ y

387

VECTOR CALCULUS

n ·j = −

Hence,

 ∂ f 1  ∂ z n ⋅ k = − n ⋅ k  ∂ y ∂ y

Hence, (ii) becomes

 ∂F1 ∂ z +  ∂F1 n ⋅k dS  ∂ z ∂ y ∂ y 

[∇ × (F1 i)]· n dS = –

...(iii)

But on surface S F1 (x, y, z) = F1 [x, y,  f 1 (x, y)]

∂F1  ∂ F1  ∂ z + ⋅ ∂ y ∂ z ∂ y

∴

= F(x, y)

...(iv)

∂F ∂ y

...(v)

=

Hence, relation (iii) with the help of relation ( v) gives ∂F ∂F dx dy ( n ·k ) dS = – ∂ y ∂ y

[∇ × (F1 i)]·  n dS = –

  S

(∇ × F1 i) ⋅ n dS



 

 ∂F − dx dy R ∂ y

=

...(vi)

where R  is projection of S on xy-plane. Now, by Green’s theorem in plane, we have

 

 





∂F dx dy , Fdx = − R ∂ y C1

where C1  is the boundary of R. As at each point (x, y) of the curve C1 the value of F is same as the value of F1 at each point (x, y,  z) on C and dx is same for both curves. Hence, we have

 

C1



F dx =

 

Hence,

C

    

C

F1 dx . 

∂F dx dy R ∂ y

F1 dx = –

...(vii)

From eqns. (vi) and (vii), we have

 S ∇ × F1i ⋅ n dS =  C F1 dx 

...(viii)

Similarly, taking projection on other planes, we have

 S ∇ × F2 j ⋅ n dS =  C F2 dy .  S ∇ × F3 k ⋅ n dS =  C F3 dz 

...(ix) ...(x)

Adding eqns. (viii), (ix), (x), we get

  S

∇ × {F1 i + F2 j + F3 k } ⋅ n dS =

 



⇒

F ⋅ dr =

C

  C

{F1 dx + F2 dy + F3 dz}

  S

 ( ∇ × F) ⋅ ndS

.

38 8

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

5.21

CARTESIAN REPRESENTATION OF STOKE'S THEOREM 

F = F 1i + F2  j + F3 k 

Let

i  ∂ curl F = ∂x F1

j ∂ ∂y F2

k  ∂ ∂z F3

 ∂F3 − ∂F2  i + ∂F1 − ∂F3   j +  ∂F2 − ∂F1  k   ∂ y ∂ z   ∂ z ∂x   ∂x ∂ y 

= So the relation

 



 



F ⋅ dr =

C

S



curl F ⋅ n dS,

is transformed into the form

  C

{F1 dx + F2 dy + F3dz } =

   

    

   

∂F3  ∂ F2 ∂F1  ∂ F3 dy dz + − − ∂ y ∂ z ∂ z ∂x

S

  dz dx +  ∂F2 −  ∂ F1  dx dy  .   ∂x ∂ y       



  Verify Stoke's theorem for F = (x2 +  y2) i – 2xy  j  taken round the rectangle  bounded by x = ± a, y = 0,  y = b. (U.P.T.U., 2002) 

 We have F ⋅ dr = {(x2 +  y2) i – 2xy  j}· {dx i + dy j} = (x2 +  y2) dx – 2xy dy

 



∴



F ⋅ dr =

C

 

C1











F ⋅ dr + F ⋅ dr + C2

 

C3





 



F ⋅ dr + F ⋅ dr C4

Y



(–a, b) D

A (a, b)

= I 1 + I 2 + I 3 + I 4 ∴

I 1 = = = = I 2 = =

     

C1

( x 2 + y 2 )dx − 2xy dy

C1

−a



( x 2 + b 2 ) dx − 0

a

 C2

  y = b  ∴ dy = 0

  x 3 + b2 x  − a   3  a 2 –   a 3 + 2b 2 a   3   ( x 2 + y 2 )dx − 2xy dy  C

C4

C X



(–a, 0)

B O

C3

(a, 0)

X

 

 

2

  0

b



(−a)2 + y 2 0 − 2(−a) y dy

  0

= 2a

b

y dy

 x = − a ∴ dx = 0

389

VECTOR CALCULUS

= 2a

       

I 3 =

C3

=

  y 2   0   2  b

( x 2 + y 2 )dx − 2xy dy

+a

−a

I 4 =

C4

  y = 0  ∴ dy = 0

x 2 dx

C3

=

= – ab2

2

x

a   x 3   2a3 dx =    =   3  − a 3

x = 0  ∴ dx = 0

−2ay dy b   y 2    y dy = −2a     2  0 0

  b

= –2a

= – ab2

 





F ⋅ dr = I1 + I 2 + I 3 + I 4   C

∴

  2a3 + 2b 2 a   − ab 2 + 2 a 3 − ab2   3   3

= –

= – 4ab2

...(i)

i ∂ curl F = ∂x 2 x + y2

j ∂ ∂y −2xy



Again,

k  ∂ ∂z 0

= – 4 yk  n = k  

∴ ∴

n · curl F = k ·(– 4 yk ) = – 4 y

   S



n ⋅ curl F dS =

         a

b

−a 0

− 4 y dx dy

 y 2 −4 −a 2 +a

=

  b dx  0

= – 2b 2 ( x) a− a = – 4ab2. From eqns. (i) and (ii), we verify Stoke’s theorem.

...(ii)

→ 

  Verify Stoke's theorem when F  =  yi +  zj + xk   and surface S  is the part of the sphere x2 +  y2 +  z2  = 1, above the xy-plane.  Stoke’s theorem is

 





F ⋅ dr =

C

  S



 (curl F) ⋅ ndS

39 0

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

Here, C is unit circle x2 +  y2 = 1,  z = 0 



F ⋅ dr = ( yi +  zj + xk ) · (dxi + dyj + dzk ) = ydx +  zdy + xdz

Also,

 



     



F ⋅ dr =   ydx +

∴

C

C

Again, on the unit circle C,  z dz = Let x = y = and

 



zdy + xdz C

=0 0 cos φ, ∴ dx = – sin φ.dφ sin φ, ∴ dy = cos φ.dφ



F ⋅ dr =

∴

C

C

=

      C

y dx

2π

0

sin φ ( − sin φ) dφ 2π

= –

0

sin 2 φ dφ

= –π 

Again,

curl F =

...(i)

i ∂ ∂x  y

j ∂ ∂y z

k  ∂ ∂z = – i –  j – k  x

Using spherical polar coordinates n = sin θ cos φ i + sin θ sin φ  j + cos θ k  

curl F · n = – (sin θ cos φ + sin θ sin φ + cos θ).

∴

 

Hence,

S



( curl F) ⋅ n dS = –

    

π/2 2 π

θ= 0 φ = 0

 

(sin θ cos φ + sin θ sin φ + cos θ ) sin θ dθ dφ

π/2

= –

θ= 0

[sin θ sin φ − sin θ cos φ + φ cos θ]20 π sin θ dθ

   

π/2

= – 2π

0

sin θ cos θ dθ

π/2

= –π

0

sin 2θ dθ

π (cos 2θ) 0π/2 2 = –π From eqns. (i) and (ii), we verify Stoke’s theorem.

=



...(ii)

  Verify Stoke’s theorem for F = xzi −  yj + x2 yk,  where S  is the surface of the region bounded by x = 0,  y = 0,  z  = 0, 2x +  y + 2 z  = 8 which is not included in the xz-plane. (U.P.T.U., 2006)  Stoke’s theorem states that

 

F ⋅ dr =

C

  S



 ∇ × F ⋅ ndS

391

VECTOR CALCULUS

Here C  is curve consisting of the straight lines  AO, OD and DA. L.H.S. = =

 

+

 

+

OD

DA

   y =

= LI 1 + LI 2 + LI 3  

0,  z = 0, F  = 0, so

F ⋅ dr = 0

 AO

O

Y B (0, 8, 0)

   

LI 2 =

D (0, 0, 4)

 AO +OD + DA

    

   

LI 1 =

 

F ⋅ dr  =

C

 AO

Z



 

 x  = 0,  y  = 0. F   = 0, so X

F ⋅ dr = 0

OD

   



A (4, 0, 0)

 

x +  z  = 4 and  y  = 0, so

F = xzi = x (4 – x) i

LI 3 =

 

F ⋅ dr =

DA

  4

0

x( 4 − x ) i ⋅ dxi =

  4

0

x ( 4 − x ) dx =

32 3

32 32  = 3 3 Here the surface S  consists of three surfaces (planes) S1 : OAB, S2 : OBD, S3 :  ABD, so that LI  = 0 + 0 +

R.H.S. = =

          ^

S

 



i ∂ ∂x xz



(∇ × F ) ⋅ n = SI1 =  

 





+

S1

∇ × F =

  

(∇ × F ) ⋅ n dS =

S2

j ∂ ∂y −y

  z  = 0,

+

S3

= SI 1 + SI 2 + SI 3 

k  ∂  = x2i + x(1 – 2 y) j ∂z x 2y  , so n = − k 

x 2 i + x (1 − 2 y ) j ⋅ ( − k )  = 0

 

 =0 (∇ × F) ⋅ ndS

S1

 Plane x  = 0, n = – i, so

∇ × F = 0

SI2 =  

 



S1 + S2 +S3

 

  S2

 =0 (∇× F) ⋅ ndS

2x +  y + 2 z = 8.

Unit normal n to the surface S3 =

∇(2x + y + 2 z) |∇(2x + y + 2 z)|

39 2

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

n =

2i + j + 2k  4 +1+ 4

 =

2i + j + 2 k  3

2 2 1 x + x (1 − 2 y ) 3 3 To evaluate the surface integral on the surface S3, project S3 on to say xz-plane i.e., projection of  ABD on xz-plane is  AOD (∇ × F ) ⋅ n =

dS = Thus

SI3 =

dx dz dx dz = 3dx dz = 13 n⋅ j

  S3

 (∇× F) ⋅ ndS

 2 x2 + x (1 − 2 y) 3 dx dz   AOD  3 3

=

 

=

    

4− x

4

x= 0  z=0

2x2 + x (1 − 2y) dz dx

since the region  AOD is covered by varying  z from 0 to 4 – x, while x varies from 0 to 4. Using the equation of the surface S3, 2x +  y + 2 z = 8, eliminate  y, then SI3 = =

     

4 4 −x

0 0

4 4− x

0 0

2x 2 + x [1 − 2 (8 − 2x − 2z)] dz dx (6x2 − 15x + 4xz) dz dx

=

6x 2 z − 15xz + 4xz 2  4−x dx   0  2 0

=

 

4

4

0

(23x 2 − 4 x 3 − 28 x ) dx =

32 3

Thus L.H.S. = L.I.  = R.H.S. = S.I. Hence Stoke’s theorem is verified.   Evaluate

  S

 (∇× F) ⋅ ndS  over the surface of intersection of the cylinders x2 +  y2

= a2, x2 +  z2 = a2  which is included in the first octant, given that F = 2 yzi – (x + 3 y – 2) j + (x2 +  z)k .  By Stoke’s theorem the given surface integral can be converted to a line integral i.e., SI =

  S

 ( ∇ × F ) ⋅ ndS =

 

F ⋅ dr = LI

C

Here C is the curve consisting of the four curves C1: x2 +  z2 = a2, y = 0; C2: x2 +  y2 = a2, z = 0, C3: x = 0,  y = a, 0 ≤  z ≤ a: C4: x = 0,  z = a, 0 ≤  y ≤ a.

393

VECTOR CALCULUS

Z 2

2

x +z =a

2

C4

C1

C3 Y 2

2

x +y =a X

2

C2

 

LI  =



F ⋅ dr =

C

 

C1 + C2 + C 3 + C 4

      

=

C1

+

C2

+

C3

+

C4

= LI 1 + LI 2 + LI 3 + LI 4       y  = 0; x2 +  z2 = a2 LI 1 = =   

   z  =

 

F ⋅ dr =

 

( a 2 − z2 ) + z dz = −

C1 0

a

 

C1

(x 2 + z) dz 2 3 a2 a − 3 2

0, x2 +  y2 = a2 LI 2 =

       C2

F ⋅ dr = a

= –

0

C2

− (x + 3y − 2) dy



a 2 − y 2 + 3 y − 2 dy

πa 2 3 2 − a + 2a = – 4 2      x  = 0,  y = a, 0 ≤  z ≤ a

LI 3 = 

  

 

C3

F ⋅ dr =

  a

0

zdz =

a2 2

x  = 0,  z = a, 0 ≤  y ≤ a LI 4 = SI  =

SI  =

      

3a 2 F ⋅ dr = ( 2 − 3 y) dy = −2a + a 2 0

S

∇×F

 ⋅ ndS =

− a2 ( 3 π + 8 a). 12

  −2a 3 a 2   LI  =    3 − 2      πa2 − 3a2 + 2a  + a2 +   −2a + 3a2    +  −   2   4 2 2  

39 4

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

  Evaluate

z  F dr  by Stoke’s theorem, where F ⋅

=

S

C  is y 2 i + x 2 j − ( x + z) k  and   





the boundary of the triangle with vertices at (0, 0, 0) (1, 0, 0) and (1, 1, 0)

( U.P.T.U., 2001 )

 Since z -coordinates of each vertex of the triangle is zero, therefore, the triangle lies in

the xy -plane and n = k  



i

curl F =





j

k 

∂

∂

∂

∂x 2

∂y 2

∂z

 y ∴



=  j



−( x + z)

x

curl F ⋅ n =  j + 2 ( x − y ) k ⋅ k 





+ 2 ( x − y)



=

B (1, 1)

2 (x − y )   y

z  F dr zz curl F n dS ⋅

=

C



Y

The equation of line OB  is  y  = x . By Stoke’s theorem

k 

⋅

  x  =

S



S

=

1 x

z  z  2 (x 0 0

− y)

dy dx

O

A (1, 0)

 

X

x L 1 F  2 x 2 I  1 y2 O 1 = z  2 Mxy − P dx = 2 z  G x − J  dx = z  x 2 dx = . 0 0 H  0 2 Q0 2 K  3 N

1

  Apply Stoke’s theorem to prove that

z ( ydx

2 = −2 2 πa , where C   is the curve given by x 2 +  y 2 + z 2 – 2ax  – 2ay   = 0, x  +  y  = 2a   and begins of the point (2 a , 0, 0).  The given curve C  is x 2 +  y 2 + z 2 – 2ax  – 2ay  = 0 x  +  y  = 2a  C

+ zdy + xdz)

(x  – a )2 + ( y  – a )2 + z 2 = b a 2 g 2 x  +  y  = 2a  which is the curve of intersection of the sphere ⇒

2

(x  – a )2 + ( y  – a )2 + z 2 = b a 2 g and the plane x  +  y  = 2a . Clearly, the centre of the sphere is ( a , a , 0) and radius is a 2 . Also, the plane passes through ( a , a , 0). Hence, the circle C   is a great circle. ∴ Radius

Now,

of circle C   = Radius of sphere =

z ( ydx C

+ zdy + xdz)

=

z ( yi C

2a

+ zj + xk ) ⋅ ( dxi + dyj + dzk )

 

395

VECTOR CALCULUS

= =

But

curl ( yi +  zj + xk ) =

    C

C



( yi + zj + xk ) ⋅ dr

 curl ( yi + zj + xk) ⋅ ndS .

i ∂ ∂x  y

j ∂ ∂y z

[Using Stoke’s theorem]

k  ∂ = – i –  j – k . ∂z x

Since S  is the surface of the plane x +  y = 2a  bounded by the circle C.  Then n =

= ∴

  ∇x + y − 2a

∇ x + y − 2a

i+ j 2

curl ( yi +  zj + xk ) · n = (– i –  j – k ) · = −

Hence, the given line integral =

1+1 2

  i + j      2  

=– 2.

 

= – = –

S

− 2 dS

2   (Area of the circle C). 2 2 2π 2 a  = −2 2 πa .

 

 



xy dx + xy 2 dy   taken round the positively oriented square with

  Evaluate

vertices (1, 0), (0, 1), (– 1, 0) and (0, – 1) by using Stoke’s theorem and verify the theorem.  We have

  C

Y 2

( xy dx + xy dy ) = = =

B (0, 1)

  C

( xy i + xy 2 j ) ⋅ (dx i + dy j )

    C



curl (xy i + xy2  j) =

O



 by Stoke’s theorem, where S  is the area of the square  ABCD.

Now,

A (1, 0)

X (–1, 0)

 curl (xy i + xy 2 j ) ⋅ ndS

i ∂ ∂x xy

x+y=1

C

( xy i + xy 2 j ) ⋅ dr

S

y–x=1

j ∂ ∂y xy 2

= ( y2 – x) k 

k  ∂ ∂z 0

X x–y=1

x + y = –1 (0, –1) D Y



 

39 6

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

curl (xy i + xy2 j ) · n^ = ( y2 – x) k · k  = ( y2 – x)

∴

∴

  S

 = curl (xy i + xy 2 j ) ⋅ ndS

= =

      S

S

S

( y 2 − x ) dS ( y 2 − x ) dx dy y 2 dx dy −

= 4

   

= 4

   

1

0

  S

1− x 2  y dx 0

1

1− x

0

0

x dx dy

dy − Sx

[By symmetry]

 y 2 dx dy − S ⋅ 0

[ x  = x-coordinate of  the C.G. of  ABCD = 0]

                 

= 4

= 4

1

1− x

0

0

1

 y 3 3

0

 y 2 dx dy 1− x

dx

0

4 1 1 (1 − x) 3 dx = . 3 0 3     The given line integral =

=

C

...(i)

( xy dx + xy2 dy ) ,

where C is the boundary of the square  ABCD. Now C can be broken up into four parts namely: (i) the line  AB  whose equation is x +  y = 1, (ii) the line BC whose equation is  y – x = 1, (iii) the line CD  whose equation is x +  y  = – 1, and (iv) the line DA whose equation is x –  y = 1. Hence, the given line integral = =

      AB

1

0

( xy dx + xy 2 dy ) +

 

 

BC

(xy dx + xy 2dy )  +

 

 

CD

( xy dx + xy 2dy ) +

  1

0

DA

(xy dx + xy 2dy )

 −1 x (1 + x) dx + 0 (y − 1) y2dy 1 0 0  0 1 1 +   x ( −x − 1) dx +  − y 2 (1 + y ) dy  +   1 0  0 x ( x − 1) dx +

x (1 − x ) dx +

1

(1 − y ) y 2 dy  +

 

 

 

−

= 2

 

 

−1

x (x − 1) dx + 2 x (1 + x ) dx + 2 0

  1

0

(1 − y) y 2 dy + 2

  0

−1

y 2 (1 + y ) dy

  0

−1

y 2 (1 + y ) dy

 

397

VECTOR CALCULUS

= 2

= 2

= 4 =

  x 2 − x 2   1 + 2   x 3 + x 2   −1 + 2  y 3 − y 4   1 + 2  y 3 + y 4   0   3 2  0   3 2  0   3 4  0   3 4  −1  1 − 1   + 2  − 1 + 1   + 2  1 − 1   + 2  1 − 1    3 2     3 2    3 4    3 4    1 − 1    3 4  

1 · 3

...(ii)

From eqns. (i) and (ii), it is evident that

 





F ⋅ dr =

C

  C



curl F ⋅ n dS

Hence, Stoke’s theorem is verified.   Verify Stoke’s theorem for the function 

F = (x + 2 y) dx + ( y + 3x) dy where C  is the unit circle in the xy-plane. 

 Let 

or

F = F1 i + F2  j + F3 k 

F · dr = (F1 i + F2  j + F3 k ) · (dx i + dy  j + dz k ) = F1dx + F2dy + F3dz Here, F1 = x + 2 y, F2 =  y + 3x, F3 = 0 Unit circle in xy-plane is x2 +  y2 = 1 x = cos φ, dx = – sin φ dφ  y = sin φ, dy = cos φ dφ.

 



Hence,



F ⋅ dr = C

   

( x + 2 y ) dx + ( y + 3x ) dy

2π

= = =

0

[ −(cos φ + 2 sin φ) sin φ dφ  + (sin φ + 3 cos φ) cos φ dφ]

     2π

0

2π

0

0

2π

=

( 3 cos2 φ − 2 sin 2 φ) dφ

      2π

=



− sin φ cos φ − 2 sin 2 φ + sin φ cos φ + 3 cos 2 φ dφ

0



( 3 cos 2 φ − 2(1 − cos2 φ) dφ



5 cos2 φ − 2  dφ

39 8

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

   

 5(1 + cos 2φ) − 2 dφ  2  0 2π  1 5  + cos 2 φ  dφ  0 2 2  2π

=

=

 

1 5 φ + sin 2 φ = 2 2

=

2π  0

1 [ 2 π + 0 ] = π. 2

i j ∂ ∂ curl F = ∂x ∂y x + 2 y y + 3x 

= i {0} –  j {0} + k 

k  ∂ ∂z 0

 ∂ ( y + 3 x) − ∂ (x + 2 y) ∂ y  ∂x 

= k   (3 – 2) = k . Hence,

   S



curl F ⋅ n dS =

= = =

         

(k ⋅ k) d S dS dx dy

xdy

2π

= =

0

1 2

cos2 φ dφ

 

2π

0

 

(1 + cos 2φ) dφ

1 sin 2 φ φ+ = 2 2

= π. So Stoke’s theorem is verified.

 2π  0

399

VECTOR CALCULUS

EXERCISE 5.6

 Evaluate

 



∇ × A ⋅ ndS  where S is the surface of the hemisphere x2 +  y2 +  z2 = 16 above

S

the xy-plane and  A = (x2 +  y – 4) i + 3xyj + (2xz +  z2) k .

 

 

 If F  = ( y2 + z2 + x2) i + ( z2 + x2 – y2) j + (x2 + y2 – z2) k   evaluate

  S





∇ × F ⋅ n dS  taken over Ans.  2πa

 

the surface S = x2 +  y2 – 2ax + az = 0,  z ≥ 0.  Evaluate

S

− 16 π

Ans.

3



∇ ×  yi + zj + xk ⋅ n dS  over the surface of the paraboloid  z = 1 – x2 –  y2, z ≥ 0. Ans.

π

 F = (2x –  y) i –  yz2 j –  y2 zk,  where S upper half surface of the sphere x2 +  y2 +  z2 = 1. Hint:

Here C, x 2 + y 2 = 1, z = 0

Ans.

π

 Using Stoke’s theorem or otherwise, evaluate

   C



2 x − y dx – yz 2 dy − y 2 z dz .

where C  is the circle x2 +  y2 = 1, corresponding to the surface of sphere of unit radius. Ans.

π

 Use the Stoke’s theorem to evaluate

   C











x + 2 y dx + x − z dy + y − z dz .

where C is the boundary of the triangle with vertices (2, 0, 0) (0, 3, 0) and (0, 0, 6) oriented in the anti-clockwise direction.

Ans.  15

 



 Verify Stoke’s theorem for the Function F  = x2 i – xy j integrated round the square in the plane  z  = 0 and bounded by the lines x = 0,  y = 0, x = a,  y = a.

Ans.  Common v alue  

–a3 2

  

 Verify Stoke’s theorem for F = (x2 +  y – 4)i + 3xy j + (2xz +  z2)k   over the surface of  hemisphere x2 +  y2 +  z2 = 16 above the xy  plane.   Ans.  Common value – 16 π  Verify Stoke’s theorem for the function F  =  zi + xj +  yk , where C is the unit circle in xy

plane bounding the hemisphere  z =

1 − x 2 − y 2  .

(U.P.T.U., 2002) Ans.

 Evaluate

  C

Common value π



F ⋅ dr   by Stoke’s theorem for F  =  yzi +  zxj + xyk   and C is the curve of inter-

section of x2 +  y2  = 1 and  y =  z2.

 

Ans.  0

400

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

5.22

GAUSS’S DIVERGENCE THEOREM

If F  is a continuously differentiable vector point function in a region V  and S is the closed surface enclosing the region V , then

   S

F ⋅ n dS =

   

diV F dV 

...(i)

V 

where n   is the unit outward drawn normal vector to the surface S.

(U.PT.U., 2006)



 Let i,  j, k   are unit vectors along X , Y , Z  axes respectively. Then F  = F1i + F2 j + F3k , where F1, F2, F3, and their derivative in any direction are assumed to be uniform, finite and continuous. Let S is a closed surface which is such that any line parallel to the coordinate axes cuts S  at the most on two points. Let  z  coordinates of these points be  z = F1 (x,  y) and  z = F2 (x, y), we have assumed that the equations of lower and upper portions S2 and S1 of S are  z = F2 (x,  y) and  z = F1 (x,  y) respectively. The result of Gauss divergence theorem ( i) incomponent form is

   S F1i + F2 j + F3 k ⋅ nds  =

 ∂F1 + ∂F2 + ∂F3  dV  V   ∂x ∂ y ∂ z   

   

...(ii) Z

Now, consider the integral I 1 = =

    

∂F3 dx dy dz V  ∂ z

n1

F2

K

dS 1 S1

S

 dz dx dy ∂ z 

r2

F1 ∂F3

R

r1

n2

S2  –K

where R is projection of S on xy-plane. I 1 = =

      R

R

R

F1 x , y











R



F3 x, y , F1 − F3 x , y , F2 dx dy F3 x , y , F1 dx dy −

  R



X



F3 x , y , F2 dx dy

For the upper portion S1 of S, dx dy =   k ⋅ n 1 ⋅ dS1 where n 1   is unit normal vector to surface dS1  in outward direction. For the lower portion S2 of S. dx dy = – k ⋅ n 1 ⋅ dS2 where n 2   is unit normal vector to surface dS2  in outward direction. Thus, we have

    R

and

R

Y

O

  F3 x, y , z dx dy F2 x , y 

   





=





= –

F3 x, y, F1 dx dy F3 x, y, F2 dx dy

S1

F3 k ⋅ n 1 dS1

S2

F3 k ⋅ n 2 dS 2

dS 2

 

x

y

401

VECTOR CALCULUS

So

  R





F3 x , y , F1 dx dy −

  R





F3 x , y , F2 dx dy

= = = I 1 =

Hence,

=

  S1





F3 k ⋅ n1 dS1 +

 



 

S2





F3 k ⋅ n 2 dS2

F3 k . n 1 dS 1 + n 2 dS 2



          S

F3 k ⋅ n dS

   nS

= n 1S1 + n 2S 2

∂F3 dx dy dz V  ∂ z

S1

F3 k ⋅ n dS

...(iii)

Similarly, projecting S on other coordinate planes, we have

   

∂F3 dx dy dz = V  ∂ y

∂F1 dx dy dz = V  ∂x Adding eqns. (iii), (iv), (v)

    S

S

F2 j ⋅ n dS

 

...(iv)

 

...(v)

F1 i ⋅ n dS

 ∂F1 +  ∂F2 +  ∂F3  dx dy dz = F1 i . n  + F2  j . n  + F3 k . n  dS  S  V  ∂x ∂ y ∂ z   ∂ +  j ∂ + k  ∂  ⋅ {F i + F  j + F k } dx dy dz ⇒  i  V  ∂x ∂ y ∂ z    F1 i + F2 j + F3 k ⋅ n dS =   S 1

⇒

 



5.23

   S

       

3



div F dV  = V 

= or

2



F ⋅ n ds

=

S S

F ⋅ n dS 

F ⋅ n dS 

div F dV  .

V 

CARTESIAN REPRESENTATION OF GAUSS’S THEOREM 

F = F1 i + F2 j + F3 k  Let where F1, F2, F3  are functions of x,  y,  z. dS = dS (cos α i + cos β  j + cos γ  k ) and where α, β, γ   are direction angles of dS. Hence, dS cos α, dS cos β, dS  cos γ   are the orthogonal projections of the elementary area dS on  yz-plane,  zx-plane and xy-plane respectively. As the mode of sub-division of surface is arbitrary, we choose a sub-division formed by planes parallel to yz-plane,  zx-plane and xy-plane. Clearly, its projection on coordinate planes will be rectangle with sides dy  and dz on  yz-plane, dz  and dx on  zx-plane, dx and dy on xy-plane.

402

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

Hence, projected surface elements are dy dz on  yz-plane, dz dx on  zx-plane and dx dy on xy-plane.

    

∴

S



F ⋅ n dS

=

      S

F1 dy dz + F2 dz dx + F3 dx dy

...(i)

By Gauss divergence theorem, we have S



F ⋅ n dS

=



div F dV .

...(ii)

V 

In cartesian coordinates, dV  =   dx dy dz. Also,





div F

= ∇· F =

Hence,

 



div F dV  = V 

∂F1  ∂ F2  ∂ F3 + + ∂x ∂ y ∂ z

  V 

∂F1  ∂ F2  ∂ F3 + + ∂x ∂ y ∂ z

 dx dy dz  

...(iii)

Hence cartesian form of Gauss theorem is,

     S

=



F1 d y dz + F2 dz dx + F3 dx dy

V 

   2x + 3z i –  xz + y j +  y2 + 2 z k   and

∂F1  ∂F2  ∂F3 + + dx dy dz . ∂x ∂ y ∂ z

 

F ⋅ n dS , where F =    Find S is the S surface of the sphere having centre at (3, – 1, 2) and radius 3. ( U.P.T.U., 2000, 2005)  Let V   be the volume enclosed by the surface S. Then by Gauss divergence theorem, we have

  S

F ⋅ n dS

= = =

       

div F dV 

V 

V 

  

∂ ∂ ∂ 2 − xz − y + 2x + 3z +  y + 2z dV  ∂x ∂ y ∂ z

V 









2 − 1 + 2 dV = 3





 

dV = 3V  

V 

But V   is the volume of a sphere of radius 3. ∴

V  =

Hence

  S

4 π3 3

 3

= 36π.

F ⋅ n dS

= 3 × 36π  = 108π .

  



   Evaluate

S

 y 2 z 2i + z 2x 2 j + z 2 y 2k ⋅ n dS , where S  is the part of the sphere

x2 +  y2 +  z2 = 1 above the xy-plane and bounded by this plane.  Let V   be the volume enclosed by the surface S. Then by divergence theorem, we have

 

  y 2 z 2i + z 2 x 2 j + z 2 y 2k ⋅ n dS = S =

    





div   y 2 z 2 i + z 2 x 2 j + z2 y2 k dV  

V 

V 

∂ 2 2 ∂ 2 2 ∂ 2 2  y z +  z x +  z y ∂x ∂ y ∂ z











  dV  

403

VECTOR CALCULUS



=

V

  

2 zy 2 dV = 2

V 

zy 2 dV 

Changing to spherical polar coordinates by putting x = r sin θ cos φ,  y = r sin θ  sin φ,  z = r cos θ dV  = r2 sin θ dr dθ dφ π  and those of φ will be 2

To cover V , the limits of r will be 0 to 1, those of θ will be 0 to 0 to 2 π. ∴

2

 

 zy 2 dV  = 2

V 

π 2 1

0

0

2

= 2

   

2π

π2

1

0

0

0

2π

π 2

0

0

1 12

 

2π

0

2

2

2

0

     

  S

2π

= 2

=    Evaluate

      r cosθr sin θsin φ r sin θ dr dθ dφ r 5 sin3 θ cos θ sin 2 φ dr dθ dφ

3

sin θ cos θ sin

sin 2 φ ⋅ dφ =

1 12

2

 

2π

0

1  r6  φ   dθ dφ  6 0

sin 2 φ dφ =

π . 12

F ⋅n dS   over the entire surface of the region above the xy-plane

 bounded by the cone  z2 = x2 +  y2 and the plane  z  = 4, if F  = 4xzi + xyz 2 j + 3zk  .  If V   is the volume enclosed by S, then V   is bounded by the surfaces  z = 0,  z = 4,  z2 = x2 +  y2. By divergence theorem, we have = =

            



 z2 − y 2

4  z

0 − z −  z 2 −y 2

= 2

4  z

 z 2 − y 2

0 – z 0

        

= 4

4

0

4

0

S

F ⋅n dS  =

  

∂ ∂ ∂ 4xz + xyz 2 + 3 z dV  = ∂x ∂ y ∂ z

V 

= 2

 

 z

− z



V 

div F dV 

V 



4 z + xz 2 + 3 dV 

4 z + xz2 + 3 dx dy dz  4 z + 3 dx dy dz, 

4 z + 3

4 z + 3



   

z 2 − y 2 dy dz = 4

 y z 2 − y 2 2

2

 

 z 2 − y 2

since

−  z 2 − y 2

    4

0

y z + sin −1 2  z

 z 0

x dx = 0

4z + 3



z 2 − y 2 dy dz

 z  dz  0

2 4   04 4 z + 3 z2 sin 1 1 dz  = 4 × π4 0  4 z 3 + 3z 2  dz   4 4 3 = π 256 + 64 = 320 π . = π  z + z 0

= 4

−

 

404

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

   By transforming to a triple integral evaluate

I  =

   S

x 3 dy dz + x 2 y dz dx + x 2 z dx dy



(U.P.T.U., 2006)

where S  is the closed surface bounded by the planes  z = 0,  z = b  and the cylinder x2 +  y2 = a2.  By divergence theorem, the required surface integral I   is equal to the volume integral

 ∂ x 3 + ∂ x 2 y + ∂ x 2 z  dV    V  ∂x   ∂ y   ∂ z   =

=

=

      b

a

 z =0  y = – a

a2 − y 2  3x 2 + x 2 + x 2  dx dy dz 2 2  x= – a − y 

a − y     2 x3  4×5 x dx dy dz = 20 dy dz    z = 0  y = 0 x = 0  z = 0  y = 0 3   x= 0 3 3 b  a 20 b a 20 20 a 2 − y 2  2 dy dz  = a 2 − y 2  2 z dy =         3  z = 0  y = 0 3  y = 0  3  z =0

      b

a

2

   

a2 − y 2

b

2

a



  a

 y = 0



2

b a −

3 2 2 y



dy .

Put  y = a sin t so that dy = a  cos t dt. ∴ I =

20 b 3

 

π 2 0





a 3 cos 3 t a cos t dt =

20 4 a b 3

  π 2 0

cos4 t dt =

20 4 3 π 5 4 = πa b . a b 3 4.2 2 4



  Verify divergence theorem for F  = (x2 – yz) i + ( y2 – zx) j + ( z2 – xy) k  taken over the rectangular parallelopiped 0 ≤ x ≤ a, 0 ≤  y ≤ b, 0 ≤  z ≤ c. [U.P.T.U. (C.O.), 2006] 



 We have div F  = ∇ · F  = ∴  Volume integral =

 

∂ 2 ∂ 2 ∂ 2 x − yz +  y − zx +  z − xy = 2x + 2y + 2z . ∂x ∂ y ∂ z





∇ ⋅ F dV =

V 



           V 









2 x + y + z dV  

a   x2 = 2      x + y + z d x dy dz = 2  2 + yx + zx dy dz  z =0  y = 0 x = 0  z = 0  y = 0   x =0 b  c b c    y 2 a2 a2 = 2  2 + ay + az dy dz = 2  z =0 2  y + a 2 + azy dz  z = 0  y = 0    y =0 c  a 2b + ab 2 + abz dz = 2 2 2   z = 0   c  a2b ab 2  z 2  = 2   z +  z + ab  = [a bc + ab c + abc ] = abc (a + b + c). 2 2 2 0  c

b

a

c

     

2

b

2

2

   Now we shall calculate

  S



F ⋅ n dS

Over the six faces of the rectangular parallelopiped. Over the face DEFG,

n

= i, x = a.



405

VECTOR CALCULUS

            

Therefore, c

=

b

b

 z = 0  y = 0

 

 

a 2 − yz i + y 2 − 2a j + z 2 − ay k ⋅ i dy dz

=

 z = 0

 a 2 b − zb  dz =  a 2 bz − z    2  4  

D

2

 = a 2 bc – c b  4 0 c

= =

    

 ABCO

c

b

 z = 0  y = 0

      

E

b2



2

2

A

O

2 2

.

G

Y

F

X

Over the face  ABCO, n = – i, x = 0. Therefore F ⋅n

B

=

2

c

=

C

   b c  2  y 2  2 a − yz dy dz =    a y − z  dz  z 0  2    y 0

 z = 0  y = 0 c

=

DEFG

Z

F ⋅n dS

 

 

0 − yz i + y j + z k ⋅ −i dy dz

dS  =

yz dy dz =

c

 z = 0

y 2z 2

b dz = c b 2  zdz = b 2 c 2   z = 0 2 4  y =0

 

Over the face  ABEF, n  =  j,  y = b. Therefore

        

 ABEF

=

c

F ⋅n dS  =

a

 z = 0 x = 0

     c

a

x 2 − bz i + b 2 − zx  j + z 2 − bx k

 z =0 x = 0

⋅ j dx dz

a 2 c2 . b − zx dx dz = b ca − 4



2

2

Over the face OGDC, n  =  – j,  y  = 0. Therefore

 

    



OGDC

c 2a2  zx dx dz = . x =0 4

c

F ⋅ n dS  =

a

 z = 0

Over the face BCDE, n  = k ,  z = c. Therefore

 



BCDE

F ⋅ n dS =

     b

a

 y = 0 x =0



2



c − xy dx dy = c 2ab −

a 2b 2 · 4

Over the face  AFGO, n  =  – k ,  z = 0. Therefore

 



 AFGO

F ⋅ n dS  =

     b

a

 y = 0 x = 0

xy dx dy =

a2b 2 · 4

Adding the six surface integrals, we get

  S



F ⋅n dS =

 a 2bc −  c 2b2 + c 2b 2   +  b 2 ca −  a 2 c 2 + a 2 c 2   +  c 2ab − a 2b 2 + a 2b 2     4 4   4 4   4 4            

= abc (a + b + c). Hence, the theorem is verified. 

  If F = 4xzi –  y2 j +  yzk   and S  is the surface bounded by x  = 0,  y  = 0,  z = 0,

x = 1,  y = 1,  z = 1, evaluate

  S



F ⋅n dS .

406

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

 By Gauss divergence theorem,

  S



F ⋅ n dS

= = =

              V 

V

V

1

= =



∇ ⋅ F dV ,   where V   is the volume enclosed by the surface S

 

1

x = 0  y = 0

2

2 z 2 − yz

y2 2 y − 2

x =0

2

  





   Evaluate 2



4 z − y dx dy dz =

1

   4z − 2y + y dV  V   1 1 1  4z − y dx dy dz x =0  y = 0 z = 0

∂ ∂ ∂  yz dV = −  y 2 + 4xz + ∂x ∂ y ∂ z

1  z = 0

                

dx dy =

1 dx =   y =0

1

0

2−

1

1

x =0

y =0

2 − y dx dy

1 3 dx = 2 2

  

1

0

dx =

3 . 2



 y 2 z 2 i + z 2 x 2 j + z 2 y2 k ⋅ n dS,   where S  is the part of the sphere

S

x  +  y  +  z   = 1, above the xy-plane and bounded by this plane. 

 We have

= y2 z2 i +  z2x2  j +  z2 y2 k 

F 

div F

∂ 2 2 ∂ 2 2 ∂  y z +  z x +  z 2 y 2 ∂x ∂ y ∂ z



=











= 2 zy2 ∴

   

Given integral =

V 

2 zy 2 dV 

[By Gauss’s divergence theorem]

where V   is the volume enclosed by the surface S, i.e., it is the hemisphere x2 +  y2 +  z2 = 1, above the xy-plane. Z

Y 2

z= 1–x –y

2

y= 1–x

2

dx dy dz dx dy

Z=0

X

O

X

O

X

dx dy

y=–1–x Y

Y

(i)

(ii)  

From the above Figure 5.28 ( i) and (ii), it is evident that limits of  z are from 0 to

1 − x2 − y2

limits of  y  are from – 1 − x2 to and limits of x  are from –1 to + 1.

1 − x 2

2

407

VECTOR CALCULUS

∴ Given integral

= 2

= 2 =

=

=

=

=

    1

 

1− x 2

1 −x 2 − y 2

x = –1  y = – 1 −x 2  z = 0

    1

  z 2   1−x − y   2    0 2

1− x 2

x = –1  y = – 1 −x

       

1− x 2

1

 zy 2dx dy dz

x = –1  y = – 1−x

2

2

2

 y 2 dx dy

1 − x2 − y 2  y2dx dy

1− x  y y 1 − x2  −  dx x = –1 3 5  – 1 x −  1 2 25 1 2 25  1 2    1 − x  − 5 1 − x   dx x =− 1  3   2 5/2  1   2  5/2 x − 1    dx  1 x − 4   − 0 5  3  3

1

8 15

8 = 15

2

       1

1 − x2

0

π 2 0

2

5



5 2

dx

2

1 − sin θ



5 2

cos θ dθ

[Putting x = sin θ]

π

8 2 = cos2 θ dθ 15 0 8 5π π ⋅ = = · 15 32 12

 





F ⋅ n dS  where F  = (x + y2) i – 2x j + 2 yz k  where S is surface bounded S  by coordinate planes and plane 2x +  y + 2 z = 6.  We know from Gauss divergence theorem,   Evaluate

  



F ⋅ n dS = S 

   

V 



div F dV 

F = (x +  y2) i – 2x  j + 2 yz k  

div F = =

 i ∂ +  j ∂ + k  ∂  ⋅ x + y 2 i − 2x j + 2yz k       ∂x ∂ y ∂ z     ∂ ∂ ∂ x + y2  + –2x +   2 yz  ∂x ∂ y ∂ z

= 1 + 2 y

         



Let

I  = = =

S

F ⋅ n dS =

V 

   

V 



1 + 2y dV 



1 + 2 y dx dy dz



div F dV  

408

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

6 − 2x − y 2 Limit of  y  is 0 to 6 – 2x Limit of x  is 0 to 3 Limit of  z  is 0 to

Hence,

I  = =

              

 6 − 2 − /2 1 + 2 y   z0 x y  dx dy 1 + 2 y dx dy dz

1 2 1 = 2 =

= =





6 − 2x + 11y − 4xy − 2 y 2 dx dy

6 2x dx 0 1 11 2 2 2 3 6 6 − 2x − 2 x 6 − 2x + 6 − 2x − 2x  6 − 2x −  6 − 2x  dx  2 2 3  1 8  − x 3 + 26x 2 − 84x + 90  dx 2 3  3 1  2 4 26 3  2 − x + x − 42x + 90x  2  3 3 0

1 = 2 =



1 + 2 y 6 − 2x − y dx dy

6 y − 2xy +

11 2 2 y − 2xy 2 − y 3 2 3

−

1 − 54 + 234 − 378 + 270 2 1 = 72 = 36 . 2    Verify Gauss divergence theorem for =

   S



x 3 − yz dy dz − 2x 2 y dz dx + z dx dy



over the surface of cube bounded by coordinate planes and the planes x =  y =  z = a 

 Let F = F1 i + F2  j + F3 k . From Gauss divergence theorem, we know

   S



F ⋅ n dS

=

   S

F1 dy dz + F2 dz dx + F3 dx dy =

Here,

F1 = x3 –  yz, F2 = – 2x2 y, F3 =  z

So,

F



= =

Hence,

   S



F ⋅ n dS

V 

= (x3 –  yz) i – 2x2 y  j +  z k 



div F

   



div F dV  

 i ∂ +  j ∂ + k  ∂  ⋅ x 3 − yz i − 2x 2 y j + z k       ∂x ∂ y ∂ z     ∂ 3 ∂ ∂ x − yz + – 2x2 y +  z  ∂x ∂ y ∂ z

= 3x2 − 2x2 + 1 = x2 + 1 =

    V 



x 2 + 1 dV 

...(i)

409

VECTOR CALCULUS

= =

                a a a

a a

 a 3  2a a  + x  3 0  a 3  a 5 3 a2 + a  = +a  3  3

= a = =

 dx dy  y0a dx

x2 + 1 0 0 a a x2 + 1 0 a 2 2

= a =

 a + 1 z 0 dx dy

x 2 + 1 dx dy dz

0 0 0 a a x2 0 0

x + 1 dx

0

...(ii)

    Outward drawn unit vector normal to face OEFG is – i and dS is dy dz. If I 1 is integral along this face,

I 1 = = = =

                        S





F ⋅ n dS =

F ⋅ − i dy dz

S

x 3 − yz dy dz

S

a a

0 0

[As x  = 0 for this face]

yz dy dz

z2 y 0 2 a

a

dy

0

a2 y2  y dy = 0 2 2

a2 = 2

a

a

0

a4 = 4

For face  ABCD, its equation is x = a and n dS = i dy dz , If I 2 is integral along this face

               

I 2 = = = = =

S

S

0 0

0

a

0

k G



x 3 − yz dy dz

a a

a

Z

F ⋅ i dy dz

D



a 3 − yz dy dz

 dy 0  a 2  a 3 a − y  dy 2 a3 z − y

 z 2  z

F

 –i

C  j

a

a

 –j a O

a

E

A B X

i

 –k

 

Y

410

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

=

a 4 y − a 2 y 2  a   2 2  0

a4 4 If I 3  is integral along face OGDA  whose equation is  y = 0

= a5 −

n dS = – j dxdz

     

Hence,

I 3 =

S

= –



F ⋅ − j dx dz

S

– 2x 2 y dx dz

= 0, as  y = 0. If I 4  is integral along face BEFC  whose equation is  y = a n dS =

Then

j dx dz

I 4 = –

  S

2 x 2 y dx dz

         a a

x 2 dx dz

=

– 2a

=

– 2a x 2 z

=

– 2a 2 x 2 dx

0 0 a

0

a 0

dx

a

0

=

– 2a

 x3 a = 3  0

2

–

2 5 a . 3

If I 5  is integral along face OABE  whose equation is  z = 0 n dS = – k dx dy

     

I 5 = =

S

F ⋅ – k dx dy



 z dx dy = 0 as z = 0.

–

S

If I 6  is integral along face CFGD  whose equation is  z = a n dS =   k dx dy

I 6 =

   

= a Total surface

S

z dx dy =

a

0

y

a 0

     a a

0 0

dx = a 2

a dx dy a

0

dx = a 3

I  = I 1 + I 2 + I 3 + I 4 + I 5 + I 6

411

VECTOR CALCULUS

=

a4 a4 2 + a5 – + 0 – a5 + 0 + a3 4 4 3

a5 + a3 3 which is equal to volume integral. Hence Gauss theorem is verified.

=

...(iii)

   Evaluate by Gauss divergence theorem

   S









xz 2 dy dz + x 2 y − z 3 dz dx + 2xy + y 2z dx dy



a2 − x 2 − y 2 .

where S is surface bounded by  z = 0 and  z = 

 Let F = F1 i + F2  j + F3 k . Cartesian form of Gauss divergence theorem is

   S



F ⋅ n dS

=

 

   

F1 dy dz + F2 dz dx + F3 dx dy =

S

2

2

3



div F dV  

V 

2

Here,

F1 = xz ; F2 = x y –  z , F3 = 2xy +  y  z.

Hence,

F



= xz2 i + (x2 y –  z3)  j + (2xy –  y2 z) k 



div F

= =

  ∂ i + ∂  j + ∂ k    ∂x ∂ y ∂ z    {xz i + (x y –  z )  j + (2xy +  y z) k } ∂ ∂ ∂ xz 2  + x2 y – z 3  +    2xy + y 2 z ∂x ∂ y ∂ z 2

= z2 + x2 +  y2.

   

Let

I  = =

Limit of  z  is 0 to

S

F ⋅ n ds =

   V 

2

3



div F dV  



x 2 + y 2 + z 2 dx dy dz

a2 − x2 − y 2

− a 2 − x 2  to Limit of x is – a to a

a2 − x 2

Limit of  y is

I  = =

    



x 2 + y 2 + z 2 dx dy dz

a −x −y z  2 2 x + y z +  3  0

=

  2 2  x + y  

=

 

3

2

2

2

dx dy

   dx dy 3   a 2 − x 2 − y 2   a 2 − x2 − y 2 x 2 + y 2 +  dx dy 3   a a 2 − x2 − y 2 +

2

−x

2

3 2 2 −y

2

412

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

= =

Let

1 3 1 3

=

1 3

=

2 3

dy =

=

= = = = = = =

  





a 2 − x 2 − y 2 3x 2 + 3y 2 + a 2 − x 2 − y 2 dx dy





2x 2 + a 2  2x 2 + a 2  

a 2 − x 2 − y 2 + 2y 2

a 2 − x 2 − y 2 2x 2 + 2y 2 + a 2 dx dy a 2 −x 2

a

− a − a 2 −x 2

a

a2 −x 2

−a 0

a 2 − x 2 − y 2 + 2y 2

 a2 − x 2 − y 2  dx dy 

a 2 − x 2 − y 2 dx dy

a 2 − x 2 sin θ

y =

I  =

           

a 2 − x 2 cos θ dθ

 2 2 2 2 2 2 2 2 x a a x a x + − θ + − 2 cos 2       sin 2 θ cos2 θ dx dθ       3 1 3 3 2 a  2 2 2 2 2 2 2 + 2 a2 − x2 2 2  dx   3 − a  2x + a  a − x  2 2 2 3   2 π  π 2 a  2 2x + a 2   a 2 − x 2  + 2 a 2 − x 2  dx   3 −a  4 16  2 4 π a × 22x 2 + a2  a 2 − x 2  + a2 − x 2   dx  3 8 0  a π 2 2x 2 a 2 − 2x 4 + a 4 − a 2x 2  + a 4 + x 4 − 22 x 2 dx 3×2 0 2 3

π 2 −a 0

a

   

 

 

π 6 π 2

      a

0

a



3a 4 − 3x 4 dx

 a x5  π 4 a x − 5  2 0 0

a 4 − x 4 dx

π 4 5 2π 5 a . × a = 2 5 5

  Using the divergence theorem, evaluate the surface integral



 yz dy dz + zx dz dx + xy dy dx , where S : x2 +  y2 +  z2  = 4.

(U.P.T.U., 2008)

S

F = F1i + F2 j + F3k  From Gauss divergence theorem, we have  Let

  

F . n dS =

S

 

S

F1 dy dz + F2 dz dx + F3 dx dy

=

   

div F dV  

V 

...(i)

413

VECTOR CALCULUS

Comparing L.H.S. of (i) with given integral, we get F1 = yz, F2 =  zx, F3 = xy F = F1i + F2  j + F3k  ⇒

So

div F = =

 i ∂ +  j ∂ + k  ∂  .{( yz)i + ( zx) j + (xy)k }   ∂x ∂ y ∂ z    ∂ ∂ ∂  yz +  zx +  xy = 0  ∂x ∂ y ∂ z

Thus

  

F = ( yz)i + ( zx) j + (xy)k 



 yz dy dz + zx dz dx + xy dy dx =

S

   

0.dV  = 0.

V 

EXERCISE 5.7

 Use divergence theorem to evaluate

 



3 3 3 F ⋅ dS  where F  = x i + y  j + z k  and S is the surface

S

2

2

2

  

2

of the sphere x  +  y  +  z  = a .  Use divergence theorem to show that surface enclosing volume V .

    

12πa 5 5

.

  



  x 2 + y 2 + z 2 dS = 6V  Where   ∇ S is any closed S

 Apply divergence theorem to evaluate SFn dS , where F  = 4x 3i − x 2yj + x 2zk   and S is the surface of the cylinder x2 +  y2 = a2  bounded by the planes  z = 0 and  z = b. .

  

 Use the divergence theorem to evaluate

S

 3ba 4 π



x dy dz + y dz dx + z dx dy , where S  is the

portion of the plane x + 2 y + 3 z  = 6 which lies in the first octant.

(U.P.T.U., 2003) .

 18

 The vector field F  = x2i +  zj +  yzk   is defined over the volume of the cuboid given by 0

 

≤ x ≤ a, 0 ≤  y ≤ b, 0 ≤ z ≤ c enclosing the surface S. Evaluate the surface integral

(U.P.T.U., 2001)  Evaluate

   S

  S

.

   

F ⋅ dS.

abc a +

b 2

   



 yzi + zxj + xyk dS  where S is the surface of the sphere x2 + y2 + z2 = a2 in the (U.P.T.U., 2004)

first octant.  Evaluate

 

S

.

 0



e x dy dz – ye x dz dx + 3z dx dy , where S  is the surface of the cylinder x2 +  y2

= c2, 0 ≤  z ≤ h.

.

 3π hc 2

414

A TEXTBOOK OF ENGINEERING MATHEMATICS—I

 Evaluate

  S

F ⋅ n ds , where F = 2xyi +  yz2 j + xzk , and S  is the surface of the region

 

 bounded by x = 0,  y = 0,  z = 0,  y  = 3 and x + 2 z = 6.

.

351 2

 

 F = 4xi – 2 y2 j +  z2k   taken over the region bounded by x2 +  y2 = 4,  z = 0 and  z = 3.

Common value 8 π

.





 F  = x 3 − yz i – 2x 2 yj + zk   taken over the entire surface of the cube 0 ≤ x ≤ a, 0 ≤  y ≤ a,

  

0 ≤  z ≤ a.

.

Common value

a5 + a3 3

  

 F  = 2xyi + yz2 j + xzk  and S is the total surface of the rectangular parallelopiped bounded

 by the coordinate planes and x = 1,  y = 2,  z = 3.

Common value 33

.

 F  = x 2 i + y 2 j + z 2 k  taken over the surface of the ellipsoid

x 2  y 2  z 2 + + = 1. a 2 b2 c 2

.

 F  = xi +  yj taken over the upper half on the unit sphere

x2 +  y2 +  z2 = 1.  Prove that  Evaluate

       dV 

  

V 

r

2

r.  n

=

S

r2

 

.

Common value 0

Common value

4π 3

 

ds .

r.  n ds , where S  : surface of cube bounded by the planes x  = –1,  y  = –1,

S

 z = –1, x = 1, y = 1,  z  = 1.

[ 24]

OBJECTIVE TYPE QUESTIONS

           If r  = xi +  yj +  zk   is position vector, then value of ∇(log r) is

(i)

r r

(iii) −

(ii)

r r2

r

(iv) None of these r3  The unit vector normal to the surface x2 y + 2xz  = 4 at (2, –2, 3) is

1 (i – 2 j + 2k ) 3 1 (iii) (i + 2 j – 2k ) 3 (i)

(ii)

1 (i – 2 j – 2k ) 3

(iv) None of these

[U.P.T.U., 2008]

415

VECTOR CALCULUS

 If r   is a position vector then the value of ∇rn is

(i) nrn–2 r (ii) nrn–2 (iii) nr2 (iv) nrn–3  If  f (x, y, z) = 3x2 y –  y3 z2, then |∇ f | at (1, –2, –1) is (i) 481 (iii)

581

 If a   is a constant vector, then grad

(i) r (iii) 0

(ii)

381

(iv)

481

r . a  is equal to (ii) − a (iv) a

 The vector rn r  is solenoidal if n  equals (i) 3 (ii) – 3 (iii) 2 (iv) 0  If r   is a position vector then div r  is equal to (i) 3 (ii) 0 (iii) 5 (iv) –1  If r   is a position vector then curl r  is equal to (i) – 5 (ii) 0 (iii) 3 (iv) –1  If F  = ∇φ, ∇2φ = – 4πρ where P  is a constant, then the value of

  

  dS is: F .n

S

(i) 4π (iii) – 4πρV 

(ii) – 4πρ (iv) V 

 If n  is the unit outward drawn normal to any closed surface S, the value of

   

div F dV 

V 

is (i) V  (iii) 0

(ii) S (iv) 2S

 If S is any closed surface enclosing a volume V   and F  = xi + 2 yj + 3 zk   then the value

of the integral

  

  dS is F .n

S

(i) 3V  (iii) 2V   The integral

  

(ii) 6V  (iv) 6S

r 5 n d S   is equal

S

(i) 0

(ii)

   

5r 3 . r dV  

V 

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A TEXTBOOK OF ENGINEERING MATHEMATICS—I

(iii)

   

5r −3 . r dV  

(iv) None of these

V 

 A vector F  is always normal to a given closed surface S in closing V   the value of the

integral

   

curl F dV  is :

V 

(i) 0 (iii) V 

(ii) 0 (iv) S

      If  f  = (bxy –  z3)i + (b – 2)x2  j + (1 – b)xz2 k   has its curl identically equal to zero then

b  = ..........  ∇ 2

  1    = ..........  r  

 div grad  f   = ..........  curl grad  f   = ..........  grad r  = ..........  grad

1  = .......... r

 If  A = 3x  yz2 i + 2x  y3  j – x2  yz k   and  f  = 3x2 –  yz then  A . ∇f    = ..........  If r  = r, then ∇ f (r) × r  = ..........  If r  = r, then

   = ..........

∇ f r ∇r

 The directional derivative of φ = xy +  yz +  zx  in the direction of the vector i + 2 j + k  at (1, 2, 0) is = ..........  The value of

 



xdy − ydx   around the circle x2 +  y2 = 1 is ..........

 If ∇2φ = 0, ∇2ψ   = 0, then 

    

φ

S

  

F . n dS   is called the .......... dF   over S.

 If S  is a closed surface, then 

    

∂ψ  ∂φ dS  = .......... − ψ  ∂n ∂n

   

∇. F dV =

V 

  

  

r . n dS   = ..........

S

A dS , then  A is equal to ..........

S

          (i) If v   is a solenoidal vector then div v = 0.

417

VECTOR CALCULUS

(ii) div v  represents the rate of loss of fluid per unit volume. (iii) If  f    is irrotational then curl  f  ≠ 0. (iv) The gradient of scalar field  f (x, y, z) at any point P represents vector normal to the surface  f   = const.  (i) The gradient of a scalar is a scalar. (ii) Curl of a vector is a scalar. (iii) Divergence of a vector is a scalar. (iv) ∇ f   is a vector along the tangent to the surface  f  = 0.  (i) The directional derivative of  f   along a is  f .  a . (ii) The divergence of a constant vector is zero vector.

(iii) The family of surfaces  f (x, y, z) = c are called level surfaces. (iv) If a  and b   are irrotational then div

a × b

= 0.

 (i) Any integral which is evaluated along a curve is called surface integral.

(ii) Green’s theorem in a plane is a special case of stoke theorem. (iii) If the surface S has a unique normal at each of its points and the direction of this normal depends continuously on the points of S, then the surface is called smooth surface. (iv) The integral

    

F . dr  is called circulation.

S

 (i) The formula



curl F . n ds =

S

 

F . dr  is governed by Stoke’s theorem.

C

(ii) If the initial and terminal points of a curve coincide, the curve is called closed curve. (iii) If n  is the unit outward drawn normal to any closed surface S, then

   

∇. n dv ≠ S .

V 

(iv) The integral

1 2

 



xdy − ydx  represents the area.

C

   





 (i) ∇. ∇ × a

(a) dφ

(ii) curl (φ  grad φ)



(iii) div a × r (iv) ∇φ . d r  (i) ∇2 r2 (ii) df /ds



(b) 0 (c) 0 (d) r  curl a (a) a . ( ∇ f ) (b) grad  f  ± grad  g