10_Areas of Bounded Regions.pdf

AREAS OF BOUNDED REGIONS

1

APPLICATIONS OF INTEGRALS AREA UNDER SIMPLE CURVE 1. The area of the region bounded by a curve y = f(x), x =a, x = b, and the x - axis. Case – I When the curve y = f(x) lies above the x-axis. Y y = f(x) The area bounded by the curve y = f(x), the x-axis

B

A

b

and the ordinates x = a and x = b is given by ò y dx a

Case – II

X’

C O

When the curve y = f(x) lies below the x-axis.

D b

X

b B

X

Y’

The area bounded by the curve y = f(x), the x-axis,

Y X’

b

a

O

C

a A

and the ordinaes x = a and x = b is given by ò ( -y ) dx a

2.

The Area of the Region Bounded by a Curve x = f(y), The Abscissae y = c, y = d and the y -axis.

ò

y=d

B

x dy

c

Case – II When the curve x = f(y) lies to the left of the y-axis. The area bounded by the curve x = f(y), the y-axis

D

Y

d

and the abscissae y = c and y = d is given by

C

Y’

Case – I When the curve x = f(y) lies to the right of the y -axis. The area bounded by the curve x = f(y), the y-axis

y=f(x)

D x=f(y)

A

C

y=c

X’

X Y’

Y y=d x=f(y) d

and the absissa y = c and y = d is given by ò ( -x ) dy c

y=c X’

X

O Y’ Y

The Area Bounded by Curve y = f(x), Ordinates x = a and x = b and x -axis is given by

a

, where a < c < b.

c

Y’

The Area Between Two Curves Case – I The area bounded by two curves y = f(x) and y = g(x), which are intersected by the ordinates x = a and

x=a

x=c

x=b

Y y=f(x)

b

a

y=g(x) X

x = b is given by ò [f ( x ) - g( x )]dx

O

4.

ò (-y)dx

Y’

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a

b

X

ò

b

ydx +

X

c

A=

y = f(x)

O

3.

X

6.1

Case – II

AREAS OF BOUNDED REGIONS Y The area bounded by two curves y = f (x) y=f(x)

b

and y = g(x) is given by ò {f ( x ) - g( x)} dx , a

Case – III

The area bounded by two curves x = f(y)

y=g(x)

Y’ Y

d

ò {f( y ) - g( y)}dx ,

y=d

and radius = g2 + f 2 - c 2. (i)

X’ O

SOME STANDARD CURVES 1. Circle (i) x2 + y2 = a2 represents a circle with centre at (0,0) and radius = a. It is symmetric about both the axes. (ii) (x–a)2 + (y – b)2 = a2 represents a circle with centre at (a, b) and radius = a. (iii) x2 + y2 + 2gx + 2fy + c = 0 represents the general equation of a circle with centre at (–g, – f)

X

6.2

y=c

x

where c and d are ordinates of the points of intersection of the two curves.

(h)g= x

c

= )f (h

and x = g(y) is given by

x=b

x=a

O

where a and b are the abscissae of the points of intersection X’ of the two curves.

Y’ Y B(0,a) a a

X’

X A(a,0)

O (0,0)

Y’

Y

Parabola L x=a y2 = 4ax, where a > 0 or y2 = 4ax, Latus rectum F whre a < 0 (standard equation) X’ X F O X’ It is symmetric about x -axis, Latus rectum x=–a where O is the vertex, L’ F is the focus, LL¢ L’ Y’ is the latus rectum : 2 y = 4ax, a > 0 LL' ^ X' X; X' X being the axis of the parabola. x2 = 4by, where b > 0 or x2 = 4by, where b < 0

(ii)

Y

O

It is symmetric about y-axis, where LL¢ is the latus rectum.

Y’ Latus rectum y=b

X’

F

O

Y

Y’

O

L’ y=–b

X

F

L

Latus rectum Y’ x2 = 4by, b < 0

O is the vertex, F is the focus, LL¢ is the latus rectum, LL' ^ YY' .

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X

y

L’

X’

X

2

x2

= 4by, b > 0

L

X

AREAS OF BOUNDED REGIONS

3 3.

Ellipse

Y B(0,b)

x2 y2 + = 1 represents an ellipse with centre at (0, 0). a 2 b2

A’(–a,0)

It is symmetric about both the axes, meeting x-axis at (±a, 0) and y-axis at (0, ±b) and a > b. Here, AA¢ = 2a = length of major axis. BB¢ = 2b = length of minor axis.

A(a,0)

X’

X

O (0,0) B’(0,–b) Y’

SOLVED PROBLESM

0Ex.1 Using integration, find the region bounded by the line 2y + x = 8 or, (2y = –x + 8) the x-axis and the lines x = 2 and x = 4. Y Sol. Here, the given line is Þ y=

2y + x = 8

1 2

(8 – x)

....(1)

B

\ the required area (shaded region) bounded by the line, the x-axis and the line x = 2 and x = 4 4

ò

=

Ex.2 Sol.

x=2

4

1 = ydx = (8 - x )dx 2 2 2

ò

4

1é x ù ê8x - ú 2ë 2 û2 2

P(x,y) C 2y + x = 8

[From (1)]

x=4 X’

A

O

1 2

X

D

Y’

= = [(32 – 8) – (16 – 2)] = 5 sq. units.

Find the area of the region bounded by the parabola y2 = 4x and y = 2x The points of intersection of y2 = 4x and y = 2x and O(0, 0) and A(1, 2) Here, we take y2 = 4x or y = 2 x as f(x) and y = 2x as g(x), where f(x) ³ g(x) in [0, 1] Y A(1,2) Therefore, Area of the shaded region

y=2x

1

ò

= [f ( x ) - g( x )]dx

O

0

(1,0)

X’

1

X

ò

= ( 2 x - 2x ) dx 0

1

é2 x2 ù é2 1ù = 2ê x 3 / 2 - ú = 2ê - ú 3 2 ë3 2 û ë û0

Ex.3 Sol.

=

y2=4x

1 sq. units 3

Y’

Find the area of the smaller region enclosed between the circle x 2 + y2 = 4 and the line x + y = 2. The points of intersection of x2 + y2 = 4 and the line x + y = 2 are A(2, 0) and B(0, 2) Here, we take x2 + y2 = 4 i.e., y = 4 - x 2 as f(x) and x + y = 2 i.e., y = 2 – x as g(x), Y where f(x) ³ g (x) in [0, 2]. B(0,2) Therefore, Area of the shaded region 2

ò

= [f ( x ) - g( x )]dx

X’

0

2

A(2,0) O

= é 4 - x 2 - (2 - x )ù dx êë úû

ò

0

2

é x 4 - x2 x2 ù æxö =ê + 2 sin -1ç ÷ - 2x + ú 2 2ú êë è2ø û0

= 2 sin–1(1) – 4 + 2 = (p – 2) sq. units

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Y’

X

x+y=2

4 Ex.4 Sol.

AREAS OF BOUNDED REGIONS Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y. Given equations to the curves are x2 + y2 =

9 4

...(1)

x2 = 4y

Equation (1) represents a circle whose centre is (0, 0) and radius

...(2) 3 . 2

Equation (2) represents a parabola whose vertex is O(0, 0) and x-axis is y-axis. From (2), we have Put y =

2

x 4

y=

x2 4

Y

in (1), we have x2 +

A(– 2 ,1/ 2)

x 9 = 16 4

x4 + 16x2 – 36 = 0 2 (x + 18) (x2 – 2) = 0 x2 = 2 or – 18 x=± 2 2 (Q x cannot be – 18)

\

x2 y= 4

Taking y =

X’ – 2

O

on f(x) and y =

ò

2

– 2

[f ( x ) - g( x )] dx

x2 4

as g(x), where f(x) ³ g (x) in [– 2 , 2 ] we have

2 2é 9 x2 ù = 2ò [ f ( x) - g( x )] dx = 2ò ê - x 2 - ú dx 0 0

ëê 4

4 ûú

2

2

ù é ù é 9 9 -2 - x2 ê 2 3 ú ú êx 9 æ 2x ö x ú 2ú 9 æ 2x ö 4 + sin -1 ç ÷ – + sin -1 ç ÷ – = 2ê 4 = 2ê ê 2 8 ú ê 6 2 8 è 8 ø 12 ú è 8 ø ú ê ú ê û0 ë û0 ë æ2 2 ö ù é æ ù é 2 ö 2 9 æ ö 2 2÷ 2 9 ÷ + sin -1 ç = 2ê - 2 + 9 sin -1 ç 2 2 ÷ú = 2ê + sin -1çç ú = ç 3 ÷ ÷ 6 4 ç ÷ 3 12 8 3 4 8 8 è ø êë è ø ûú è øúû ëê

Ex.5 Sol.

X

2

Y’

1 2

9 - x2 4

Required area =

B( 2 ,1/ 2) 4x2 + 4y2 = 9

4

Þ Þ Þ Þ

=

x2=4y

sq. units

Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y2 = 1. Given equations to the curves are x2 + y2 = 1 ...(1) Y (x – 1)2 + y2 = 1 ...(2) 2 2 (x – 1) + (1 – x ) = 1 Þ

x2 – 2x + 1 + 1 – x2 = 1 Þ 2x = 1 Þ x =

Requried area 1 ù é1/ 2 = 2ê y of circle (2) dx + y of circle (1) dx ú úû êë 0 1/ 2

ò

ò

1 2

X’

1 ù é1/ 2 = 2ê 1 - ( x - 1)2 dx + 1 - x 2 dx ú úû êë 0 1/ 2

ò

ò

O

Y’

(1,0)

x=1/2 x=1

1/ 2 1 ìé 2 ù ù üï é x 1- x2 1 1 ï ê ( x - 1) 1 - ( x - 1) -1 -1 + sin x)ú ý = 2í + sin ( x - 1)ú + ê 2 2 2 2 ú úû ï ê ï êë ë û0 1/ 2 þ î

ìï é 1 ù é 1 1 1 1 1 1 1 æ 1ö æ 1 öù üï = 2í ê 1 - - sin -1ç ÷ú ý 1 - + sin -1ç - ÷ - 0 - sin -1( -1)ú + ê0 + sin -1(1) 4 2 4 2 2 4 2 è 2ø è 2 øúû ïþ ïî êë 4 úû êë

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X

AREAS OF BOUNDED REGIONS

5

1 3 1 3 æ 1ö æ 1ö + sin -1 ç – ÷ - sin -1(-1) + sin -1(1) - sin -1ç ÷ 2 4 2 2 4 è ø è 2ø

=-

3 æ pö æ pö p 3 p + ç- ÷ – ç– ÷ + 4 è 6ø è 2ø 2 4 6

=-

æ 3 2p ö÷ = ç– + ç 2 3 ÷ø è

Ex.6 Sol.

sq. units.

Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 3. Given equations are y = x2 + 2 ...(1) y=x ...(2) x=0 ...(3) x=3 ...(4) Equation (1) represetns a parabola with vertex A(0, 2) Y y = x2 + 2 and axis is y-axis. y=x

Equation (2) represents a line passing through (0, 0) and (1, 1) A(0,2) Equation (3) represents y-axis Equation (4) represent a line parallel to y-axis and passX’ O ing through B(3, 0)

= Ex.7 Sol.

ò

0

ò

3

0

0

3

3

( x 2 + 2)dx - xdx

=

0

3

é x3 é x2 ù ù ê + 2x ú - ê ú ë3 û0 ë 2 û0

Y’

æ9 ö 21 = (9 + 6) - ç 2 + 0 ÷ = sq. units. è

ø

2

a

2 2 Required area = ò0 éêë a - ( x - a) - ax ùûú dx

ò

a

0

Y

A(a,a)

a

é 2 1 2 2 a 3/2 ù 2 -1æ x - a ö x ú ÷ê a - ( x - a) + a sin ç 2 3 è a ø ûú 0 ëê

2 ù é1 – 2 2 1 2æ - p ö é1 ù = ê a 2 sin -1(0) - a 2 ú - ê a 2 sin -1( -1)ú = a - a ç ÷ 3 û ë2 2 è 2 ø 3 ë2 û

Sol.

x=3

Find the area of the region | (x,y) : x2 + y2 £ 2ax, y2 £ 2ax, a > 0, x ³ 0, y ³ 0|

=

Ex.8

X x=0

ò y(of parabola )dx - ò y (of line y = x) dx

Required area = 3

3

B(3,0)

æ

C

X’

ö

X

Y’

2 2 pa = çç 4 - 3 a ÷÷ sq. units. è ø 2

Using the method of integration, find the area of the triangle ABC, coordinates of whose vertices are A (2, 0), B(4, 5) and C (6, 3). Required area = area of the region ABC = area of the region APB + area of the region PQCB – Area of the region AQC Y 6 5x - 10 dx + (9 - x )dx 2 4 2 4

ò

=

ò

=

ù é ù 5 é x2 x2 ù 3 é x2 ê - 2x ú + ê9x - ú - ê - 2xú 2ë 2 2 4 2 û2 ë û4 ë û2

4

= 7 sq. units.

6

B(4,5)

æ 3x - 6 ö ç ÷ dx 2è 4 ø

ò

6

C(6,3)

6

A(2,0) X’ O Y’

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P

Q

X

AREAS OF BOUNDED REGIONS

6

UNSOLVED PROBLEMS EXERCISE – I Q.1

Using integration, find the area of the region bounded by the lines y = 4x + 5, y = 5 – x and 4y – x = 5.

Q.2

Find the area of the region bounded by the curve x2 = 4y and the lines x = 4y – 2

Q.3

Find the area of the region bounded by the curve y = x 2 + 2, y = x, x = 0 and x = 3

Q.4

Find the area of the region enclosed between the parabola y2 = x and the line x + y = 2 in the first quadrant.

Q.5

Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line x =

Q.6

Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32.

Q.7

Find the area of the region bounded by the curve y = 1 - x 2 , the line y = x and the positive x-axis.

Q.8

Find the area of the smaller region bounded by the ellipse

Q.9

Find the area included between the curves y2 = 4ax and x2 = 4ay, a > 0

Q.10

Draw the rough sketch of y2 = x + 1 and y2 = – x + 1 and determine the area enclosed by the two curves.

Q.11

Calculate the area of the region enclosed between the circle x2 + y2 = 4 and (x – 2)2 + y2 = 4

Q.12

Using integration, find the area of the circle x2 + y2 = 16 which is exterior to the parabola y2 = 6x

Q.13

Sketch the region common to the circle x2 + y2 = 16 and the parabola x2 = 6y. Also find the area of the region, using integration

Q.14

Sketch the graph of y = |x + 3|. Evaluate ò | x + 3 | dx What does the value of the integral represent on the

a 2

x y x2 y2 + = 1 and the line + = 1 a b a 2 b2

0

–6

graphs ? Q.15

Using integration find the area of the region bounded by the following curves after making a rough sketch : y = 1 + |x – 1|, x = – 3, x = 3 and y = 0

Q.16

Find the area of the region {(x, y) : x2 + y2 £ 1 £ x + y}

Q.17

Find the area of the region {(x, y) : x2 £ y £ |x|}

Q.18

Find the area of the region : {(x, y) : 0 £ y £ (x2 + 1), 0 £ y £ (x + 1)}, 0 £ x £ 2}

Q.19

Find the area of the region : {(x,y) : x2 + y2 £ 2ax}, y2 ³ ax, x ³ 0, y ³ 0

Q.20

Find the area bounded by the curve y = sin x between x = o and x = 2p.

ANSWER KEY 1. 9.

15 2

9 8

2.

16a2 3

10. æp

1ö

15. 16 16. ç 4 - 2 ÷ è ø

3. 8 3

21 2

4.

7 6

5.

a 2 æç p - 1ö÷ 2 è2 ø

8p

æ ö 11. ç 3 - 2 3 ÷ è ø

17.

1 3

12.

18.

23 6

6. 4p

4p + 3 8p - 3

7. 13.

æ pa2

p 8

16p + 4 3 3 2a2 ö

ç ÷ 19. ç 4 - 3 ÷ è ø

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8.

abp 4

14. 9

20. 4

–

ab 2

AREAS OF BOUNDED REGIONS

7

BOARD PROBLES Q.1

Using integration, find the area of the region, given : {(x, y) : 0 £ y £ x2 + 1 ; 0 £ y £ x + 1, 0 £ x £ 2}

[C.B.S.E. 2001]

Q.2

Using integration, find the area of DPQR where P is (2, 1) Q is (3, 4) and R is (5, 2) [C.B.S.E. 2001]

Q.3

Using integration, find the area of the region bounded by the line 2y + x = 8, x-axis and the line x = 2 sand x = 4. [C.B.S.E. 2002]

Q.4

Find the area of the region given by {(x, y) : x2 £ y £ |x|}.

Q.5

Using integration, find the area of triangle ABC, whose vertices are A(3,0), B(4, 5) and C(5, 1) [C.B.S.E. 2003]

Q.6

Find the area of the following region : {(x, y) : y2 £ 4x, 4x2 + 4y2 £ 9}.

Q.7

Using integration, find the area of the region in the first quadrant enclosed by the x-axis and the line y = x and the circle x2 + y2 = 32. [C.B.S.E. 2004]

Q.8

Find the area of the region lying between the parabola y2 = 4ax and x2 = 4ay, where a > 0 [C.B.S.E. 2004]

Q.9

Using integration, find the area of the smaller region bounded by the curve

x2 y2 + =1 16 9

and the straight line

[C.B.S.E. 2002]

[C.B.S.E. 2003]

x y + =1 4 3

[C.B.S.E. 2005]

Q.10

Find the area enclosed by the parabola y2 = x and the line x + y = 2 and the x-axis. [C.B.S.E. 2005]

Q.11

Using integration, find the area of the smaller region bounded by the ellipse and the straight line

x2 y2 + =1 a 2 b2

x y + =1 a b

[C.B.S.E. 2006]

Q.12

Using integration, find the area of the region enclosed between two circles x2 + y2 = 1 and (x – 1)2 + y2 =1. [C.B.S.E. 2006]

Q.13

Using integration, find the area of the region enclosed between the circles x2 + y2 = 1 and (x – 1)2 + y2 = 1. [C.B.S.E. 2007]

Q.14

Find the area of that part of the circle x2 + y2 = 16 which is exterior to the parabola y2 = 6x. [C.B.S.E. 2007]

Q.15

Find the area of the region bounded by the parabolas y2 = 4ax and x2 = 4ay.

Q.16

Using integration, find the area of the region enclosed between the circles x2 + y2 = 4 and (x – 2)2 + y2 = 4. [C.B.S.E. 2008]

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[C.B.S.E. 2008]

8 Q.17

AREAS OF BOUNDED REGIONS Find the area of the region included between the parabola y2 = x and the line x + y = 2. [C.B.S.E. 2009]

Q.18

Using the method of integration, find the area of the region bounded by the lines 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0 [C.B.S.E. 2009]

Q.19

Find the are of the circle 4x2 + 4y2 = 9 which is interior to the parabola x 2 = 4y. [C.B.S.E. 2010]

Q.20

Using integration, find the area of the triangle ABC, coordinates of whose vertices are A(4, 1), B(6, 6) and C(8, 4). [C.B.S.E. 2010]

Q.21

Sketch the graph of y = |x + 3| and evaluate the area under the curve y = |x + 3| above x-axis and between x = – 6 to x = 0. [C.B.S.E. 2011]

Q.22

Find the area of the region : {(x, y) : x2 + y2 £ 4, x + y ³ 2}.

[C.B.S.E. 2012]

Q.23

Find the area of the region bourded by the parabola y = x 2 and y = |x|

[C.B.S.E. 2013]

ANSWER KEY 1.

23 6

7. 4p

12.

2p 3

21. 9

–

3 2

1 3

2. 4

3. 5

4.

8. 16/3

9. 3(p – 2)

10. 7/6

æ 2p 3ö 4 13. çç 3 - 2 ÷÷ 14. (8p – 3 ) 3 è ø

22.

5. 9/2

6. 11.

15.

16 2 a 3

1 3 2

+

9p 8

ab (p - 2) 4 8p

æ ö 16. ç 3 - 2 3 ÷ è ø

(p – 2)

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–

9 4

–1

sin

1 3