Understanding Multivariable Calculus
February 13, 2017 | Author: rashadwcartwright | Category: N/A
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Topic Science & Mathematics
Understanding Multivariable Calculus: Problems, Solutions, and Tips Course Workbook Professor Bruce H. Edwards University of Florida
Subtopic Mathematics
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Copyright © The Teaching Company, 2014
Printed in the United States of America This book is in copyright. All rights reserved. Without limiting the rights under copyright reserved above, no part of this publication may be reproduced, stored in or introduced into a retrieval system, or transmitted, in any form, or by any means ͼelectronic, mechanical, photocopying, recording, or otherwiseͽ, without the prior written permission of The Teaching Company.
Bruce H. Edwards, Ph.D. Professor of Mathematics University of Florida
P
rofessor Bruce H. Edwards has been a Professor of Mathematics at the University of Florida since 1976. He received his B.S. in Mathematics from Stanford University in 1968 and his Ph.D. in Mathematics from Dartmouth College in 1976. From 1968 to 1972, he was D3HDFH&RUSVYROXQWHHULQ&RORPELDZKHUHKHWDXJKWPDWKHPDWLFVͼin Spanishͽ at Universidad Pedagógica y Tecnológica de Colombia.
Professor Edwards’s early research interests were in the broad area of pure mathematics called algebra. His dissertation in quadratic forms was titled “Induction Techniques and Periodicity in Clifford Algebras.” Beginning in 1978, Professor Edwards became interested in applied mathematics while working summers for NASA at the Langley Research Center in Virginia. This work led to his research in numerical analysis and the solution of differential equations. During his sabbatical year, 1984 to 1985, he worked on two-point boundary value problems with Professor Leo Xanthis at the Polytechnic of Central London. Professor Edwards’s current research is focused on the algorithm called CORDIC that is used in computers and graphing calculators for calculating function values. Professor Edwards has coauthored a number of mathematics textbooks with Professor Ron Larson of Penn State Erie, The Behrend College. Together, they have published leading texts in calculus, applied calculus, OLQHDUDOJHEUD¿QLWHPDWKHPDWLFVDOJHEUDWULJRQRPHWU\DQGSUHFDOFXOXV Over the years, Professor Edwards has received many teaching awards at the University of Florida. He was named Teacher of the Year in the College of Liberal Arts and Sciences in 1979, 1981, and 1990. In addition, he was named the College of Liberal Arts and Sciences Student Council Teacher of the Year and the University of Florida Honors Program Teacher of the Year in 1990. He also served as the Distinguished Alumni Professor for the UF Alumni Association from 1991 to 1993. The winners of this two-year award are selected by graduates of the university. The Florida Section of the Mathematical Association of America awarded Professor Edwards the Distinguished Service Award in 1995 for his work in mathematics education for the state of Florida. His textbooks have been honored with various awards from the Text and Academic Authors Association. 3URIHVVRU(GZDUGVKDVWDXJKWDZLGHUDQJHRIPDWKHPDWLFVFRXUVHVDWWKH8QLYHUVLW\RI)ORULGDIURP¿UVW\HDU calculus to graduate-level classes in algebra and numerical analysis. He particularly enjoys teaching calculus to freshmen because of the beauty of the subject and the enthusiasm of the students.
i
Professor Edwards has been a frequent speaker at both research conferences and meetings of the National Council of Teachers of Mathematics. He has spoken on issues relating to the Advanced Placement calculus examination, especially on the use of graphing calculators. Professor Edwards has taught four other Great Courses:
ii
x
Mathematics Describing the Real World: Precalculus and Trigonometry;
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Understanding Calculus: Problems, Solutions, and Tips;
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Understanding Calculus II: Problems, Solutions, and Tips; and
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Prove It: The Art of Mathematical Argument. Ŷ
Table of Contents
INTRODUCTION Professor Biography ................................................................................................................................i Course Scope .........................................................................................................................................1 LESSON GUIDES LESSON 1 A Visual Introduction to 3-D Calculus .....................................................................................................3 LESSON 2 Functions of Several Variables ...............................................................................................................7 LESSON 3 Limits, Continuity, and Partial Derivatives ............................................................................................11 LESSON 4 Partial Derivatives—One Variable at a Time ........................................................................................15 LESSON 5 Total Differentials and Chain Rules ......................................................................................................19 LESSON 6 Extrema of Functions of Two Variables ................................................................................................22 LESSON 7 Applications to Optimization Problems .................................................................................................26 LESSON 8 Linear Models and Least Squares Regression.....................................................................................29 LESSON 9 Vectors and the Dot Product in Space..................................................................................................32 LESSON 10 The Cross Product of Two Vectors in Space ........................................................................................36 LESSON 11 Lines and Planes in Space ...................................................................................................................40 LESSON 12 Curved Surfaces in Space ....................................................................................................................44 LESSON 13 Vector-Valued Functions in Space........................................................................................................48 LESSON 14 Kepler’s Laws—The Calculus of Orbits ................................................................................................52 LESSON 15 Directional Derivatives and Gradients ..................................................................................................55
iii
Table of Contents
LESSON 16 Tangent Planes and Normal Vectors to a Surface ................................................................................58 LESSON 17 Lagrange Multipliers—Constrained Optimization .................................................................................61 LESSON 18 Applications of Lagrange Multipliers .....................................................................................................64 LESSON 19 Iterated Integrals and Area in the Plane ...............................................................................................67 LESSON 20 Double Integrals and Volume ...............................................................................................................71 LESSON 21 Double Integrals in Polar Coordinates ..................................................................................................75 LESSON 22 Centers of Mass for Variable Density ...................................................................................................79 LESSON 23 Surface Area of a Solid .........................................................................................................................83 LESSON 24 Triple Integrals and Applications ...........................................................................................................87 LESSON 25 Triple Integrals in Cylindrical Coordinates ............................................................................................91 LESSON 26 Triple Integrals in Spherical Coordinates ..............................................................................................95 LESSON 27 Vector Fields—Velocity, Gravity, Electricity ..........................................................................................99 LESSON 28 Curl, Divergence, Line Integrals ........................................................................................................ 104 LESSON 29 More Line Integrals and Work by a Force Field................................................................................. 108 LESSON 30 Fundamental Theorem of Line Integrals.............................................................................................112 LESSON 31 Green’s Theorem—Boundaries and Regions.....................................................................................117 LESSON 32 Applications of Green’s Theorem ...................................................................................................... 122 LESSON 33 Parametric Surfaces in Space ........................................................................................................... 126
iv
Table of Contents
LESSON 34 Surface Integrals and Flux Integrals .................................................................................................. 130 LESSON 35 Divergence Theorem—Boundaries and Solids ................................................................................. 136 LESSON 36 Stokes’s Theorem and Maxwell’s Equations ..................................................................................... 140 SUPPLEMENTAL MATERIAL Solutions ............................................................................................................................................ 144 Glossary ............................................................................................................................................ 193 Summary of Differentiation Formulas ................................................................................................ 214 Summary of Integration Formulas ..................................................................................................... 216 Quadric Surfaces ............................................................................................................................... 218 Bibliography ....................................................................................................................................... 221
v
vi
Understanding Multivariable Calculus: Problems, Solutions, and Tips
Scope:
T
he goal of this course is to complete your understanding and appreciation of calculus by seeing how calculus is extended to three dimensions. Many of the ideas of elementary calculus in the plane generalize naturally to space, whereas other concepts will be brand new. Most concepts will be introduced using illustrative examples, and you will see how multivariable calculus plays a fundamental role in all of science and engineering. You will also gain a new appreciation for the achievements of higher mathematics. cos x sin x @S 44
2 2.
([DPSOH 2
6NHWFKWKHUHJLRQZKRVHDUHDLVUHSUHVHQWHGE\WKHLWHUDWHGLQWHJUDO ³
0
³
4 y2
dx dy.
6ROXWLRQ We know that y2x6RWKHUHJLRQLV bounded on the left by x y 2 y x and on the right by x
2
)XUWKHUPRUHyVRWKHUHJLRQLVDV shown in )LJXUH .
4 Figure 19.2
Lesson 19: Iterated Integrals and Area in the Plane
([DPSOH 6NHWFKWKHUHJLRQRILQWHJUDWLRQUHSUHVHQWHGE\WKHLWHUDWHGLQWHJUDO ³
2
0
³
2 x
2
e y dy dx. Then, evaluate the integral
by reversing the order of integration. 4
6ROXWLRQ
y=x The region of integration is shown in )LJXUH . Reversing the order, we have
2
2
0
x
³³
2
e y dy dx
2
³³ 0
y
0
2
e y dx dy.
2
Although the original integral could not be evaluated using the fundamental theorem of calculus, the new integral can easily be evaluated using substitution. 2
The answer is 1 §¨1 14 2© e
68
·. ¸ ¹
4
Figure 19.3
6WXG\7LSV x
Iterated integrals are similar to partial derivatives in that you integrate with respect to one variable ZKLOHKROGLQJWKHRWKHUYDULDEOH¿[HG)RUH[DPSOHLI f x x, y
2 xy , then f x , y
³ f x, y dx ³ 2 xy dx x
x2 y C y .
Notice that the “constant of integration” is a function of y. x
Iterated integrals are usually written without brackets or parentheses. For instance, the iterated LQWHJUDOLQ([DPSOHLVXVXDOO\ZULWWHQDVIROORZV 4
³ ª«¬ ³ 2
x
x
1
2 xy dy º dx »¼
4
³³ 2
x
1
2 xy dy dx.
Representative rectangles can be very useful in describing the region of integration.
3LWIDOOV x
For area computations, the outer limits of integration must be constants. For instance, the following x 4 iterated integral is incorrect: ³ ³ 2 dx dy. 0
x
y
.HHSLQPLQGWKDWWKHYDULDEOHRILQWHJUDWLRQFDQQHYHUDSSHDUDVDOLPLWRILQWHJUDWLRQ)RUH[DPSOH x the following integral is incorrect: ³ y dx. 0
3UREOHPV x
Evaluate the integral
³ x 2 y dy.
Evaluate the integral
³
0
1
2y
y dx. x
Evaluate the iterated integral
1
2
0
0
³ ³ x y dy dx. S
2
Evaluate the iterated integral
³ ³
Evaluate the iterated integral
³³
1
0
0
3
y
1
0
y cos x dy dx. 4 dx dy. x2 y 2
8VHDQLWHUDWHGLQWHJUDOWR¿QGWKHDUHDRIWKHUHJLRQERXQGHGE\ x y
2, x DQGy
69
Evaluate the iterated integral
1
1 y 2
0
1 y 2
³³
dx dy.
Then, reverse the order of integration and evaluate the resulting iterated integral.
Evaluate the iterated integral
2
³³ 0
1 x
dy dx.
2
Then, reverse the order of integration and evaluate the resulting iterated integral. 1
2
0
2x
³³
Evaluate the iterated integral
³³
Lesson 19: Iterated Integrals and Area in the Plane
Evaluate the iterated integral
70
2
4
0
y2
2
4e y dy dx. x sin x dx dy.
Double Integrals and Volume Lesson 20
Topics x
Double integrals and volume.
x
Properties of double integrals.
x
Average value.
'H¿QLWLRQVDQG7KHRUHPV x
Properties of double integrals:
³³ cf x, y dA R
c ³³ f x , y dA R
³³ ª¬ f x, y g x, y º¼ dA ³³ f x, y dA ³³ g x, y dA. R
x
R
R
Let f be integrable over the plane region R of area A. The DYHUDJHYDOXH of f over R is 1 f x , y dA. A ³³ R
6XPPDU\ We continue our study of integration of functions of two variables. We show that the volume of a solid can be represented by a double integral. These double integrals have many of the same properties as single integrals. Although the motivation for double integrals was area and volume, we will see in upcoming lessons that there are many more applications of such integrals. We end the lesson with the familiar topic of average value. ([DPSOH Calculate the volume below the surface z íy and above the rectangle given by 0 d x dd y d 2.
z
6ROXWLRQ The volume is given by the double integral
4
2
0
0
³³ f ( x, y ) dA ³ ³ 6 2 y dy dx.
z = xíy
R
We evaluate the integral as follows: 4
2
4
0
0
0
³ ³ 6 2 y dy dx ³
2
ª¬6 y y 2 º¼ dx 0
³
4
0
8 dx
4
>8 x @0
32 . y
ͼ6HH)LJXUHͽ ([DPSOH
x Figure 20.1
The double integral for the volume under the surface z VLQ y2 VLQ ͼy2ͽDQGDERYHWKHUHJLRQERXQGHGE\ 2 1 y x , x DQGy LV V ³ ³ x sin y 2 dy dx. Reverse the order of integration. 0 2 2 6ROXWLRQ y
The region of integration is a triangle, and the given integral uses YHUWLFDOUHSUHVHQWDWLYHUHFWDQJOHVͼ6HH)LJXUHͽ
y x
If instead we use horizontal representative rectangles, we obtain the integral V
2
³³ 0
1 x
sin y 2 dy dx
2
1
2y
0
0
³³
x 2 2y (2, 1)
1
sin y 2 dx dy.
x 2
1RWLFHWKDWWKH¿UVWLQWHJUDOFDQQRWEHGRQHHDVLO\ZKHUHDVWKH second integral is straightforward. The answer is cos1 1 | 0.4597.
Figure 20.2
Lesson 20: Double Integrals and Volume
([DPSOH Find the average value of f x , y DQGͼͽ
1 xy over the rectangular region RZLWKYHUWLFHVͼͽͼͽͼͽ 2
6ROXWLRQ 7KHDUHDRIWKHUHJLRQLVî 7KHDYHUDJHYDOXHLV 1 f x , y dA A ³³ R
1 4 3 1 xy dy dx . 12 ³0 ³0 2
7KLVLQWHJUDOLVHDV\WRHYDOXDWHDQGWKH¿QDODQVZHULV 3 . 2
72
6WXG\7LSV x
Double integrals do not only represent areas and volumes. We will see many other applications in upcoming lessons.
x
It is very helpful to draw the region of integration together with a representative rectangle.
x
&RPSXWHUVDQGJUDSKLQJFDOFXODWRUVFDQHYDOXDWHGRXEOHLQWHJUDOV6RPHWLPHVWKHDQVZHUPLJKWEHDQ DSSUR[LPDWLRQ7KHIROORZLQJDUHWZRUHVXOWVIURPDFDOFXODWRU 1
2y
0
0
³³ 2
³ ³ 0
1 x
sin y 2 dx dy
cos(1) 1
sin y 2 dy dx
0.4596976941.
2
3LWIDOOV x
Remember that the outer limits of integration must be constants, and the variable of integration can QHYHUDSSHDUDVDOLPLWRILQWHJUDWLRQ)RUH[DPSOHWKHIROORZLQJGRXEOHLQWHJUDOLVLQFRUUHFWIRUWZR 2 1 reasons: ³ ³x sin y 2 dx dy. y
x
2
,Q([DPSOHWKHJLYHQLQWHJUDOFDQQRWEHHYDOXDWHGE\WKHIXQGDPHQWDOWKHRUHPRIFDOFXOXVEHFDXVH the integrand, sin y2, does not have an elementary antiderivative.
3UREOHPV Find the volume of the solid bounded by the surface z
y DQGDERYHWKHUHFWDQJOHxy 2
Find the volume of the solid bounded by the surface z íxy and above the triangle bounded by y x, y DQGx
6HWXSWKHGRXEOHLQWHJUDOIRUWKHYROXPHRIWKHVROLGLQWKH¿UVWRFWDQWERXQGHGE\z xy, z y x, and x
6HWXSWKHGRXEOHLQWHJUDOIRUWKHYROXPHRIWKHVROLGERXQGHGE\x2 + y2 + z2 r2. 6HWXSWKHGRXEOHLQWHJUDOIRUWKHYROXPHRIWKHVROLGLQWKH¿UVWRFWDQWERXQGHGE\y íx2 and z íx2.
Evaluate the iterated integral Evaluate the iterated integral
1
1
³³ 0
2
2
e x dx dy by switching the order of integration.
2
³³ 0
2
y
2
x2
y cos y dy dx by switching the order of integration. 2
Find the average value of the function f ͼx, yͽ xRYHUWKHUHFWDQJOHZLWKYHUWLFHVͼͽͼͽͼͽ DQGͼͽ
Find the average value of the function f ͼx, yͽ VLQͼx + yͽRYHUWKHUHFWDQJOHZLWKYHUWLFHVͼͽͼʌͽ ͼʌ, ʌͽDQGͼʌͽ 2
y
Lesson 20: Double Integrals and Volume
:K\LVWKHH[SUHVVLRQ ³0 ³0 x y dy dx invalid?
Double Integrals in Polar Coordinates Lesson 21
Topics x
Polar coordinates.
x
Double integrals in polar coordinates.
'H¿QLWLRQVDQG7KHRUHPV x
Conversion formulas: x r cos ș, y r sin ș x2 y2
x
y . x
r 2 , tan T
Double integrals in polar coordinates: Let R be a planar region consisting of all points x, y r cos T , r sin T satisfying gͼșͽrg2ͼșͽĮșȕDQGͼȕíĮͽʌ. Then,
E
³³ f x, y dA ³D ³ R
g 2 T g1 T
f r cos T , r sin T r dr dT .
6XPPDU\ In this lesson, we develop double integrals in polar coordinates. This conversion is especially useful if the UHJLRQRILQWHJUDWLRQRUWKHLQWHJUDQGLVHDVLO\H[SUHVVHGLQSRODUFRRUGLQDWHV:HEHJLQZLWKDUHYLHZRI polar coordinates and then develop the formula for a double integral in polar coordinates. In this case, the differential of area, dA, becomes r dr Gș'RQ¶WIRUJHWWKHH[WUDr factor. We illustrate these ideas with area and YROXPHH[DPSOHV y
([DPSOH 2
Use polar coordinates to describe the region in )LJXUH. 6ROXWLRQ The region is a quarter circle of radius 2:
1
^ r, T : 0 d r d 2, 0 d T d S2 `.
x 1
2 Figure 21.1
75
([DPSOH 2
³³
Evaluate the double integral
0
4 y2
0
y dx dy by converting to polar coordinates.
6ROXWLRQ The region is a quarter circle of radius 2. In polar coordinates, the integral becomes
4 y2
2
³³ 0
0
S
y dx dy
2
³ ³ r sin T r dr dT . 2
0
0
This integral is easy to evaluate because the limits of integration are constants: S
S
2
3
º 2ªr ³0 ³0 r sin T r dr dT ³0 «¬ 3 sin T »¼ 0 dT 2
2
8 S 2 sinT dT 3 ³0 8 ª0 1 º ¼ 3¬
8 cos T S 2 > @0 3 8. 3
([DPSOH
ʌ 2
6HWXSWKHGRXEOHLQWHJUDOLQSRODUFRRUGLQDWHVWR¿QGWKH area of the region bounded by the polar graph r FRV ș. ͼ6HH)LJXUHͽ
r = 3 cos 3θ
θ=π 6
Lesson 21: Double Integrals in Polar Coordinates
6ROXWLRQ 7KHJUDSKLVDURVHFXUYHZLWKSHWDOV
3
θ = −π 6
2QHSHWDOLVGH¿QHGE\ S d T d S , 6 6 where 0 d r d 3cos 3T . 6RWKHWRWDODUHDLV A 3³
S
6
S
6
³
3cos 3T
0
r dr dT
9S . 4 Figure 21.2
([DPSOH 6HWXSWKHGRXEOHLQWHJUDOIRUWKHYROXPHRIWKHLFHFUHDPFRQHERXQGHGDERYHE\WKHKHPLVSKHUH z
76
2 x 2 y 2 and bounded below by the cone z
x2 y2 .
z
6ROXWLRQ :H¿UVWGHWHUPLQHZKHUHWKHVXUIDFHVLQWHUVHFWE\HTXDWLQJ the equations: 2 x2 y2 2 x2 y2 2 x2 y2
z=
x2 y2
z=
x2 y2
2 − x2 − y2
x2 + y2
x2 + y2 = 1
2x2 2 y2 1.
y x
ͼ6HH)LJXUHͽ
Figure 21.3
In polar coordinates, the equations are
z
2 x2 y2
2 r 2 and z
x2 y2
r.
The volume is therefore
V
2S
1
0
0
³ ³
ª 2 r 2 r º r dr dT . ¬ ¼
The evaluation requires substitution, and you obtain
V
4 2 1 3
S.
6WXG\7LSV x
The area of a polar sector is A r¨r¨ș. Hence, the differential of area in rectangular coordinates, dA dy dx dx dy becomes r dr Gș in polar coordinates.
x
In polar coordinates, area is given by
³³ dA ³³ r dr dT . R
x
R
The r value can be negative in polar coordinates.
3LWIDOO x
Remember that the differential of area in polar coordinates is dA r dr Gș. Don’t forget the r.
77
3UREOHPV 9 x2
3
Evaluate the iterated integral
³³
Evaluate the iterated integral
³ ³
Evaluate the iterated integral
0
0
2
x dy dx by converting to polar coordinates.
4 x2
2 0 2
³³ 0
0
2 x x2
x
2
y 2 dy dx by converting to polar coordinates.
xy dy dx by converting to polar coordinates.
8VHDGRXEOHLQWHJUDOLQSRODUFRRUGLQDWHVWR¿QGWKHDUHDRIWKHUHJLRQHQFORVHGE\WKHJUDSKRIWKH equation r FRV ș.
8VHDGRXEOHLQWHJUDOLQSRODUFRRUGLQDWHVWR¿QGWKHDUHDRIWKHUHJLRQHQFORVHGE\WKHJUDSKVRIWKH equations r DQGr
8VHDGRXEOHLQWHJUDOLQSRODUFRRUGLQDWHVWR¿QGWKHDUHDERXQGHGE\WKHWKUHHOHDYHGURVHFXUYH r VLQ ș.
8VHDGRXEOHLQWHJUDOLQSRODUFRRUGLQDWHVWR¿QGWKHYROXPHRIWKHVROLGLQWKH¿UVWRFWDQWERXQGHG by z xy and x2 + y2
8VHDGRXEOHLQWHJUDOLQSRODUFRRUGLQDWHVWR¿QGWKHYROXPHRIWKHVROLGERXQGHGE\ z
x 2 y 2 , z
and x2 + y2
8VHDGRXEOHLQWHJUDOLQSRODUFRRUGLQDWHVWR¿QGWKHYROXPHRIWKHVROLGLQVLGHWKHKHPLVSKHUH z
16 x 2 y 2 and outside the cylinder x2 + y2
6HWXSWKHGRXEOHLQWHJUDOLQSRODUFRRUGLQDWHVIRUWKHDUHDLQVLGHWKHFLUFOHr FRV ș and outside the Lesson 21: Double Integrals in Polar Coordinates
circle r
78
Centers of Mass for Variable Density Lesson 22
Topics x
Mass.
x
Moments.
x
Centers of mass for variable density.
'H¿QLWLRQVDQG7KHRUHPV x
If the planar lamina given by the region R has variable density ȡͼx, yͽWKHQWKHPDVV is
m
³³ U x, y dA . R
x
The PRPHQWVRIPDVV with respect to the xDQGyD[HVDUH
Mx
³³ y U x, y dA, M R
x
y
³³ x U x, y dA . R
§ My Mx · If m is the mass of the lamina, the FHQWHURIPDVVLVͼଲxଲ y ͽ ¨ , ¸. © m m ¹
6XPPDU\ In this lesson, we apply our knowledge of double integrals to the calculation of mass and centers of mass. The formula for mass is the double integral of the density function. The formulas for the moments with UHVSHFWWRWKHD[HVDUHPXFKVLPSOHUWKDQWKHFRUUHVSRQGLQJIRUPXODVLQ y HOHPHQWDU\FDOFXOXV,QVRPHH[DPSOHVSRODUFRRUGLQDWHV\LHOGHDVLHU y=3 integrals than Cartesian coordinates. (0, 3)
(2, 3)
3
([DPSOH 2
)LQGWKHPDVVRIWKHWULDQJXODUODPLQDZLWKYHUWLFHVͼͽͼͽ DQGͼͽLIWKHGHQVLW\DWWKHSRLQWͼx, yͽLVȡͼx, yͽ x + y. ͼ6HH)LJXUHͽ
R 2y 3
x 1
(0, 0) 1
x 2
3 Figure 22.1
79
6ROXWLRQ The boundaries of the triangular region are x y DQGx
2y . 3
Using a horizontal representative rectangle, the mass is m
2y
3
3 ³³ 2 x y dA ³0 ³0 2 x y dx dy
10.
R
([DPSOH )LQGWKHFHQWHURIPDVVRIWKHODPLQDFRUUHVSRQGLQJWRWKHSDUDEROLFUHJLRQyíx2 if the density at the SRLQWͼx, yͽLVFRQVWDQWȡͼx, yͽ 6ROXWLRQ The mass is m
2
³ ³
4 x2
2 0
2
ª x3 º «¬ 4 x 3 »¼ 2
2
4 x2
2
0
³ > y@
1 dy dx
2
³ 4 x dx
dx
2
2
ª§ 8· § 8 ·º «¬¨© 8 3 ¸¹ ¨© 8 3 ¸¹ »¼
32 . 3
By symmetry, the center of mass lies on the yD[LVVRMy The moment about the xD[LVLV M x
Lesson 22: Centers of Mass for Variable Density
6RZHKDYHଲ y
256
Mx m
32
15
3
2
³ ³
4 x2
2 0
y dy dx
256 . 15
8 , DQGWKHFHQWHURIPDVVLVͼଲxଲ y ͽ § 0, 8 · . ¨ ¸ 5 © 5¹
([DPSOH 6ROYHWKHSUHYLRXVH[DPSOHDVVXPLQJWKDWWKHGHQVLW\LVQRWFRQVWDQWEXWUDWKHUJLYHQE\ȡͼx, yͽ y. 6ROXWLRQ 7KHFRPSXWDWLRQVDUHYHU\VLPLODUWRWKHSUHYLRXVH[DPSOH The mass is m
2
³ ³
4 x2
2 0
2 y dy dx
The moment about the xD[LVLV M x
80
512 . By symmetry, M y 15 2
³ ³
4 x2
2 0
y 2 y dy dx
8192 . 105
y
6Rଲ y
Mx = m
8192
105 512 15
= 16 . 7
4
)LQDOO\WKHFHQWHURIPDVVLVͼ ଲ xଲ y ͽ §¨ 0, 16 ·¸ . © 7 ¹
16 7 8 5
Notice that the balancing point has moved up a bit from the previous H[DPSOHͼ6HH)LJXUHͽ
x í
6WXG\7LSV
2 Figure 22.2
x
Usually, density is mass per unit of volume. But for planar laminas, density is mass per unit of surface area.
x
The formulas for center of mass are equivalent to those studied in elementary calculus.
x
The setup of the problem is the most important step. Calculating the resulting integrals can be done by hand or by using computers and graphing calculators.
x
1RWLFHKRZZHWRRNDGYDQWDJHRIV\PPHWU\LQ([DPSOHVDQG
3LWIDOO x
The formula for Mx involves y, and the formula for My involves x6LPLODUO\WKHIRUPXODIRUଲ x involves yDQGWKHIRUPXODIRUଲ y involves x.
3UREOHPV )LQGWKHPDVVRIWKHVTXDUHODPLQDERXQGHGE\xDQGyLIWKHGHQVLW\LVȡͼx, yͽ xy. )LQGWKHPDVVRIWKHODPLQDERXQGHGE\xDQGy 1 x 2 if the density is ȡͼx, yͽ xy. )LQGWKHPDVVDQGFHQWHURIPDVVRIWKHWULDQJXODUODPLQDZLWKYHUWLFHVͼͽͼͽDQGͼͽLIWKH density is ȡͼx, yͽ y.
)LQGWKHPDVVDQGFHQWHURIPDVVRIWKHWULDQJXODUODPLQDZLWKYHUWLFHVͼͽͼͽDQGͼͽLIWKH density is ȡͼx, yͽ x.
Find the mass and center of mass of the lamina bounded by y x , y DQGx LIWKHGHQVLW\LV ȡͼx, yͽ y.
Find the mass and center of mass of the lamina bounded by y x2, y DQGx LIWKHGHQVLW\LV ȡͼx, yͽ xy.
Find the mass and center of mass of the lamina bounded by x2 + y2 xDQGy if the density is
Lesson 22: Centers of Mass for Variable Density
ȡͼx, yͽ ͼx2 + y2ͽ
82
Surface Area of a Solid Lesson 23
Topics x
6XUIDFHDUHDRIVROLGVLQVSDFH
x
The differential of arc length and the differential of surface area.
x
6XUIDFHDUHDLQSRODUFRRUGLQDWHV
'H¿QLWLRQVDQG7KHRUHPV x
Let the function f represent a smooth curve on the interval > a , b @ . The DUFOHQJWK between a and b is s
x
b
2
2S ³ f x 1 ª¬ f c x º¼ dx. a
b
2
2S ³ x 1 ª¬ f c x º¼ dx . a
The GLIIHUHQWLDORIDUFOHQJWK is 2
1 ª¬ f c x º¼ dx .
ds x
2
1 ª¬ f c x º¼ dx.
If a piece of arc length is rotated about the yD[LVWKHVXUIDFHDUHD of the resulting surface of revolution is A
x
b
a
If a piece of arc length is rotated about the xD[LVWKHVXUIDFHDUHD of the resulting surface of revolution is A
x
³
For a surface given by z f ͼx, yͽGH¿QHGRYHUDUHJLRQR in the xySODQHWKHVXUIDFHDUHD is S
³³
2
2
1 ª¬ f x x , y º¼ ª¬ f y x , y º¼ dA.
R
x
The GLIIHUHQWLDORIVXUIDFHDUHD is dS
2
2
1 ª¬ f x x , y º¼ ª¬ f y x , y º¼ dA.
6XPPDU\ The formula for surface area is similar to that of arc length. Both involve an important differential: the differential of arc length and the differential of surface area. After a brief review of arc length and surfaces of revolution, we present the general formula for surface area of graphs of functions of two variables. ,QVRPHH[DPSOHVZHZLOOVHHWKDWSRODUFRRUGLQDWHVDUHXVHIXOLQVLPSOLI\LQJWKHFRPSXWDWLRQV ([DPSOH 2
Find the surface area of the plane z íxíyLQWKH¿UVWRFWDQW ͼ6HH)LJXUHͽ
z (0, 0, 2)
z íxíy 1
6ROXWLRQ We have fx ͼx, yͽ íf y ͼx, yͽ íDQG
(2, 0, 0) x2
2
2
1 f x f y dA
dS
1 1 1 dA
1 1
(0, 2, 0) y
2
(1, 1, 0)
Figure 23.1
3 dA .
6RWKHVXUIDFHDUHDLV S
³³
2
2
1 ª¬ f x x , y º¼ ª¬ f y x , y º¼ dA
³³
R
3 dA
R
3³
Using a vertical representative rectangle, S
2
0
³
2 x
0
3 ³³ dA. R
dy dx 2 3.
([DPSOH
Lesson 23: Surface Area of a Solid
6HWXSWKHGRXEOHLQWHJUDOIRUWKHVXUIDFHDUHDRIWKHSRUWLRQRIWKHVXUIDFHf ͼx, yͽ íx2 + y that lies above WKHWULDQJXODUUHJLRQZLWKYHUWLFHVͼͽͼíͽDQGͼͽ 6ROXWLRQ The partial derivatives are fx ͼx, yͽ íx and f y ͼx, yͽ +HQFH S
2
1 ª¬ f x x , y º¼ ª¬ f y x , y º¼ dA
³³
1 4 x 2 1 dA
³³
2 4 x 2 dA.
R
R
R
2
³³
y
:HQRZQHHGWR¿QGWKHERXQGVIRUWKHUHJLRQGHWHUPLQHGE\WKH WKUHHSRLQWVͼ6HH)LJXUHͽ :HVHHWKDWxxíyíx. Hence, the integral for surface area becomes S
³³
2 4 x 2 dA
1
1 x
0
x 1
³³
R
ln 3 2
y íx R x x íyíx x
2 4 x 2 dy dx.
1
7KLVLQWHJUDOLVGLI¿FXOWWRHYDOXDWH$FDOFXODWRUJLYHV S
1
í
y = xí
2 | 1.618. 3
([DPSOH
Figure 23.2
x 2 y 2 that lies above the circular region x2 + y2
)LQGWKHVXUIDFHDUHDRIWKHLFHFUHDPFRQH z 6ROXWLRQ x and f y x y2
The partial derivatives are f x
2
§ 1 ¨ ¨ ©
dS
1
2
x x2 y2
· § ¸ ¨ ¸ ¨ ¹ ©
y . The differential of surface area is x y2 2
2
y x2 y2
· ¸ dA ¸ ¹
2 x2 y dA x2 y2 x2 y2
2x2 2 y2 dA x2 y2
2( x 2 y 2 ) dA x2 y2
2 dA.
+HQFHWKHVXUIDFHDUHDRIWKHLFHFUHDPFRQHLV S
³³ dS ³³ R
2 ³³ dA
2 dA
R
2 Area of circle
2 S S 2.
R
6WXG\7LSV x
Notice the similarity between the differential of arc length and the differential of surface area: 2
ds 1 ª¬ f c x º¼ dx differential of arc length 2
2
dS 1 ª¬ f x x , y º¼ ª¬ f y x , y º¼ dA differential of surface area.
85
x
6XUIDFHDUHDLVWKHGRXEOHLQWHJUDORIWKHGLIIHUHQWLDORIVXUIDFHDUHD S
³³ dS . R
x
z @ dy dx ³ ³ 2 dy dx ³ > 2 y @ dx ³ 6 dx 4
3
0
0
2
4
0
4
> 6 x @0 ͼ6HH)LJXUHͽ
4
24. x
3 y
Figure 24.1
87
z
([DPSOH
1
6NHWFKWKHVROLGZKRVHYROXPHLVUHSUHVHQWHGE\WKHWULSOHLQWHJUDO 1 2 1 y ³ ³ ³ dz dx dy. Then, rewrite the integral in the order dy dz dx. 0
0
z íy
0
6ROXWLRQ
1
The limits of integration determine the shape of the solid. ͼ6HH)LJXUHͽ 1
2
1 y
Because z íy y íz, ³ ³ ³ dz dx dy 0 0 0 %RWKLQWHJUDOVJLYHWKHVDPHYROXPHRI
y
2
2
1
1 z
0
0
0
³³³
dy dz dx.
x Figure 24.2
([DPSOH 6HWXSWKHWULSOHLQWHJUDOIRUWKHYROXPHRIWKHVROLGUHJLRQERXQGHG below by the surface z x2 + y2 and above by z íx2íy2. ͼ6HH)LJXUHͽ
2
6ROXWLRQ
1
:HPXVW¿UVW¿QGWKHLQWHUVHFWLRQRIWKHWZRSDUDERORLGVE\VHWWLQJ the equations equal to each other:
z
í í
í
z
2 x2 y2
x2 y2 2
2x2 2 y2 x2 y2
y
1.
1
1
x
The region of integration is the unit circle. The volume is
Lesson 24: Triple Integrals and Applications
V
1
³ ³
1 x 2
1 1 x
2
³
2 x2 y2 x2 y2
Figure 24.3
dz dy dx.
7KLVLQWHJUDOLVGLI¿FXOWDQGWKHDQVZHULVʌ,QWKHQH[WOHVVRQZHZLOOVHHKRZWRVROYHWKHSUREOHPXVLQJ cylindrical coordinates. ([DPSOH )LQGWKHPDVVRIWKHXQLWFXEHLQWKH¿UVWRFWDQWJLYHQWKDWWKHGHQVLW\DWWKHSRLQWͼx, y, zͽLVWKHVTXDUHRILWV distance to the origin.
88
6ROXWLRQ The density is ȡͼx, y, zͽ kͼx2 + y2 + z2ͽ+HQFHWKHPDVVLVJLYHQE\ 1
1
1
0
0
0
³³³ U x, y , z dV ³ ³ ³ k x
m
2
y 2 z 2 dz dy dx.
Q
7KLVLQWHJUDOLVQRWGLI¿FXOWWRHYDOXDWHDQGWKH¿QDODQVZHULVk. 6WXG\7LSV x
-XVWDVZLWKGRXEOHLQWHJUDOVZHRIWHQRPLWLQWHJUDQGVRI)RULQVWDQFHLQ([DPSOH 4 3 2 4 3 2 ³ ³ ³ 1 dz dy dx ³ ³ ³ dz dy dx. 0
0
0
0
0
0
x
7KHUHDUHVL[RUGHUVRILQWHJUDWLRQIRUWULSOHLQWHJUDOVLQ&DUWHVLDQFRRUGLQDWHV dz dy dx, dz dx dy, dy dz dx, dy dx dz, dx dy dz, dx dz dy.
x
It is worth repeating that the setup of a problem is more important than the actual integrations.
3LWIDOO x
Remember that the variable of integration cannot appear as a limit of integration. The following triple 1 2 1 z integral is incorrect: ³ ³ ³ dz dx dy. 0
0
0
3UREOHPV 5
2
1
0
0
0
Evaluate the triple integral
³ ³ ³ dy dx dz. What does this represent?
Evaluate the triple integral
³ ³ ³ x y z dx dz dy.
Evaluate the triple integral
³ ³ ³
3
2
1
0
0
0
1
1
1
1 1 1
x 2 y 2 z 2 dx dy dz.
6HWXSWKHWULSOHLQWHJUDOIRUWKHYROXPHRIWKHVROLGLQWKH¿UVWRFWDQWERXQGHGE\WKHFRRUGLQDWHSODQHV and the plane z íxíy.
6HWXSWKHWULSOHLQWHJUDOIRUWKHYROXPHRIWKHVROLGERXQGHGE\z íx2, z y DQGy x. 6HWXSWKHWULSOHLQWHJUDOIRUWKHYROXPHRIWKHVROLGERXQGHGE\z íx2íy2 and z 6HWXSWKHLQWHJUDOIRUWKHPDVVRIWKHVROLGERXQGHGE\xy + 6z x y DQGz LIWKH density is ȡͼx, y, zͽ
89
6HWXSWKHLQWHJUDOIRUWKHPDVVRIWKHVROLGERXQGHGE\xy + 5z x y DQGz LIWKH density is ȡͼx, y, zͽ y.
Rewrite the iterated integral
Lesson 24: Triple Integrals and Applications
Rewrite the iterated integral
90
y2
1
0
0
1 0
4
4 x
0
0
³³ ³ ³³
2
dz dy dx using the order dy dz dx. 12 3x 6y
³
0
4
dz dy dx using the order dy dx dz.
Triple Integrals in Cylindrical Coordinates Lesson 25
Topics x
Cylindrical coordinates.
x
Conversion formulas.
x
Triple integrals in cylindrical coordinates.
x
The differential of volume in cylindrical coordinates.
x
Applications of triple integrals in cylindrical coordinates.
'H¿QLWLRQVDQG7KHRUHPV x
Let P ͼx, y, zͽEHDSRLQWLQVSDFH,WVF\OLQGULFDOFRRUGLQDWHVDUHͼr, ș, zͽZKHUHͼr, șͽDUHWKHSRODU coordinates of the projection of the point onto the xySODQH7KHz coordinate is the same.
x
Conversion formulas: x r cos ș, y r sin ș, z z r2 x2 + y2, tan ș
x
y , z z. x
The GLIIHUHQWLDORIYROXPH in cylindrical coordinates is dV U࣠G]࣠GU࣠Gș.
6XPPDU\ 7KHF\OLQGULFDOFRRUGLQDWHV\VWHPLVWKHWKUHHGLPHQVLRQDOJHQHUDOL]DWLRQRISRODUFRRUGLQDWHV These coordinates are especially useful for representing cylindrical surfaces and surfaces of revolution. The conversion formulas are similar to the conversion formulas between polar coordinates and Cartesian FRRUGLQDWHV:H¶OOVWXG\H[DPSOHVRIWULSOHLQWHJUDOVLQF\OLQGULFDOFRRUGLQDWHVDQGQRWHWKDWWKHGLIIHUHQWLDO RIYROXPHKDVWKHH[WUDr factor, dV U࣠G]࣠GU࣠Gș. ([DPSOH Convert the point r , T , z
4, 56S , 3 to Cartesian coordinates.
6ROXWLRQ r cosT
We use the conversion formulas: x and z
4cos 5S 6
§ 3· 4¨ ¸ 2 © ¹
2
The Cartesian coordinates of the point are x , y , z
2 3, y
4sin 5S 6
r sin T
4 1 2
2,
3, 2, 3 .
([DPSOH Convert the point x , y , z
1,
3, 2 to cylindrical coordinates.
6ROXWLRQ We have r2 x2 + y2 r DQG tan T Of course, z
y x
3 1
7KHUHDUHPDQ\SRVVLEOHF\OLQGULFDOFRRUGLQDWHV)RUH[DPSOH r , T , z or r , T , z 2, 4S , 2 . 3
S nS .
3, which gives T
3
2, S3 , 2 z
([DPSOH The surface ș cLVDYHUWLFDOSODQHͼ6HH)LJXUHͽ ([DPSOH
θ =c
Find the volume of the solid bounded below by z x + y and above by z íx2íy2ͼ6HH)LJXUHͽ 2
Lesson 25: Triple Integrals in Cylindrical Coordinates
y
x 2
Figure 25.1
6ROXWLRQ 2
The intersection of the two paraboloids is obtained by setting the equations equal to each other:
z
z íx2íy2 x2 + y2 x2 + 2y2 x2 + y2 1
Converting to cylindrical coordinates, z íx2íy2 ír2 and z x2 + y2 r2. The volume is V
³³³ dV Q
2S
1
2 r 2
0
0
r2
³ ³³
r dz dr dT .
í í
í
y 1
1
x Figure 25.2
92
This integral is easy to evaluate: V
2S
1
2 r 2
0
r2
³ ³ > rz @ 0
2S
1
0
0
dr dT
³ ³ ª¬r 2 r 2S
1
0
0
³ ³ ª¬2r 2r ³
2S
0
³
2S
0
2
r 2 º¼ dr dT
3
º¼ dr dT
1
ª 2 r4 º «¬ r 2 »¼ dT 0 1 dT 2
2S
ª1T º «¬ 2 »¼ 0
S.
6WXG\7LSV x
1RWLFHWKDWF\OLQGULFDOFRRUGLQDWHVDUHWKHQDWXUDOH[WHQVLRQRI&DUWHVLDQFRRUGLQDWHVWR three dimensions.
x
When converting from one coordinate system to another, you can always check your answer by converting back to the original coordinates.
x
,WLVKHOSIXOWRLGHQWLI\WKHFRRUGLQDWHV\VWHPUHSUHVHQWLQJDJLYHQSRLQW)RULQVWDQFHLQ([DPSOH write r , T , z 4, 5S , 3 and x , y , z 2 3, 2, 3 . 6
3LWIDOOV x
The cylindrical coordinates of a point are not unique. In particular, the rYDOXHFDQEHSRVLWLYHRU QHJDWLYH$QGWKHUHDUHLQ¿QLWHO\PDQ\FKRLFHVIRUWKHDQJOHș.
x
'RQ¶WIRUJHWWKHH[WUDr in the differential of volume, dV U࣠G]࣠GU࣠Gș.
3UREOHPV &RQYHUWWKHSRLQWͼr, ș, zͽ ͼíʌíͽWRUHFWDQJXODUFRRUGLQDWHV &RQYHUWWKHSRLQWͼr, ș, zͽ ͼ S , ͽWRUHFWDQJXODUFRRUGLQDWHV 4
&RQYHUWWKHSRLQWͼx, y, zͽ 2 2 , 2 2 , 4 to cylindrical coordinates. Find an equation in cylindrical coordinates for the rectangular equation x Find an equation in rectangular coordinates for the cylindrical equation r2 + z2
Find an equation in rectangular coordinates for the cylindrical equation r VLQș. Verify that V
2³
2S
0
Convert the integral
R1
³ ³ R2 2
³ ³
R12 r 2
0 4 x2
2 4 x
2
3
r dz dr dT
³
4 x2 y 2
4S R 2 R 2 2 . 1 2 3
x dz dy dx to cylindrical coordinates.
6HWXSWKHWULSOHLQWHJUDOLQF\OLQGULFDOFRRUGLQDWHVIRUWKHYROXPHRIWKHVROLGERXQGHGDERYH by z x and below by z x2 + 2y2.
6HWXSWKHWULSOHLQWHJUDOLQF\OLQGULFDOFRRUGLQDWHVIRUWKHYROXPHRIWKHVROLGLQVLGHWKHVSKHUH
Lesson 25: Triple Integrals in Cylindrical Coordinates
x2 + y2 + z2 DQGDERYHWKHXSSHUQDSSHRIWKHFRQHz2 x2 + y2.
Triple Integrals in Spherical Coordinates Lesson 26
Topics x
6SKHULFDOFRRUGLQDWHV
x
Conversion formulas.
x
Triple integrals in spherical coordinates.
x
The differential of volume in spherical coordinates.
x
Applications of triple integrals in spherical coordinates.
'H¿QLWLRQVDQG7KHRUHPV x
Let P ͼx, y, zͽEHDSRLQWLQVSDFH,WVVSKHULFDOFRRUGLQDWHVDUHͼȡ, ș, ͽZKHUHȡ is the distance from P to the origin, ș is the same angle as used in cylindrical coordinates, and is the angle between the positive zD[LVDQGWKHOLQHVHJPHQW OP ʌ.
x
Conversion formulas: r ȡ sin , z ȡ cos x r cos ș ȡ sin cos ș y r sin ș ȡ sin sin ș ȡ2 x2 + y2 + z2 ȡ tan ș
y x
cos z
U
x
x2 y 2 z 2
z . x2 y2 z 2
The GLIIHUHQWLDORIYROXPH in spherical coordinates is dV ȡ2 sin ࣠Gȡ࣠G࣠Gș.
95
6XPPDU\ 6SKHULFDOFRRUGLQDWHVDUHVLPLODUWRWKHORQJLWXGHDQGODWLWXGHFRRUGLQDWHVRQ(DUWK7KH¿UVWFRRUGLQDWHLVD GLVWDQFHDQGWKHRWKHUWZRFRRUGLQDWHVDUHDQJOHV:HEHJLQE\GH¿QLQJVSKHULFDOFRRUGLQDWHVLQVSDFHDQG GHYHORSWKHLUFRQYHUVLRQIRUPXODV$IWHUORRNLQJDWVRPHH[DPSOHVRIVXUIDFHVLQVSKHULFDOFRRUGLQDWHVZH apply them to the calculation of volumes and mass. For spherical coordinates, the differential of volume is a bit complicated: dV ȡ2 sin ࣠Gȡ࣠G࣠Gș. ([DPSOH § 4, S , S · to Cartesian coordinates. ¨ ¸ © 6 4¹
Convert the point U , T , I 6ROXWLRQ
We use the conversion formulas:
x
U sin I cos T
4sin S cos S 4 6
§ ·§ 3 · 4 ¨ 2 ¸¨ ¸ © 2 ¹© 2 ¹
y
U sin I sin T
4sin S sin S 4 6
§ · 4 ¨ 2 ¸ §¨ 1 ·¸ 2 © ¹© 2 ¹
z
U cos I
4 cos S 4
§ · 4¨ 2 ¸ © 2 ¹
6
2
2 2.
Hence, the Cartesian coordinates are x , y , z
2 , 2 2 .
6,
Lesson 26: Triple Integrals in Spherical Coordinates
([DPSOH &RQYHUWWKHSRLQWͼx, y, zͽ ͼͽWRVSKHULFDOFRRUGLQDWHV z
6ROXWLRQ x2 y2 z 2
We have U and cos I
z
U
0I
Therefore, U , T , I
42 0 0
4, tan T
S.
y x
0 4
0 T
0,
2
§ 4, 0, S · ͼ6HH)LJXUHͽ ¨ ¸ 2¹ ©
y
( x , y , z ) = ( 4, 0, 0 ) x
( e, θ , φ ) = ⎛⎜ 4, 0, π2 ⎞⎟ ⎝
⎠
Figure 26.1
96
z
([DPSOH The surface ȡ cLVDVSKHUHFHQWHUHGDWWKHRULJLQͼ6HH)LJXUHͽ ([DPSOH
c )LQGWKHYROXPHRIWKHLFHFUHDPFRQHERXQGHGDERYHE\WKHXSSHUKDOIRI the sphere x2 + y2 + z2 DQGEHORZE\ z x 2 y 2 ͼ6HH)LJXUHͽ
y
x Figure 26.2
6ROXWLRQ z
The intersection of the two surfaces is obtained by setting the equations equal to each other:
x2 y2 2
2
x y z
z z2 2
x2 y2
z=
x2 y2
x y x y 2
2
2
2
2 − x2 − y2
z=
2
x2 + y2
x2 + y2 = 1
1 z 1.
y x
Converting to spherical coordinates,
x2 y2 z 2
2
U2 U
Figure 26.3
2 and z
U cos I 1
$OVRșʌ7KHLFHFUHDPFRQHLVJLYHQE\ 0 d U d
V
2S
S
0
0
4
³³³ dV ³ ³ ³
0
2
U 2 sin I d U dI dT
Q
2 cos I cos I
1 I 2
S. 4
2 ,0 d I d S ,0 d T d 2S . The volume is 4
4S ª 2 1º . ¼ 3 ¬
You are asked to verify this integration in Problem 7. 6WXG\7LSV x
6SKHULFDOFRRUGLQDWHVDUHHVSHFLDOO\XVHIXOIRUVSKHUHVZKLFKKDYHDFHQWHURIV\PPHWU\
x
,WLVKHOSIXOWRLGHQWLI\WKHFRRUGLQDWHV\VWHPUHSUHVHQWLQJDJLYHQSRLQW)RULQVWDQFHLQ([DPSOH write U , T , I §¨ 4, S , S ·¸ and x , y , z 6 , 2 , 2 2 . © 6 4¹
97
3LWIDOOV x
)RUVSKHULFDOFRRUGLQDWHVʌ and ȡ)XUWKHUPRUHș is the same angle as in polar coordinates for r
x
7KHUHFDQEHFRQIXVLRQLIWKHFRRUGLQDWHV\VWHPLVQRWPDGHH[SOLFLW)RULQVWDQFHͼʌͽLVWKH origin in spherical coordinates, but it is a point on the yD[LVLQUHFWDQJXODUFRRUGLQDWHV
x
Don’t forget the complicated differential of volume in spherical coordinates, dV ȡ2 sin ࣠Gȡ࣠G࣠Gș.
3UREOHPV Convert the point U , T , I §¨12, S , 0 ·¸ to rectangular coordinates. 4 ©
¹
Convert the point U , T , I §¨ 5, S , 3S ·¸ to rectangular coordinates. 4 4 ©
Convert the point x, y , z
¹
2, 2
3, 4 to spherical coordinates.
Find an equation in spherical coordinates for the rectangular equation z Find an equation in rectangular coordinates for the spherical equation S . 6
Convert the integral Verify that V
2
³ ³
4 x
2
2
2 4 x 2
2S
S
0
0
³
2 4 x y
2
4
³³³ dV ³ ³ ³
0
2
2
x dz dy dx to spherical coordinates.
U 2 sin I dU dI dT
Lesson 26: Triple Integrals in Spherical Coordinates
Q
4S ª 2 1º . ¼ 3 ¬
6HWXSWKHWULSOHLQWHJUDOLQVSKHULFDOFRRUGLQDWHVIRUWKHYROXPHRIWKHVROLGLQVLGHWKHVSKHUH x2 + y2 + z2 RXWVLGH z
x 2 y 2 , and above the xySODQH
6HWXSWKHWULSOHLQWHJUDOLQVSKHULFDOFRRUGLQDWHVIRUWKHYROXPHRIWKHWRUXVJLYHQE\ȡ VLQ . 6HWXSWKHWULSOHLQWHJUDOLQVSKHULFDOFRRUGLQDWHVIRUWKHPDVVRIWKHVSKHUHRIUDGLXVLIWKHGHQVLW\LV proportional to the distance of the point to the zD[LV
98
Vector Fields—Velocity, Gravity, Electricity Lesson 27
Topics x
9HFWRU¿HOGV
x
5RWDWLRQDQGUDGLDOYHFWRU¿HOGV
x
7KHJUDGLHQWDVDYHFWRU¿HOG
x
*UDYLWDWLRQDO¿HOGV
x
(OHFWULFIRUFH¿HOGV
x
&RQVHUYDWLYHYHFWRU¿HOGV
x
&DOFXODWLQJWKHSRWHQWLDOIRUDFRQVHUYDWLYHYHFWRU¿HOGLQWKHSODQH
'H¿QLWLRQVDQG7KHRUHPV x
A YHFWRU¿HOG over a planar region R is a function F that assigns a vector )ͼx, yͽWRHDFKSRLQWLQR.
x
A YHFWRU¿HOG over a solid region Q is a function F that assigns a vector )ͼx, y, zͽWRHDFKSRLQWLQQ.
x
1HZWRQV¶VODZRIJUDYLWDWLRQVWDWHVWKDWWKHIRUFHRIDWWUDFWLRQH[HUWHGRQDSDUWLFOHRIPDVVm ORFDWHGDWͼx, y, zͽE\DSDUWLFOHRIPDVV m2ORFDWHGDWͼͽLV F x, y , z
Gm1m2 u. x2 y2 z 2
Here, G is the gravitational constant, and uLVWKHXQLWYHFWRUIURPWKHRULJLQWRͼx, y, zͽ x
&RXORPE¶VODZVWDWHVWKDWWKHIRUFHH[HUWHGRQDSDUWLFOHZLWKHOHFWULFFKDUJHqORFDWHGDWͼx, y, zͽE\ c q1 q2 a particle of charge q2ORFDWHGDWͼͽLV F x , y , z u. 2 r
x
$YHFWRU¿HOGF is FRQVHUYDWLYHLIWKHUHH[LVWVDGLIIHUHQWLDEOHIXQFWLRQfͼWKHSRWHQWLDOͽVXFKWKDW F f .
x
Theorem: Let M and NKDYHFRQWLQXRXV¿UVWSDUWLDOGHULYDWLYHVRQDQRSHQGLVNR7KHYHFWRU¿HOG given by ) ͼx, yͽ Mi + Nj is conservative if and only if wN wM . wx wy
99
6XPPDU\ ,QWKLVOHVVRQZHGHYHORSWKHFRQFHSWRIDYHFWRU¿HOG,QWKHSODQHDYHFWRU¿HOGLVDIXQFWLRQWKDWDVVLJQVD YHFWRUWRHDFKSRLQWLQWKHSODQH6LPLODUO\LQVSDFHDYHFWRU¿HOGDVVLJQVDYHFWRUWRHDFKSRLQWLQVSDFH:H EHJLQE\GHVFULELQJW\SLFDOYHFWRU¿HOGVLQFOXGLQJURWDWLRQDO¿HOGVJUDYLWDWLRQDO¿HOGVDQGHOHFWULF¿HOGV,Q IDFWWKHJUDGLHQWRIDIXQFWLRQLVDYHFWRU¿HOG:HGH¿QHZKDWLVPHDQWE\DFRQVHUYDWLYHYHFWRU¿HOGRQHWKDW LVWKHJUDGLHQWRIDSRWHQWLDOIXQFWLRQ0RVWRIWKHLPSRUWDQW¿HOGVLQDSSOLFDWLRQVDUHFRQVHUYDWLYH:HGHYHORS DWHVWIRUFRQVHUYDWLYHYHFWRU¿HOGVLQWKHSODQHDQGVKRZKRZWR¿QGWKHSRWHQWLDOIXQFWLRQ ([DPSOH 6NHWFKVRPHYHFWRUVLQWKHYHFWRU¿HOGJLYHQE\) ͼx, yͽ íyi + xj. 6ROXWLRQ
y :H¿UVWFRPSXWHVRPHSRLQWVWRJHWDQLGHDRIWKH¿HOG ) ͼx, yͽ ) ͼͽ ) ͼͽ ) ͼͽ
Vector field: F(x, y) = íyi + xj
íyi + xj ij j íi + 0j íi íi + 2j.
x
%\SORWWLQJWKHVHYHFWRUVZHREWDLQWKHIROORZLQJH[DPSOHRID URWDWLRQYHFWRU¿HOGͼ6HH)LJXUHͽ Figure 27.1
Lesson 27: Vector Fields—Velocity, Gravity, Electricity
([DPSOH 6NHWFKVRPHYHFWRUVLQWKHYHFWRU¿HOG) ͼx, yͽ xi + yj.
y
6ROXWLRQ :H¿UVWFRPSXWHVRPHSRLQWVWRJHWDQLGHDRIWKH¿HOG í
) ͼx, yͽ xi + yj ) ͼͽ i + j ) ͼííͽ íiíj ) ͼíͽ íi + 2j. %\SORWWLQJWKHVHYHFWRUVZHREWDLQWKHIROORZLQJH[DPSOHRID UDGLDOYHFWRU¿HOGͼ6HH)LJXUHͽ
x íí
Figure 27.2
([DPSOH Thegradient of a function fLVDYHFWRU¿HOG )RUH[DPSOHLI f x , y
x2 y
y2 , then f x , y 2
f x x , y i f y x , y j 2 xyi x 2 y jLVDYHFWRU¿HOG
$WHDFKSRLQWͼx, yͽLQWKHSODQHWKHJUDGLHQWDVVLJQVDYHFWRU ([DPSOH 7KHYHFWRU¿HOG)ͼx, yͽ ͼxyͽiͼx + 2yͽj is conservative with potential f ͼx, yͽ x2xy + y2 because F f . ([DPSOH 7KHYHFWRU¿HOG)ͼx, yͽ x2yi + xyj is not conservative because M
x 2 y , wM wy
x2 , N
xy , wN wx
y , and x2y.
([DPSOH )LQGDSRWHQWLDOIRUWKHFRQVHUYDWLYHYHFWRU¿HOG)ͼx, yͽ xyiͼx2íyͽj. 6ROXWLRQ 1RWHWKDWWKH¿HOGLVFRQVHUYDWLYHEHFDXVH wM wy
2 x and wN wx
2 x.
:HQHHGWR¿QGDIXQFWLRQf ͼx, yͽVXFKWKDWf x , y 2 xyi x 2 y j. That is, fx࣠ͼx, yͽ xy and f y࣠ͼx, yͽ x2íy. ,QWHJUDWLQJWKH¿UVWHTXDWLRQ
f x, y
³ f x, y dx ³ 2 xy dx x
x 2 y g y .
Integrating the second equation,
f x, y
³ f x, y dy ³ x y
2
y dy
x2 y
y2 h ( x ). 2
From these two versions of the function f, we have
f x, y
x2 y
y2 K. 2
6WXG\7LSV x
$YHFWRU¿HOGDVVLJQVDvector to each point in the domain.
x
*UDYLWDWLRQDO¿HOGVDQGHOHFWULF¿HOGVKDYHWKHVDPHIRUPDQGDUHH[DPSOHVRILQYHUVHVTXDUH¿HOGV. k u. Given U xi + yj + zkWKHYHFWRU¿HOGFLVDQLQYHUVHVTXDUH¿HOGLI F x , y , z 2 r
x
(YHU\LQYHUVHVTXDUH¿HOGLVFRQVHUYDWLYH+HQFHJUDYLWDWLRQDO¿HOGVDQGHOHFWULFIRUFH¿HOGV are conservative.
x
$IWHUFDOFXODWLQJWKHSRWHQWLDOIRUDFRQVHUYDWLYHYHFWRU¿HOG\RXFDQFKHFN\RXUDQVZHUE\WDNLQJ the gradient of the potential. § · y2 )RULQVWDQFHLQ([DPSOH f x , y ¨ x 2 y K¸ 2 © ¹ YHFWRU¿HOG
2 xyi x 2 y j, which is the original
3LWIDOOV x
)RUFRQVHUYDWLYHYHFWRU¿HOGV F f , keep in mind that f is a function of two or three variables, whereas FLVDYHFWRU¿HOG
x
1RWLFHLQ([DPSOHWKDWWKH³FRQVWDQWVRILQWHJUDWLRQ´DUHIXQFWLRQVRIWKH³RWKHU´YDULDEOH7KH¿QDO answer has a true constant.
Lesson 27: Vector Fields—Velocity, Gravity, Electricity
3UREOHPV 'HVFULEHWKHYHFWRU¿HOG) ͼx, yͽ i + j and compute F . 'HVFULEHWKHYHFWRU¿HOG) ͼx, y, zͽ i + j + k and compute F . 'HWHUPLQHZKHWKHURUQRWWKHYHFWRU¿HOG) ͼx, yͽ y2ͼyixjͽLVFRQVHUYDWLYH 'HWHUPLQHZKHWKHURUQRWWKHYHFWRU¿HOG) ͼx, yͽ
1 i j is conservative. x2 y 2
)LQGWKHFRQVHUYDWLYHYHFWRU¿HOGIRUWKHSRWHQWLDOIXQFWLRQf ͼx, yͽ x 2 1 y 2 . 4
)LQGWKHFRQVHUYDWLYHYHFWRU¿HOGIRUWKHSRWHQWLDOIXQFWLRQf ͼx, y, zͽ xyz. 2
)LQGWKHFRQVHUYDWLYHYHFWRU¿HOGIRUWKHSRWHQWLDOIXQFWLRQ f ͼx, y, zͽ z ye x .
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y x i 2 j. 2 x y x y2 2
)LQGDSRWHQWLDOIXQFWLRQIRUWKHFRQVHUYDWLYHYHFWRU¿HOG)ͼx, yͽ x2y2i + 2xyj.
Curl, Divergence, Line Integrals Lesson 28
Topics x
7KHFXUORIDYHFWRU¿HOG
x
7KHGLYHUJHQFHRIDYHFWRU¿HOG
x
/LQHLQWHJUDOVRIIXQFWLRQVGH¿QHGRQFXUYHV
x
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TheFXUORIWKHYHFWRU¿HOG)ͼx, y, zͽLV § curl F x , y , z ¨ wP wN © wy wz
u F x, y , z
Lesson 28: Curl, Divergence, Line Integrals
· § wN wM wP wM ¸ i wx wz j ¨ wx wy ¹ © i w wx M
j w wy N
· ¸k ¹
k w . wz P
x
8QGHUVXLWDEOHK\SRWKHVHVDYHFWRU¿HOGLVFRQVHUYDWLYHLIDQGRQO\LIFXUO F
x
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2
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,IDYHFWRU¿HOGLVFRQVHUYDWLYHWKHQWKHOLQHLQWHJUDOEHWZHHQDQ\WZRSRLQWVLVVLPSO\WKH GLIIHUHQFHRIWKHSRWHQWLDOIXQFWLRQHYDOXDWHGDWWKRVHSRLQWV7KHOLQHLQWHJUDOLVLQGHSHQGHQWRI SDWK)RUH[DPSOHWKHYHFWRU¿HOGLQ([DPSOHLVFRQVHUYDWLYHZLWKSRWHQWLDOf ͼx, yͽ xy%\WKH IXQGDPHQWDOWKHRUHPRIOLQHLQWHJUDOV
³
C
x
F < dr
³
C
f < dr
f 1, 1 f 0, 0 (12 )(1) 0 1 0 1.
1RWHWKHVLPLODULWLHVEHWZHHQWKHIXQGDPHQWDOWKHRUHPRIFDOFXOXVDQGWKHIXQGDPHQWDOWKHRUHPRI OLQHLQWHJUDOV
³
b
a
³
C
³
f ( x ) dx F < dr
b
a
³
C
F c x dx
f < dr
F (b ) F ( a )
f x b , y b f x a , y a .
r u u @ r u uc r c u u. dt
FXUORIDYHFWRU¿HOG ͼͽ $YHFWRU¿HOGWKDWPHDVXUHVDQRWKHUYHFWRU¿HOG¶VWHQGHQF\WRURWDWHͼwhen curl = 0, D¿HOGLVLUURWDWLRQDOͽ8VHGDVDWHVWIRUFRQVHUYDWLYHYHFWRU¿HOGV&DOFXODWHGXVLQJDFURVVSURGXFWRIWKH GLIIHUHQWLDORSHUDWRUZLWKWKHYHFWRU¿HOG curl F u F. )RUWKHYHFWRU¿HOG)ͼx, y, zͽ, § wP wN ¨ wy wz ©
curl F x , y , z
· § wP wM · § wN wM ¸ i ¨© wx wz ¸¹ j ¨ wx wy ¹ ©
u F x, y , z
i w wx M
j w wy N
· ¸k ¹
k w . wz P
curvature: A measure of how much a curve bends, K
Tc( t ) . r c( t )
The curvature of y = f ͼxͽ is K
y cc ª1 y c 2 º ¬ ¼
3
. 2
See Calculus II, Lesson 36.
FXUYHͼͽ: $SODQDUFXUYHLVGH¿QHGE\WKHIXQFWLRQVx = f ͼtͽ, y = J࣠ͼtͽ, and z = K࣠ͼtͽ. A curve is called simple if it does not cross itself.
Glossary
F\OLQGHUͼͽ: ,QHOHPHQWDU\JHRPHWU\DF\OLQGHUUHVXOWVZKHQOLQHVSHUSHQGLFXODUWRDFLUFOHJHQHUDWHDWXEH shape, but LQKLJKHUPDWKHPDWLFVDF\OLQGHUͼRUF\OLQGULFDOVXUIDFHͽFDQUHIHUWRDQ\VXUIDFHFUHDWHGZKHQ DQ\JHQHUDWLQJFXUYHLQDSODQHͼnot just a circleͽLVH[WHQGHGLQWRDWKLUGGLPHQVLRQE\OLQHVLQWHUVHFWLQJ that curve and orthogonal to its plane.
F\OLQGULFDOFRRUGLQDWHVͼͽ: The three-dimensional generalization of polar coordinates: x y r sin t , z z.
196
r cos t ,
cycloid: 7KHFXUYHWUDFHGRXWE\DSRLQWRQWKHFLUFXPIHUHQFHRIDFLUFOHUROOLQJDORQJDOLQH See Calculus II, Lesson 28.
GH¿QLWHLQWHJUDO: Let fEHGH¿QHGRQWKHLQWHUYDO>࣠D, E࣠@3DUWLWLRQWKHLQWHUYDOLQWRn equal subintervals of length 'x b a , x0 a , x1 , x2 , }, xn 1 , xn b. n n
Assume that the following limit exists: lim ¦ f ci 'x , where xi 1 d ci d xi . n of
i 1
7KHQWKLVOLPLWLVWKHGH¿QLWHLQWHJUDO of f from a to b and is denoted
b
³ f x dx. See Calculus II, Lesson 3. a
GHOͼͽ: See differential operator.
delta xͼ¨ xͽͼͽ: 7KHV\PERO ¨x is read “delta x” and denotes Dͼsmallͽ change in x. Some textbooks use h LQVWHDGRI¨x.
GHQVLW\ͼͽ: 8VXDOO\PDVVSHUXQLWYROXPHEXWIRUSODQDUODPLQDVGHQVLW\LVPDVVSHUXQLWRIVXUIDFHDUHD
derivative: ,QHOHPHQWDU\FDOFXOXVWKHGHULYDWLYH of f at xLVJLYHQE\WKHIROORZLQJOLPLWLILWH[LVWV f c x
lim
'x o 0
f x 'x f x . 'x
Notations for the derivative of y = f ͼxͽ: f c x ,
dy , y c, d ª¬ f x º¼ , D > y @ . dx dx
7KHGH¿QLWLRQVRIVORSHDQGWKHGHULYDWLYHDUHEDVHGRQWKHGLIIHUHQFHTXRWLHQWIRUVORSH slope
change in y change in x
'y . 'x
In multivariable calculus, WKHIXQFWLRQVDUHRIWZRͼor moreͽ variables, and we use partial derivatives: f x x, y
lim
'x o 0
f x 'x , y f x , y , f y x, y 'x
lim
'y o 0
f x , y 'y f x , y . 'y
197
GHWHUPLQDQWQRWDWLRQͼͽ: ,QWKLVFRXUVHZHXVHDîGHWHUPLQDQWIRUPRQO\WRKHOSXVUHPHPEHUDQG calculate the cross product RIWZRYHFWRUV7HFKQLFDOO\DGHWHUPLQDQWLVDVLQJOHUHDOQXPEHUREWDLQHGE\ using determinant notation, but in that sense, this course has no determinants—which are a topic covered in linear algebra. See cross product.
differentiable ͼͽ: In multivariable calculus, a function z = f ͼx, yͽLVGLIIHUHQWLDEOHDWWKHSRLQWͼx0, y0ͽLI¨z can be written in the form 'z f x x0 , y0 'x f y x0 , y0 'y H 1'x H 2 'y , where İ1 and İ2 tend to zero as 'x , 'y o 0, 0 . 'LIIHUHQWLDELOLW\DWDSRLQWRQDVXUIDFHLPSOLHVWKDWWKHVXUIDFHFDQEHDSSUR[LPDWHG E\DWDQJHQWSODQHDWWKDWSRLQW
differential: ,QHOHPHQWDU\FDOFXOXVZHOHWy = f ͼxͽ be a differentiable function. Then, dx ¨x is called the differential of x. The differential of y is dy f c x dx. For multivariable calculus, see total differential. See Calculus II, Lesson 2.
differential equation: A differential equation in x and y is an equation that involves x, y, and derivatives of y. The order RIDGLIIHUHQWLDOHTXDWLRQLVGHWHUPLQHGE\WKHKLJKHVWRUGHUGHULYDWLYHLQWKHHTXDWLRQ dy $¿UVWRUGHUOLQHDUGLIIHUHQWLDOHTXDWLRQFDQEHZULWWHQLQWKHVWDQGDUGIRUP P x y Q x . dx See Calculus II, Lessons 4–6.
GLIIHUHQWLDORSHUDWRUͼGHOͽͼͽͼͽ: “del,” or “grad,” or “nabla.”
w , wx
w , or wy
w . 8VHGLQFXUOGLYHUJHQFH3URQRXQFHG wz
GLUHFWLRQDOGHULYDWLYHͼͽ $JHQHUDOL]DWLRQRIWKHFRQFHSWRISDUWLDOGHULYDWLYHWKDWFDQEHXVHGWR¿QGWKH VORSHDZD\IURPDSRLQWLQDQ\JLYHQGLUHFWLRQ
GLUHFWLRQQXPEHUVͼ ͽ: Component numbers in a direction vector.
Glossary
GLVNͼͽ: Two-dimensional analog for intervals along the x-axis in beginning calculus. An open disk that is the interior of a circle. Compare with planar lamina.
GLYHUJHQFHRIDYHFWRU¿HOGͼͽ: $VFDODUWKDWPHDVXUHVRXWZDUGÀX[SHUXQLWYROXPHWKHWHQGHQF\ RIDYHFWRU¿HOGWRGLYHUJHIURPDJLYHQSRLQW3RVLWLYHGLYHUJHQFHLVDsource, negative divergence is a sink, and divergence = 0 is divergence free or incompressible. Calculated using a dot product of the GLIIHUHQWLDORSHUDWRUZLWKWKHYHFWRU¿HOG div F < F. 198
GLYHUJHQFHWKHRUHPͼͽ: A generalization of Green’s theorem that relates a ÀX[LQWHJUDORYHUWKHERXQGDU\ of a solid with a triple integral over the entire solid:
³³ F < N dS ³³³ divF dV ³³³ a, b@6HHCalculus II, Lesson 15.
201
integrating factor: For a linear differential equation, the integrating factor is u Lesson 6.
e³
P ( x ) dx
. See Calculus II,
integration by partial fractions: An algebraic technique for splitting up complicated algebraic expressions— LQSDUWLFXODUUDWLRQDOIXQFWLRQV²LQWRDVXPRIVLPSOHUIXQFWLRQVZKLFKFDQWKHQEHLQWHJUDWHGHDVLO\ using other techniques. See Calculus II, Lesson 13.
integration by parts: ³ u dv
uv ³ v du. See Calculus II, Lesson 10.
integration by substitution: Let F be an antiderivative of f. If u = J࣠ͼxͽ, then du
³ f g x g c x dx
F g x C because
³ f u du
g c x dx , so we have
F u C.
See Calculus II, Lesson 3.
iterated integrals ͼ ͽ: Repeated simple integrals, such as double integrals and triple integrals. The inside limits of integration can be variable with respect to the outer variable of integration, but the outside limits of integration must be constant with respect to both outside limits of integration.
inverse functions 7KRVHZKRVHJUDSKVDUHV\PPHWULFDFURVVWKHOLQHy = x. A function g is the inverse function of the function f if f ͼJ࣠ͼxͽͽ = x for all x in the domain of g and J࣠ͼf ͼxͽͽ = x for all x in the domain of f. The inverse of f is denoted f 1 . Reviewed in Calculus II, Lesson 1.
LQYHUVHVTXDUH¿HOGV ͼͽ: Fields where the force decreases in proportion with the square of distance. k u. Given r xi yj zk , WKHYHFWRU¿HOGFLVDQLQYHUVHVTXDUH¿HOGLI F x , y , z 2 r
Glossary
inverse trigonometric functions: These inverse functions DUHGH¿QHGE\UHVWULFWLQJWKHGRPDLQRIWKHRULJLQDO function, as follows.
202
y
arcsin x
sin 1 x sin y
y
arccos x
cos 1 x cos y
x , for 1 d x d 1 and S d y d S . 2 2 x , for 1 d x d 1 and 0 d y d S .
y
arctan x
tan 1 x tan y
x , for f x f and S y S . 2 2
y
arcsec x
sec 1 x sec y
x , for x t 1, 0 d y d S , and y z S . 2
Reviewed in Calculus II, Lesson1.
.HSOHU¶VODZVͼͽ: 1ͽ The orbit of each planet is an ellipse, with the Sun at one of the two foci; 2ͽ a line joining a planet and the Sun sweeps out equal areas during equal intervals of time; 3ͽ the square of the RUELWDOSHULRGRIDSODQHWLVGLUHFWO\SURSRUWLRQDOWRWKHFXEHRIWKHVHPLPDMRUD[LVRIWKHRUELW
/DJUDQJHPXOWLSOLHUͼͽ: A scalar, ȜXVHGLQDSRZHUIXOWHFKQLTXHJLYHQE\/DJUDQJH¶VWKHRUHPIRUVROYLQJ optimization problems that have constraints.
ODPLQDͼͽ $WKLQÀDWSODWHRIPDWHULDOXVXDOO\RIXQLIRUPGHQVLW\
/DSODFH¶VSDUWLDOGLIIHUHQWLDOHTXDWLRQͼͽ 'HVFULEHVWKHVWHDG\VWDWHWHPSHUDWXUHGLVWULEXWLRQLQSODWHVRU 2 2 solids. w z2 w z2 0. $IXQFWLRQWKDWVDWLV¿HVWKLVHTXDWLRQLVVDLGWREHharmonic. wx wy
ODZRIFRQVHUYDWLRQRIHQHUJ\ͼͽ ,QDFRQVHUYDWLYHIRUFH¿HOGWKHVXPRISRWHQWLDODQGNLQHWLFHQHUJLHVRI an object remain constant from point to point.
OHDVWVTXDUHVUHJUHVVLRQOLQHͼͽ: 8VHGWR¿WDOLQHWRDVHWRISRLQWVLQWKHSODQH:RUNVEHVWZKHQWKHGDWDLV QHDUO\OLQHDU'HULYHGE\PLQLPL]LQJWKHVXPRIWKHVTXDUHVRIWKHGLIIHUHQFHVEHWZHHQWKHGDWDDQGWKH line. If f ͼxͽ = ax + b, then the values of a and bDUHJLYHQE\ n
n
n
i 1
i 1
i 1
n ¦ xi yi ¦ xi ¦ yi a
n § n · n ¦ xi 2 ¨ ¦ xi ¸ i 1 ©i 1 ¹
2
,b
n n 1 § y a x ·. i i ¸ ¦ ¦ ¨ n© i 1 i 1 ¹
level curve ͼͽ: Also known as a contour lineWKHVHWRIDOOSRLQWVLQWKHSODQHVDWLVI\LQJf ͼx, yͽ = c, when z = f ͼx, yͽ and c is a constant. Contrast with trace, which is the intersection of a surface with a plane.
203
level surface ͼͽ: Although a function in three variables f ͼx, y, zͽ cannot itself be graphed, it is possible to graph a level surface, the set of all points in space where that function equals a constant, f ͼx, y, zͽ = c.
L’Hôpital’s rule: A technique for evaluating indeterminate forms for limits such as 0 or f , where no f 0 guaranteed limit exists. See Calculus II, Lesson 14.
limit 'H¿QHGLQIRUPDOO\LIf ͼxͽEHFRPHVDUELWUDULO\FORVHWRDVLQJOHQXPEHUL as x approaches c from either VLGHZHVD\WKDWWKHOLPLWRIf ͼxͽ as x approaches c is L, which we write as lim f x L. x oc
Also, the equation lim f x x oc
f means that f ͼxͽ increases without bound as x approaches c.
0RUHIRUPDOO\ /HWfEHDIXQFWLRQGH¿QHGRQDQRSHQLQWHUYDOFRQWDLQLQJc ͼH[FHSWSRVVLEO\DWcͽ, and let L be a real number. The statement lim f x L means that for each İ > 0, there exists a į > 0 such that if x oc 0 x c G , then f ( x ) L H . See Calculus II, Lesson 1. 7KHGH¿QLWLRQIRUDOLPLWLQPXOWLYDULDEOHFDOFXOXVLVVLPLODUWRWKDWLQHOHPHQWDU\FDOFXOXVH[FHSWWKDWZH XVHRSHQGLVNVͼDQGDSSURDFKIURPDQ\GLUHFWLRQͽLQVWHDGRIXVLQJRSHQLQWHUYDOVͼDSSURDFKLQJIURPRQO\ two directionsͽ:HVD\WKDW lim f x , y L if for eachİ > 0, there existsį > 0 such that x , y o x0 , y0
f x , y L H whenever 0<
2
x x0 y y 0
2
G.
OLQHDUPRGHOͼͽ: Given a set of data, a linear model is a function y = ax + bWKDWFORVHO\¿WVWKHGDWD
OLQHLQWHJUDOͼ ͽ: Integration over a piecewise smooth curve, which can be used to calculate the mass of DWKLQZLUHͼwhere the value of the integral does not depend on the orientation of the curveͽRUZRUNE\D IRUFH¿HOGͼwhere the orientation of the path does matter and determines the sign of the answerͽ. Green’s theoremLPSOLHVWKDWWKHOLQHLQWHJUDODURXQGDQ\FORVHGFXUYHZLWKLQDFRQVHUYDWLYHYHFWRU¿HOGLV]HUR See fundamental theorem of line integrals.
Glossary
OLQHVLQVSDFHͼ ͽ 'H¿QHGXVLQJDSRLQWDQGDGLUHFWLRQYHFWRUWKHVHDUHTXLWHGLIIHUHQWIURPOLQHVLQWKHSODQH ͼGH¿QHGXVLQJVORSHDQGy-interceptͽ. See planes in space and parametric equations.
204
log rule for integration:
1
³ x dx
ln x C .
For a logarithmic function to base a, when a > 0 and a log a x See Calculus II, Lesson 3.
1 ln x. See Calculus, Lesson 27. ln a
mass ͼͽ 7KHGRXEOHLQWHJUDORIWKHGHQVLW\IXQFWLRQ
0D[ZHOO¶VHTXDWLRQVͼͽ: Four partial differential equations, and their integral forms, describing the LQWHUDFWLRQRIHOHFWULFDQGPDJQHWLF¿HOGV7KHLUGLIIHUHQWLDOIRUPVDUHDVIROORZV
)RUDQHOHFWULF¿HOGE and a surface S enclosing a charge Q,
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