Understanding Multivariable Calculus

Citation preview

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

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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&RORPELDZKHUHKHWDXJKWPDWKHPDWLFVͼ௘in Spanish௘ͽ 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;

x

Understanding Calculus: Problems, Solutions, and Tips;

x

Understanding Calculus II: Problems, Solutions, and Tips; and

x

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 y2”x”6RWKHUHJLRQLV bounded on the left by x y 2 œ y x and on the right by x 

2

)XUWKHUPRUH”y”VRWKHUHJLRQLVDV 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

ͼ௘6HH)LJXUH௘ͽ ([DPSOH

x Figure 20.1

The double integral for the volume under the surface z VLQ y2 VLQ ͼ௘y2௘ͽDQGDERYHWKHUHJLRQERXQGHGE\ 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 YHUWLFDOUHSUHVHQWDWLYHUHFWDQJOHVͼ௘6HH)LJXUH௘ͽ

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 DQGͼ௘௘ͽ

1 xy over the rectangular region RZLWKYHUWLFHVͼ௘௘ͽͼ௘௘ͽͼ௘௘ͽ 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 DQGDERYHWKHUHFWDQJOH”x””y” 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 ͼ௘x, y௘ͽ xRYHUWKHUHFWDQJOHZLWKYHUWLFHVͼ௘௘ͽͼ௘௘ͽͼ௘௘ͽ DQGͼ௘௘ͽ

 Find the average value of the function f ͼ௘x, y௘ͽ VLQͼ௘x + y௘ͽRYHUWKHUHFWDQJOHZLWKYHUWLFHVͼ௘௘ͽͼ௘ʌ௘ͽ ͼ௘ʌ, ʌ௘ͽDQGͼ௘ʌ௘ͽ 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 ”gͼ௘ș௘ͽ”r”g2ͼ௘ș௘ͽĮ”ș”ȕDQG”ͼ௘ȕíĮ௘ͽ”ʌ. 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. ͼ௘6HH)LJXUH௘ͽ

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

ͼ௘6HH)LJXUH௘ͽ

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 ȡ௘ͼ௘x, y௘ͽWKHQWKHPDVV 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 FHQWHURIPDVVLVͼ௘ଲxଲ 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

)LQGWKHPDVVRIWKHWULDQJXODUODPLQDZLWKYHUWLFHVͼ௘௘ͽͼ௘௘ͽ DQGͼ௘௘ͽLIWKHGHQVLW\DWWKHSRLQWͼ௘x, y௘ͽLVȡ௘ͼ௘x, y௘ͽ x + y. ͼ௘6HH)LJXUH௘ͽ

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 )LQGWKHFHQWHURIPDVVRIWKHODPLQDFRUUHVSRQGLQJWRWKHSDUDEROLFUHJLRQ”y”íx2 if the density at the SRLQWͼ௘x, y௘ͽLVFRQVWDQWȡ௘ͼ௘x, y௘ͽ  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 , DQGWKHFHQWHURIPDVVLVͼ௘ଲxଲ y ௘ͽ  § 0, 8 · . ¨ ¸ 5 © 5¹

([DPSOH 6ROYHWKHSUHYLRXVH[DPSOHDVVXPLQJWKDWWKHGHQVLW\LVQRWFRQVWDQWEXWUDWKHUJLYHQE\ȡ௘ͼ௘x, y௘ͽ 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[DPSOHͼ௘6HH)LJXUH௘ͽ

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\”x”DQG”y”LIWKHGHQVLW\LVȡ௘ͼ௘x, y௘ͽ xy.  )LQGWKHPDVVRIWKHODPLQDERXQGHGE\”x”DQG”y” 1  x 2 if the density is ȡ௘ͼ௘x, y௘ͽ xy.  )LQGWKHPDVVDQGFHQWHURIPDVVRIWKHWULDQJXODUODPLQDZLWKYHUWLFHVͼ௘௘ͽͼ௘௘ͽDQGͼ௘௘ͽLIWKH density is ȡ௘ͼ௘x, y௘ͽ y.

 )LQGWKHPDVVDQGFHQWHURIPDVVRIWKHWULDQJXODUODPLQDZLWKYHUWLFHVͼ௘௘ͽͼ௘௘ͽDQGͼ௘௘ͽLIWKH density is ȡ௘ͼ௘x, y௘ͽ x.

 Find the mass and center of mass of the lamina bounded by y  x , y DQGx LIWKHGHQVLW\LV ȡ௘ͼ௘x, y௘ͽ y.



 Find the mass and center of mass of the lamina bounded by y x2, y DQGx LIWKHGHQVLW\LV ȡ௘ͼ௘x, y௘ͽ xy.

 Find the mass and center of mass of the lamina bounded by x2 + y2 ”xDQG”y if the density is

Lesson 22: Centers of Mass for Variable Density

ȡ௘ͼ௘x, y௘ͽ ͼ௘x2 + y2௘ͽ

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 ͼ௘x, y௘ͽGH¿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 ͼ௘6HH)LJXUH௘ͽ

z (0, 0, 2)

z íxíy 1

6ROXWLRQ We have fx ͼ௘x, y௘ͽ íf y ͼ௘x, y௘ͽ 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 ͼ௘x, y௘ͽ íx2 + y that lies above WKHWULDQJXODUUHJLRQZLWKYHUWLFHVͼ௘௘ͽͼ௘í௘ͽDQGͼ௘௘ͽ 6ROXWLRQ The partial derivatives are fx ͼ௘x, y௘ͽ íx and f y ͼ௘x, y௘ͽ +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 WKUHHSRLQWVͼ௘6HH)LJXUH௘ͽ :HVHHWKDW”x”xí”y”íx. 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 í”y”íx 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 ͼ௘6HH)LJXUH௘ͽ

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. ͼ௘6HH)LJXUH௘ͽ 1

2

1 y

Because z íy œ 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. ͼ௘6HH)LJXUH௘ͽ

2

6ROXWLRQ

1

:HPXVW¿UVW¿QGWKHLQWHUVHFWLRQRIWKHWZRSDUDERORLGVE\VHWWLQJ the equations equal to each other:

z

í í

í

z

2  x2  y2

x2  y2 Ÿ 2

2x2  2 y2 Ÿ 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\DWWKHSRLQWͼ௘x, y, z௘ͽLVWKHVTXDUHRILWV distance to the origin.

88

6ROXWLRQ The density is ȡ௘ͼ௘x, y, z௘ͽ k௘ͼ௘x2 + y2 + z2௘ͽ+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&DUWHVLDQFRRUGLQDWHVdz 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 ȡ௘ͼ௘x, y, z௘ͽ 

89

 6HWXSWKHLQWHJUDOIRUWKHPDVVRIWKHVROLGERXQGHGE\xy + 5z x y DQGz LIWKH density is ȡ௘ͼ௘x, y, z௘ͽ 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 ͼ௘x, y, z௘ͽEHDSRLQWLQVSDFH,WVF\OLQGULFDOFRRUGLQDWHVDUHͼ௘r, ș, z௘ͽZKHUHͼ௘r, ș௘ͽDUHWKHSRODU 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   Ÿ 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 cLVDYHUWLFDOSODQHͼ௘6HH)LJXUH௘ͽ ([DPSOH

θ =c

Find the volume of the solid bounded below by z x + y and above by z íx2íy2ͼ௘6HH)LJXUH௘ͽ 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 Ÿ  x2 + 2y2 Ÿ 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  &RQYHUWWKHSRLQWͼ௘r, ș, z௘ͽ ͼ௘íʌí௘ͽWRUHFWDQJXODUFRRUGLQDWHV  &RQYHUWWKHSRLQWͼ௘r, ș, z௘ͽ ͼ௘ S , ௘ͽWRUHFWDQJXODUFRRUGLQDWHV 4

 &RQYHUWWKHSRLQWͼ௘x, y, z௘ͽ   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 ͼ௘x, y, z௘ͽEHDSRLQWLQVSDFH,WVVSKHULFDOFRRUGLQDWHVDUHͼ௘ȡ, 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 Ÿ 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 ‫࣠׋‬Gȡ࣠G‫࣠׋‬Gș.

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 ‫࣠׋‬Gȡ࣠G‫࣠׋‬Gș. ([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 &RQYHUWWKHSRLQWͼ௘x, y, z௘ͽ ͼ௘௘ͽWRVSKHULFDOFRRUGLQDWHV z

6ROXWLRQ x2  y2  z 2

We have U and cos I

z

U

0ŸI

Therefore,  U , T , I

42  0  0

4, tan T

S.

y x

0 4

0 ŸT

0,

2

§ 4, 0, S · ͼ௘6HH)LJXUH௘ͽ ¨ ¸ 2¹ ©

y

( x , y , z ) = ( 4, 0, 0 ) x

( e, θ , φ ) = ⎛⎜ 4, 0, π2 ⎞⎟ ⎝

⎠

Figure 26.1

96

z

([DPSOH The surface ȡ cLVDVSKHUHFHQWHUHGDWWKHRULJLQͼ௘6HH)LJXUH௘ͽ ([DPSOH

c )LQGWKHYROXPHRIWKHLFHFUHDPFRQHERXQGHGDERYHE\WKHXSSHUKDOIRI the sphere x2 + y2 + z2 DQGEHORZE\ z x 2  y 2 ͼ௘6HH)LJXUH௘ͽ

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 Ÿ z2 2

Ÿ x2  y2

z=

x2  y2

x  y x  y 2

2

2

2



2 − x2 − y2

z=

2

x2 + y2

x2 + y2 = 1

1 Ÿ z 1.

y x

Converting to spherical coordinates,

x2  y2  z 2

2

U2 Ÿ U

Figure 26.3

2 and z

U cos I Ÿ 1

$OVR”ș”ʌ7KHLFHFUHDPFRQHLVJLYHQE\ 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 Ÿ cos I

1 ŸI 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

)RUVSKHULFDOFRRUGLQDWHV”‫”׋‬ʌ and ȡ•)XUWKHUPRUHș is the same angle as in polar coordinates for r•

x

7KHUHFDQEHFRQIXVLRQLIWKHFRRUGLQDWHV\VWHPLVQRWPDGHH[SOLFLW)RULQVWDQFHͼ௘ʌ௘ͽLVWKH 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 ‫࣠׋‬Gȡ࣠G‫࣠׋‬Gș.

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

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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 )ͼ௘x, y௘ͽWRHDFKSRLQWLQR.

x

A YHFWRU¿HOG over a solid region Q is a function F that assigns a vector )ͼ௘x, y, z௘ͽWRHDFKSRLQWLQQ.

x

1HZWRQV¶VODZRIJUDYLWDWLRQVWDWHVWKDWWKHIRUFHRIDWWUDFWLRQH[HUWHGRQDSDUWLFOHRIPDVVm ORFDWHGDWͼ௘x, y, z௘ͽE\DSDUWLFOHRIPDVV m2ORFDWHGDWͼ௘௘ͽLV F  x, y , z

Gm1m2 u. x2  y2  z 2

Here, G is the gravitational constant, and uLVWKHXQLWYHFWRUIURPWKHRULJLQWRͼ௘x, y, z௘ͽ x

&RXORPE¶VODZVWDWHVWKDWWKHIRUFHH[HUWHGRQDSDUWLFOHZLWKHOHFWULFFKDUJHqORFDWHGDWͼ௘x, y, z௘ͽE\ c q1 q2 a particle of charge q2ORFDWHGDWͼ௘௘ͽLV F  x , y , z u. 2 r

x

$YHFWRU¿HOGF is FRQVHUYDWLYHLIWKHUHH[LVWVDGLIIHUHQWLDEOHIXQFWLRQfͼ௘WKHSRWHQWLDO௘ͽVXFKWKDW F ’f .

x

Theorem: Let M and NKDYHFRQWLQXRXV¿UVWSDUWLDOGHULYDWLYHVRQDQRSHQGLVNR7KHYHFWRU¿HOG given by ) ͼ௘x, y௘ͽ 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\) ͼ௘x, y௘ͽ íyi + xj. 6ROXWLRQ

y :H¿UVWFRPSXWHVRPHSRLQWVWRJHWDQLGHDRIWKH¿HOG ) ͼ௘x, y௘ͽ ) ͼ௘௘ͽ ) ͼ௘௘ͽ ) ͼ௘௘ͽ

Vector field: F(x, y) = íyi + xj

íyi + xj ij j íi + 0j íi íi + 2j.

x

%\SORWWLQJWKHVHYHFWRUVZHREWDLQWKHIROORZLQJH[DPSOHRID URWDWLRQYHFWRU¿HOGͼ௘6HH)LJXUH௘ͽ Figure 27.1

Lesson 27: Vector Fields—Velocity, Gravity, Electricity

([DPSOH 6NHWFKVRPHYHFWRUVLQWKHYHFWRU¿HOG) ͼ௘x, y௘ͽ xi + yj.

y

6ROXWLRQ :H¿UVWFRPSXWHVRPHSRLQWVWRJHWDQLGHDRIWKH¿HOG í

) ͼ௘x, y௘ͽ xi + yj ) ͼ௘௘ͽ i + j ) ͼ௘íí௘ͽ íiíj ) ͼ௘í௘ͽ íi + 2j. %\SORWWLQJWKHVHYHFWRUVZHREWDLQWKHIROORZLQJH[DPSOHRID UDGLDOYHFWRU¿HOGͼ௘6HH)LJXUH௘ͽ





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

$WHDFKSRLQWͼ௘x, y௘ͽLQWKHSODQHWKHJUDGLHQWDVVLJQVDYHFWRU ([DPSOH 7KHYHFWRU¿HOG)ͼ௘x, y௘ͽ ͼ௘xy௘ͽiͼ௘x + 2y௘ͽj is conservative with potential f ͼ௘x, y௘ͽ x2xy + y2 because F ’f . ([DPSOH 7KHYHFWRU¿HOG)ͼ௘x, y௘ͽ x2yi + xyj is not conservative because M

x 2 y , wM wy

x2 , N

xy , wN wx

y , and x2y.

([DPSOH )LQGDSRWHQWLDOIRUWKHFRQVHUYDWLYHYHFWRU¿HOG)ͼ௘x, y௘ͽ xyiͼ௘x2íy௘ͽj. 6ROXWLRQ 1RWHWKDWWKH¿HOGLVFRQVHUYDWLYHEHFDXVH wM wy

2 x and wN wx

2 x.

:HQHHGWR¿QGDIXQFWLRQf ͼ௘x, y௘ͽVXFKWKDW’f  x , y 2 xyi   x 2  y j. That is, fx࣠ͼ௘x, y௘ͽ xy and f y࣠ͼ௘x, y௘ͽ 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) ͼ௘x, y௘ͽ i + j and compute F .  'HVFULEHWKHYHFWRU¿HOG) ͼ௘x, y, z௘ͽ i + j + k and compute F .  'HWHUPLQHZKHWKHURUQRWWKHYHFWRU¿HOG) ͼ௘x, y௘ͽ y2ͼ௘yixj௘ͽLVFRQVHUYDWLYH  'HWHUPLQHZKHWKHURUQRWWKHYHFWRU¿HOG) ͼ௘x, y௘ͽ 

1  i  j is conservative. x2  y 2

 )LQGWKHFRQVHUYDWLYHYHFWRU¿HOGIRUWKHSRWHQWLDOIXQFWLRQf ͼ௘x, y௘ͽ  x 2  1 y 2 . 4

 )LQGWKHFRQVHUYDWLYHYHFWRU¿HOGIRUWKHSRWHQWLDOIXQFWLRQf ͼ௘x, y, z௘ͽ xyz. 2

 )LQGWKHFRQVHUYDWLYHYHFWRU¿HOGIRUWKHSRWHQWLDOIXQFWLRQ f ͼ௘x, y, z௘ͽ  z  ye x .



 )LQGDSRWHQWLDOIXQFWLRQIRUWKHFRQVHUYDWLYHYHFWRU¿HOG)ͼ௘x, y௘ͽ yi + xj.  )LQGDSRWHQWLDOIXQFWLRQIRUWKHFRQVHUYDWLYHYHFWRU¿HOG)ͼ௘x, y௘ͽ 

y x i 2 j. 2 x y x  y2 2

 )LQGDSRWHQWLDOIXQFWLRQIRUWKHFRQVHUYDWLYHYHFWRU¿HOG)ͼ௘x, y௘ͽ x2y2i + 2xyj.



Curl, Divergence, Line Integrals Lesson 28

Topics x

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x

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x

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x

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'H¿QLWLRQVDQG7KHRUHPV x

TheFXUORIWKHYHFWRU¿HOG)ͼ௘x, y, z௘ͽLV § 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

TheGLYHUJHQFHRIWKH¿HOG) ͼ௘x, y, z௘ͽLV div F  x , y , z ’ i  j@t @

2

1RWLFHWKDWDOOWKUHHWHFKQLTXHVJLYHWKHVDPHDQVZHU:KLFKPHWKRGGR\RXSUHIHU" 6WXG\7LSV x

,IDYHFWRU¿HOGLVFRQVHUYDWLYHWKHQWKHOLQHLQWHJUDOEHWZHHQDQ\WZRSRLQWVLVVLPSO\WKH GLIIHUHQFHRIWKHSRWHQWLDOIXQFWLRQHYDOXDWHGDWWKRVHSRLQWV7KHOLQHLQWHJUDOLVLQGHSHQGHQWRI SDWK)RUH[DPSOHWKHYHFWRU¿HOGLQ([DPSOHLVFRQVHUYDWLYHZLWKSRWHQWLDOf ͼ௘x, y௘ͽ 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 ͼ௘௘ͽ$YHFWRU¿HOGWKDWPHDVXUHVDQRWKHUYHFWRU¿HOG¶VWHQGHQF\WRURWDWHͼ௘when curl = 0, D¿HOGLVLUURWDWLRQDO௘ͽ8VHGDVDWHVWIRUFRQVHUYDWLYHYHFWRU¿HOGV&DOFXODWHGXVLQJDFURVVSURGXFWRIWKH GLIIHUHQWLDORSHUDWRUZLWKWKHYHFWRU¿HOG curl F ’ u F. )RUWKHYHFWRU¿HOG)௘ͼ௘x, y, z௘ͽ, § 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 ͼ௘x௘ͽ is K

y cc ª1   y c 2 º ¬ ¼

3

. 2

See Calculus II, Lesson 36.

FXUYHͼ௘௘ͽ: $SODQDUFXUYHLVGH¿QHGE\WKHIXQFWLRQVx = f ͼ௘t௘ͽ, y = J࣠ͼ௘t௘ͽ, and z = K࣠ͼ௘t௘ͽ. A curve is called simple if it does not cross itself.

Glossary

F\OLQGHUͼ௘௘ͽ: ,QHOHPHQWDU\JHRPHWU\DF\OLQGHUUHVXOWVZKHQOLQHVSHUSHQGLFXODUWRDFLUFOHJHQHUDWHDWXEH shape, but LQKLJKHUPDWKHPDWLFVDF\OLQGHUͼ௘RUF\OLQGULFDOVXUIDFH௘ͽFDQUHIHUWRDQ\VXUIDFHFUHDWHGZKHQ DQ\JHQHUDWLQJFXUYHLQDSODQHͼ௘not just a circle௘ͽLVH[WHQGHGLQWRDWKLUGGLPHQVLRQE\OLQHVLQWHUVHFWLQJ that curve and orthogonal to its plane.

F\OLQGULFDOFRRUGLQDWHVͼ௘௘ͽ: 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

GHOͼ௘௘ͽ: See differential operator.

delta xͼ௘¨ x௘ͽͼ௘௘ͽ: 7KHV\PERO ¨x is read “delta x” and denotes Dͼ௘small௘ͽ change in x. Some textbooks use h LQVWHDGRI¨x.

GHQVLW\ͼ௘௘ͽ: 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 ͼ௘x௘ͽ: 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, WKHIXQFWLRQVDUHRIWZRͼ௘or more௘ͽ 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

GHWHUPLQDQWQRWDWLRQͼ௘௘ͽ: ,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 ͼ௘௘ͽ: In multivariable calculus, a function z = f ͼ௘x, y௘ͽLVGLIIHUHQWLDEOHDWWKHSRLQWͼ௘x0, y0௘ͽLI¨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 ͼ௘x௘ͽ 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.

GLIIHUHQWLDORSHUDWRUͼ௘GHO௘ͽͼ௘’௘ͽͼ௘௘ͽ: ’ “del,” or “grad,” or “nabla.”

w ,’ wx

w , or ’ wy

w . 8VHGLQFXUOGLYHUJHQFH3URQRXQFHG wz

GLUHFWLRQDOGHULYDWLYHͼ௘௘ͽ$JHQHUDOL]DWLRQRIWKHFRQFHSWRISDUWLDOGHULYDWLYHWKDWFDQEHXVHGWR¿QGWKH VORSHDZD\IURPDSRLQWLQDQ\JLYHQGLUHFWLRQ

GLUHFWLRQQXPEHUVͼ௘௘ͽ: Component numbers in a direction vector.

Glossary

GLVNͼ௘௘ͽ: 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¿HOGͼ௘௘ͽ: $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

GLYHUJHQFHWKHRUHPͼ௘௘ͽ: 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 = J࣠ͼ௘x௘ͽ, 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 functions7KRVHZKRVHJUDSKVDUHV\PPHWULFDFURVVWKHOLQHy = x. A function g is the inverse function of the function f if f ͼ௘J࣠ͼ௘x௘ͽ௘ͽ = x for all x in the domain of g and J࣠ͼ௘f ͼ௘x௘ͽ௘ͽ = 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 ͼ௘௘ͽ: 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 œ sin y

y

arccos x

cos 1 x œ 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 œ tan y

x , for  f  x  f and  S  y  S . 2 2

y

arcsec x

sec 1 x œ sec y

x , for x t 1, 0 d y d S , and y z S . 2

Reviewed in Calculus II, Lesson1.

.HSOHU¶VODZVͼ௘௘ͽ: 1௘ͽ The orbit of each planet is an ellipse, with the Sun at one of the two foci; 2௘ͽ a line joining a planet and the Sun sweeps out equal areas during equal intervals of time; 3௘ͽ the square of the RUELWDOSHULRGRIDSODQHWLVGLUHFWO\SURSRUWLRQDOWRWKHFXEHRIWKHVHPLPDMRUD[LVRIWKHRUELW

/DJUDQJHPXOWLSOLHUͼ௘௘ͽ: A scalar, ȜXVHGLQDSRZHUIXOWHFKQLTXHJLYHQE\/DJUDQJH¶VWKHRUHPIRUVROYLQJ optimization problems that have constraints.

ODPLQDͼ௘௘ͽ$WKLQÀDWSODWHRIPDWHULDOXVXDOO\RIXQLIRUPGHQVLW\

/DSODFH¶VSDUWLDOGLIIHUHQWLDOHTXDWLRQͼ௘௘ͽ'HVFULEHVWKHVWHDG\VWDWHWHPSHUDWXUHGLVWULEXWLRQLQSODWHVRU 2 2 solids. w z2  w z2 0. $IXQFWLRQWKDWVDWLV¿HVWKLVHTXDWLRQLVVDLGWREHharmonic. wx wy

ODZRIFRQVHUYDWLRQRIHQHUJ\ͼ௘௘ͽ,QDFRQVHUYDWLYHIRUFH¿HOGWKHVXPRISRWHQWLDODQGNLQHWLFHQHUJLHVRI an object remain constant from point to point.

OHDVWVTXDUHVUHJUHVVLRQOLQHͼ௘௘ͽ: 8VHGWR¿WDOLQHWRDVHWRISRLQWVLQWKHSODQH:RUNVEHVWZKHQWKHGDWDLV QHDUO\OLQHDU'HULYHGE\PLQLPL]LQJWKHVXPRIWKHVTXDUHVRIWKHGLIIHUHQFHVEHWZHHQWKHGDWDDQGWKH line. If f ͼ௘x௘ͽ = 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 ͼ௘௘ͽ: Also known as a contour lineWKHVHWRIDOOSRLQWVLQWKHSODQHVDWLVI\LQJf ͼ௘x, y௘ͽ = c, when z = f ͼ௘x, y௘ͽ and c is a constant. Contrast with trace, which is the intersection of a surface with a plane.

203

level surface ͼ௘௘ͽ: Although a function in three variables f ͼ௘x, y, z௘ͽ 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 ͼ௘x, y, z௘ͽ = 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 ͼ௘x௘ͽEHFRPHVDUELWUDULO\FORVHWRDVLQJOHQXPEHUL as x approaches c from either VLGHZHVD\WKDWWKHOLPLWRIf ͼ௘x௘ͽ 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 ͼ௘x௘ͽ increases without bound as x approaches c.

0RUHIRUPDOO\/HWfEHDIXQFWLRQGH¿QHGRQDQRSHQLQWHUYDOFRQWDLQLQJc ͼ௘H[FHSWSRVVLEO\DWc௘ͽ, 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 XVHRSHQGLVNVͼ௘DQGDSSURDFKIURPDQ\GLUHFWLRQ௘ͽLQVWHDGRIXVLQJRSHQLQWHUYDOVͼ௘DSSURDFKLQJIURPRQO\ two directions௘ͽ: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