Vector Calculus

CHAPTER 9.10~9.17

Ch9.10~9.17_2

Contents

9.10 Double Integrals9.11 Double Integrals in Polar Coordinates9.12 Green’s Theorem9.13 Surface Integrals9.14 Stokes’ Theorem9.15 Triple Integrals9.16 Divergence Theorem9.17 Change of Variables in Multiple Integrals

Ch9.10~9.17_3

9.10 Double Integrals

Recall from Calculus Region of Type I

See the region in Fig 9.71(a) R: a x b, g1(y) y g2(y)

Region of Type IISee the region in Fig 9.71(b)

R: c y d, h1(x) x h2(x)

Ch9.10~9.17_4

Fig 9.71

Ch9.10~9.17_5

Iterated Integral

For Type I:

(4)

For Type II:

(5)

xddyyxfxddyyxfb

a

xg

xg

b

a

xg

xg )(

)(

)(

)(

2

1

2

1),(),(

yddxyxfdyxdyxfd

c

yh

yh

d

c

yh

yh )(

)(

)(

)(

2

1

2

1),(),(

Ch9.10~9.17_6

Let f be continuous on a region R.

(i) For Type I:(6)

(ii) For Type II:(7)

THEOREM 9.12Evaluation of Double Integrals

b

a

xg

xgR

xddyyxfdAyxf)(

)(

2

1),(),(

d

c

yh

yhR

dyxdyxfdAyxf)(

)(

2

1),(),(

Ch9.10~9.17_7

Note:

Volume =

where z = f(x, y) is the surface.

R

dAyxf ),(

Ch9.10~9.17_8

Example 1

Evaluate over the region bounded by y = 1, y = 2, y = x, y = −x + 5. See Fig 9.73.

SolutionThe region is Type II

dAeR

yx 3

2

1

42552

1

3

2

1

5 33

)( dyeedye

dydxedAe

yyy

y

yx

y

y

yx

R

yx

64.2771

41

21

41

21

41

21 4789

2

1

425

eeeeee yy

Ch9.10~9.17_9

Fig 9.73

Ch9.10~9.17_10

Example 2

Evaluate over the region in the first quadrant bounded by y = x2, x = 0, y = 4.

SolutionFrom Fig 9.75(a) , it is of Type I

However, this integral can not be computed.

Ry dAxe

2

2

0

4

2

22

x

y

R

y dxdyxedAxe

Ch9.10~9.17_11

Fig 9.75(a) Fig 9.75(b)

Ch9.10~9.17_12

Example 2 (2)

Trying Fig 9.75(b), it is of Type II

4

00

2

4

0 0

2

22

2dye

x

dydxxedAxe

yy

y y

R

y

)1(41

41

21 16

4

0

4

0

22

eedyye yy

Ch9.10~9.17_13

Method to Compute Center of Mass

The coordinates of the center of mass are

(10)where

(11)are the moments. Besides, (x, y) is a variable density function.

,),(R

y dAyxxM R

x dAyxyM ),(

mM

ym

Mx xy ,

Ch9.10~9.17_14

Example 3

A lamina has shape as the region in the first quadrant that is bounded by the y= sin x, y = cos x between x = 0 and x = 4. Find the center of mass if (x, y) = y.

SolutionSee Fig 9.76.

Ch9.10~9.17_15

Example 3 (2)

4/

0

cos

sin

2

4/

0

cos

sin

2

dxy

dxdyydAym

x

x

x

xR

41

2sin41

2cos21

)sin(cos21

4/

0

4/

0

4/

0

22

xdxx

dxxx

Ch9.10~9.17_16

Example 3 (3)

162

2cos81

2sin41

2cos21

21

4/

0

4/

0

4/

0

cos

sin

2

4/

0

cos

sin

xxx

dxxx

dxxy

dxdyxydAxyM

x

x

x

xR

y

Ch9.10~9.17_17

Example 3 (4)

4/

0

33

4/

0

cos

sin

22

)sin(cos31

dxxx

dxdyydAyMx

xR

x

18425

cos31

cossin31

sin31

)]cos1(sin)sin1([cos31

4/

0

33

4/

0

22

xxxx

dxxxxx

Ch9.10~9.17_18

Example 3 (5)

Hence

29.04/1

16/)2( m

Mx y

68.04/1

18/)425( m

My x

Ch9.10~9.17_19

Moments of Inertia

(12)

are the moments of inertia about the x-axis and y-axis, respectively.

R

x dAyxyI ),(2

R

y dAyxxI ),(2

Ch9.10~9.17_20

Example 4

Refer to Fig 9.77. Find Iy of the thin homogeneous disk of mass m. Fig 9.77

Ch9.10~9.17_21

Example 4 (2)

Solution Since it is homogeneous, the density is the constant (x, y) = m/r2.

r

r

xr

xrR

y dxdyxr

mdA

r

mxI

22

222

222

2/

2/

222

2222

cossin2

2

dmr

dxxrxr

m r

r

Ch9.10~9.17_22

Example 4 (3)

22/

2/

2

2/

2/

22

41

)4cos1(4

2sin2

mrdmr

dmr

Ch9.10~9.17_23

Radius of Gyration

Defined by

(13)In Example 4,

mI

Rg

2/)4/(/ 2 rmrmmIR yg

Ch9.10~9.17_24

9.11 Double Integrals in Polar Coordinates

Double Integral Refer to the figure.

The double integral is

)(

)(

2

1),(),(

g

gR

ddrrrfdArf

Ch9.10~9.17_25

Refer to the figure.

The double integral is

b

a

rh

rh

R

drdrrf

dArf

)(

)(

2

1),(

),(

Ch9.10~9.17_26

Example 1

Refer to Fig 9.83. Find the center of mass wherer = 2 sin 2 in the first quadrant and is proportional to the distance from the pole.

Fig 9.83

Ch9.10~9.17_27

Example 1 (2)

Solution We have: 0 /2, = kr, then

2/

0

2sin2

0)(||

ddrrrkdArkm

R

2/

0

32/

0

2sin2

0

3

2sin38

3

dkdr

k

kk

dk

916

2cos61

2cos21

38

2sin)2cos1(38

2/

0

3

2/

0

2

Ch9.10~9.17_28

Example 1 (3)

Since x = r cos andthen

R

y dArxkM ||

2/

0

4

2/

0

2sin2

0

42/

0

2sin2

0

3

cos2sin4

cos4

cos

dk

dr

kddrrM y

2/

0

54

2/

0

44

cossin64

coscossin164

dk

dk

Ch9.10~9.17_29

Example 1 (4)

Similarly, y = r sin , then

2/

0

864

2/

0

224

cos)sinsin2(sin64

cos)sin1(sin64

dk

d

kk315512

sin91

sin72

sin51

642/

0

975

kddrrkM x 315512

sin2/0

2sin20

2

3532

9/16315/51

kk

yx

Ch9.10~9.17_30

Change of Variables

Sometimes we would like to change the rectangular coordinates to polar coordinates for simplifying the question. If and then

(3)

Recall: x2 + y2 = r2 and

)(

)(

2

1)sin,cos(),(

g

gR

ddrrrrfdAyxf

ryx 22

,),()(0 21 grg 20

Ch9.10~9.17_31

Example 2

Evaluate

SolutionFrom the graph is shown in Fig 9.84.Using x2 + y2 = r2, then 1/(5 + x2 + y2 ) = 1/(5 + r2)

2

0

8

22

2

5

1x

xxddy

yx

20,8 2 xxyx

Ch9.10~9.17_32

Fig 9.84

Ch9.10~9.17_33

Example 2 (2)

Thus the integral becomes

2/

4/

8

0 2

2/

4/

8

0 2

2

0

8

22

5

221

5

1

5

1

d

r

drrddrr

r

dxdyyx

x

x

2/

4/

8

0

2/

4/

2 )5ln13(ln21

)5ln(21

ddr

513

ln842

)15ln13(ln21

Ch9.10~9.17_34

Example 3

Find the volume of the solid that is under and above the region bounded by

x2 + y2 – y = 0. See Fig 9.85.

SolutionFig 9.85

221 yxz

Ch9.10~9.17_35

Example 3 (2)

We find that

and the equations becomeand r = sin . Now

R

dAyxV 221

21 rz

2/

0

sin

0

2/122 )1(21

ddrrrdArVR

2/

0

2/32 ])sin1(1[32

d

Ch9.10~9.17_36

Example 3 (3)

60.094

3sin

31

sin32

]cos)sin1(1[32

]cos1[32

])(cos1[32

2/

0

3

2/

0

2

2/

0

32/

0

2/32

d

dd

Ch9.10~9.17_37

Area

If f(r, ) = 1, then the area is

)(

)(

2

1

g

gR

ddrrdAA

Ch9.10~9.17_38

9.12 Green’s Theorem

Along Simply Closed Curves For different orientations for simply closed curves, please refer to Fig 9.88.

Fig 9.88(a) Fig 9.88(b) Fig 9.88(c)

Ch9.10~9.17_39

Notations for Integrals Along Simply Closed Curves

We usually write them as the following forms

where and represents in the positive and negative directions, respectively.

( , ) ( , ) ,

( , ) ( , ) ,

( , ) ,

C

c

C

P x y dx Q x y dy

P x y dx Q x y dy

F x y ds

C C

Ch9.10~9.17_40

Partial ProofFor a region R is simultaneously of Type I and Type II,

IF P, Q, P/y, Q/x are continuous on R, which is bounded by a simply closed curve C, then

THEOREM 9.13Green’s Theorem in the Plane

CR

Q PPdx Qdy dA

x y

dycyhxyhR

bxaxgyxgR

),()(:

),()(:

21

21

Ch9.10~9.17_41

Fig 9.89(a) Fig 9.89(b)

Ch9.10~9.17_42

Partial Proof

Using Fig 9.89(a), we have

b

a

b

a

xg

xgR

dxxgxPxgxP

dxdyyP

dAyP

))](,())(,([ 12

)(

)(

2

1

C

a

b

b

a

dxyxP

dxxgxPdxxgxP

),(

))(,())(,( 21

Ch9.10~9.17_43

Partial Proof

Similarly, from Fig 9.89(b),

From (2) + (3), we get (1).

d

c

d

c

yh

yhR

dyyyhQyyhQ

dydxxQ

dAxQ

)]),(()),(([ 12

)(

)(

2

2

c

d

d

cdyyyhQdyyyhQ )),(()),(( 12

= ( , )C

P x y dx

Ch9.10~9.17_44

Note:

If the curves are more complicated such as Fig 9.90, we can still decompose R into a finite number of subregions which we can deal with.

Fig 9.90

Ch9.10~9.17_45

Example 1

Evaluate where C is shown in Fig 9.91.

C

dyxyxdyx )2()( 22

Ch9.10~9.17_46

Example 1 (2)

SolutionIf P(x, y) = x2 – y2, Q(x, y) = 2y – x, then

andThus

yyP 2/ 1/ xQ

42011

)()(1

0

3461

0

22

3 dxxxxxdxyy

x

x

1

0

22

2

3 )21()21(

)2()(

x

xR

C

dxdyydAy

dyxydxyx

Ch9.10~9.17_47

Example 2

Evaluate where C is the circle (x – 1)2 + (y – 5)2 = 4 shown in Fig 9.92.

C

y dyexxdyx )2()3(35

Ch9.10~9.17_48

Example 2 (2)

SolutionWe have P(x, y) = x5 + 3y andthenHence

Since the area of this circle is 4, we have

3

2),( yexyxQ

,3/ yP 2/ xQ

4)2()3(35 C

y dyexdxyx

RR

C

y dAdAdyexdxyx )32()2()3(35

Ch9.10~9.17_49

Example 3

Find the work done by F = (– 16y + sin x2)i + (4ey + 3x2)j along C shown in Fig 9.93.

Ch9.10~9.17_50

Example 3 (2)

SolutionWe have

Hence from Green’s theorem

In view of R, it is better handled in polar coordinates, since R: 4/34/,10 r

R

dAxW )166(

C

y

C

dyxedxxy

W

)34()sin16(

．

22

drF

Ch9.10~9.17_51

Example 3 (3)

drr

ddrrrW

1

0

4/3

4/

23

4/3

4/

1

0

)8cos2(

)16cos6(

4)8cos2(

4/3

4/ d

Ch9.10~9.17_52

Example 4

The curve is shown in Fig 9.94. Green’s Theorem is no applicable to the integral

since P, Q, P/x, Q/y are not continuous at the region.

Cdy

yx

xdx

yx

y2222

Ch9.10~9.17_53

Fig 9.94

Ch9.10~9.17_54

Region with Holes

Green’s theorem cal also apply to a region with holes. In Fig 9.95(a), we show C consisting of two curves C1 and C2. Now We introduce cross cuts as shown is Fig 9.95(b), R is divided into R1 and R2. By Green’s theorem:

(4)

C

CC

RRR

dyQdxP

dyQdxPdyQdxP

dAyP

xQ

dAyP

xQ

dAyP

xQ

21

21

Ch9.10~9.17_55

Fig 9.95(a) Fig 9.95(b)

The last result follows from that fact that the line integrals on the crosscuts cancel each other.

Ch9.10~9.17_56

Example 5

Evaluate

where C = C1 C2 is shown in Fig 9.96.

SolutionBecause

Cdy

yx

xxd

yx

y2222

2222 ),(,),(yx

xyxQ

yx

yyxP

222

22

222

22

)(,

)( yx

xyxQ

yx

xyyP

Ch9.10~9.17_57

Example 5 (2)

are continuous on the region bounded by C, then

0)()( 222

22

222

22

2222

dAyx

xy

yx

xy

dyyx

xdx

yx

y

R

C

Ch9.10~9.17_58

Fig 9.96

Ch9.10~9.17_59

Conditions to Simply the Curves

As shown in Fig 9.97, C1 and C2 are two nonintersecting piecewise smooth simple closed curves that have the same orientation. Suppose that P and Q have continuous first partial derivatives such that P/y = Q/x in the region R bounded between C1 and C2, then we have

021

CCdyQdxPdyQdxP

21 CC

dyQdxPdyQdxP

Ch9.10~9.17_60

Fig 9.97

Ch9.10~9.17_61

Example 6

Evaluate the line integral in Example 4.

SolutionWe find P = – y / (x2 + y2) and Q = x / (x2 + y2) have continuous first partial derivatives in the region bounded by C and C’. See Fig 9.98.

Ch9.10~9.17_62

Fig 9.98

Ch9.10~9.17_63

Example 6 (2)

Moreover,

we have

xQ

yx

xyyP

222

22

)(

CCdy

yx

xdx

yx

ydy

yx

xdx

yx

y22222222

Ch9.10~9.17_64

Example 6 (3)

Using x = cos t, y = sin t, 0 t 2 , then

Note: The above result is true for every piecewise smooth simple closed curve C with the region in its interior.

2

)cos(sin

)](coscos)sin(sin[

2

0

2

0

22

2

0

2222

dt

dttt

dttttt

dyyx

xdx

yx

yC

Ch9.10~9.17_65

9.13 Surface Integrals

Let f be a function with continuous first derivatives

fx, fy on a closed region. Then the area of the surface

over R is given by

(2)

DEFINITION 9.11Surface Area

R

yx AdyxfyxfSA 22 )],([)],([1)(

Ch9.10~9.17_66

Example 1

Find the surface area of portion of x2 + y2 + z2 = a2 and is above the xy-plane and within x2 + y2 = b2, where 0 < b < a.

SolutionIf we definethen

Thus

where R is shown in Fig 9.103.

222),(),,( yxayxfyxfz

,),(222 yxa

xyxfx

222),(

yxa

yyxf y

222

222 )],([)],([1

yxa

ayxfyxf yx

R

dAyxa

aSA

222)(

Ch9.10~9.17_67

Fig 9.103

Ch9.10~9.17_68

Example 1 (2)

Change to polar coordinates:

2

0 0

2/122 )()(b

ddrrraaSA

2

0

22

2

0 02/122

)(

)(

dbaaa

draab

)(2 22 baaa

Ch9.10~9.17_69

Differential of Surface Area

The function

is called the differential of surface area.

dAyxfyxfdS yx22 )],([)],([1

Ch9.10~9.17_70

Let G be a function of three variables defined over a region of space containing the surface S. Then the surface integral of G over S is given by

(4)

DEFINITION 9.12Surface Integral

n

kkkkk

PS

SzyxGdSzyxG1

***

0) , ,(lim),,(

Ch9.10~9.17_71

Method of Evaluation

(5)

where we define z = f(x, y) is the equation of S projects onto a region R of the xy-plane.

R

yx

S

dAyxfyxfyxfyxG

dSzyxG

22 )],([)],([1)),(,,(

),,(

Ch9.10~9.17_72

Projection of S Into Other Planes

If we define y = g(x, z) is the equation of S projects onto a region R of the xz-plane, then

(6)

Similarly, if x = h(y, z) is the equation of S projects onto a region R of the yz-plane, then

(7)

R

zx

S

dAzxgzxgzzxgxG

dSzyxG

22 )],([)],([1)),,(,(

),,(

R

zy

S

dAzyhzyhzyzyhG

dSzyxG

22 )],([)],([1),),,((

),,(

Ch9.10~9.17_73

Mass of a Surface

Let (x, y, z) be the density of a surface, then the mass m of the surface is

(8) S

dSzyxm ),,(

Ch9.10~9.17_74

Example 2

Find the mass of the surface of z = 1 + x2 + y2 in the first octant for 1 z 5 if the density at a point is proportional to its distance from the xy-plane.

Solution The projection graph is shown in Fig 9.104. Now, since ρ(x, y, z) = kz and z = 1 + x2 + y2, then

R

S

dAyxyxk

dSkzm

2222 441)1(

Ch9.10~9.17_75

Fig 9.104

Ch9.10~9.17_76

Example 2 (2)

Change to polar coordinates

2/

0

2

0

22 41)1(

ddrrrrkm

2/

0

2

0

2/1232/12 ])41()41([

ddrrrrrk

2/

2

2

0

2/522/3222/32 )41(120

1)41(

121

)41(121

drrrrk

kk

2.19403

12017

12)17(5

2

2/52/3

Ch9.10~9.17_77

Example 3

Evaluate , where S is the portion of y = 2x2 + 1 in the first octant bounded by x = 0, x = 2, z = 4 and z = 8.

Solution The projection graph on the xz-plane is shown in Fig 9.105.

S dSxz2

Ch9.10~9.17_78

Example 3 (2)

Let y = g(x, z) = 2x2 + 1. Since gx(x, z) = 4x and gz(x, z) = 0, then

2

0

8

4

222 161 dxdzxxzdSxzS

2

0

2/122

0

8

4

23

)161(3

448161

3dxxxdxxx

z

3.1627]165[928

)161(928 2/3

2

0

2/32 x

Ch9.10~9.17_79

Orientable Surface

A surface is said to be orientable or an oriented surface if there exists a continuous unit normal vector function n, where n(x, y, z) is called the orientation of the surface. Eg: S is defined by g(x, y, z) = 0, then

n = g / ||g||(9)

where is the gradient. kji

zg

yg

xg

g

Ch9.10~9.17_80

Fig. 9.106

Ch9.10~9.17_81

Fig 9.107

Ch9.10~9.17_82

Example 4

Consider x2 + y2 + z2 = a2, a > 0. If we define g(x, y, z) = x2 + y2 + z2 – a2, then

Thus the two orientations are

where n defines outward orientation, n1 = − n defines inward orientation. See Fig 9.108.

, 222 kji zyxg azyxg 2444|||| 222

, kjinaz

ay

ax kjinn

az

ay

ax 1

Ch9.10~9.17_83

Fig 9.108

Ch9.10~9.17_84

Computing Flux

We have

(10)

See Fig 9.109.

S

dS)．(flux nF

Ch9.10~9.17_85

Example 5

Let F(x, y, z) = zj + zk represent the flow of a liquid. Find the flux of F through the surface S given by that portion of the plane z = 6 – 3x – 2y in the first octant oriented upward.

SolutionRefer to the figure.

Ch9.10~9.17_86

Example 5 (2)

We define g(x, y, z) = 3x + 2y + z – 6 = 0. Then a unit normal with a positive k component (it should be upward) is

Thus

With R the projection of the surface onto the xy-plane, we have

kjin141

142

143

||||

gg

SS

dSzdS 3141

)．(flux nF

18)236(3

)14)(236(3141

flux

2

0

2/33

0

x

R

dxdyyx

dAyx

Ch9.10~9.17_87

9.14 Stokes’ Theorem

Vector Form of Green’s Theorem If F(x, y) = P(x, y)i + Q(x, y)j, then

Thus, Green’s Theorem can be written as

k

kji

FF

yP

xQ

Pzyx

00

curl

(1) ．)curl(．． R

CCdAdsd kFTFrF

Ch9.10~9.17_88

Let S be a piecewise smooth orientable surface bounded by a piecewise smooth simple closed curve C. Let F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k be a vector field for which P, Q, R, are continuous and havecontinuous first partial derivatives in a region of 3-space containing S. If C is traversed in the positive direction, then

where n is a unit normal to S in the direction of the orientation of S.

THEOREM 9.14Stokes’ Theorem

S

CCdSdSd nFTFrF ．)(curl)( ．．

Ch9.10~9.17_89

Example 1

Let S be the part of the cylinder z = 1 – x2 for 0 x 1, −2 y 2. Verify Stokes’ theorem if F = xyi + yzj + xzk.

Fig 9.116

Ch9.10~9.17_90

Example 1 (2)

Solution See Fig 9.116. Surface Integral: From F = xyi + yzj + xzk, we find

14

2 is normalupper the

cylinder, thedefines 01 If

curl

2

x

xgg

xzzyx

xzy

xzyzxyzyx

kin

kji

kji

F

Ch9.10~9.17_91

Example 1 (3)

(7) 2)4()2(

)2(14

2

:9.13 Sec of (5) Using

14

2) (curl Therefore

1

0

1

0

2

2

2

2

dxxdydxxxy

dAxxydSx

xxy

dSx

xxydS

RS

SS

nF

Ch9.10~9.17_92

Example 1 (4)

1511

)222(

)2)(1(0)1(22 so

,2,0,1,2:on

00)0()0( so

,0,0,0,1:on

write we:Integral Line

0

1

42

22

22

1

1

1

2 431

dxxxx

xdxxxxxdx

xdzdyxzyC

dyyy

dzdxzxC

C

C

C C CCC

Ch9.10~9.17_93

Example 1 (5)

1519

)222(

)2)(1(0)1(22 so

,2,0,1,2:on

000 so

,0,0,1,0:on

1

0

42

22

24

2

2

3

4

3

dxxxx

xdxxxxxdx

xdzdyxzyC

ydyydy

dzdxzxC

C

C

(7). with agreeswhich

21519

01511

0 Hence C xzdzyzdyxydx

Ch9.10~9.17_94

Example 2

Evaluate

where C is the trace of the cylinder x2 + y2 = 1 in the plane y + z = 2. Orient C counterclockwise as viewed from above. See Fig 9.117.

C

ydzxdyzdx

Ch9.10~9.17_95

Fig 9.117

Ch9.10~9.17_96

Example 2 (2)

Solution

The given orientation of C corresponding to an upward orientation of the surface S.

kji

kji

F

kjiF

yxzzyx

yxz

curl

then , If

Ch9.10~9.17_97

Example 2 (3)

Thus if g(x, y, z) = y + z – 2 = 0 defines the plane, then the upper normal is

2222

)2

12

1()(

(2), from Hence

2

12

1

RS

SC

dAdS

dSd

gg

kjkjirF

kjn

Ch9.10~9.17_98

9.15 Triple Integrals

Let F be a function of three variables defined over a

Closed region D of space. Then the triple integral of F

over D is given by

(1)

DEFINITION 9.13The Triple Integral

* * *

0 1

( , , ) lim ( , , )n

k k k kP kD

F x y z dV F x y z V

Ch9.10~9.17_99

Evaluation by Iterated Integrals:

See Fig 9.123.

(2) )()(

n region the I Type a is if Thus,

)()(

then,),(by below bounded is and

),(by above bounded is region theIf

)(

)(

),(

),(

),(

),(

1

2

2

1

2

1

2

1

b

a

xg

xg

yxf

yxfD

R

yxf

yxfD

dzdydxx,y,zFdVx,y,zF

R

dAdzx,y,zFdVx,y,zF

yxfz

yxfzD

Ch9.10~9.17_100

Fig 9.123

Ch9.10~9.17_101

Applications

D

D

dVzyxmD

zyx

dVVD

zyxF

),,( is solid theof

mass then thedensity, is ),,( If

is solid the

of volume then the,1),,( If

:Mass

:Volume

),,( ,),,(

,),,(

Dyz

Dxz

Dxy

dVzyxxMdVzyxyM

dVzyxzM

:MomentFirst

Ch9.10~9.17_102

centroid. thecalled is mass

ofcenter theconstant, is ),,( If

,,

are mass ofcenter theof scoordintae The

zyx

m

Mz

mM

ym

Mx xyxzyz

:Centroid

:Mass ofCenter

Ch9.10~9.17_103mI

R

I

dVzyxyxI

dVzyxzxI

dVzyxzyI

g

Dz

Dy

Dx

isgyration of radius then the

axis,given aabout solid theof inertia ofmoment a is If

),,()(

),,()(

,),,()(

22

22

22

:Gyration of Radius

:Moment Second

Ch9.10~9.17_104

Example 1

Find the volume of the solid in the first octant bounded by z = 1 – y2, y = 2x and x = 3.

Fig 9.125(a) Fig 9.125(b)

Ch9.10~9.17_105

Example 1 (2)

Solution Referring to Fig 9.125(a), the first integration with respect to z is from 0 to 1 – y2. From Fig 9.125(b), we see that the projection of D in the xy-plane is a region of Type II. Hence

815

21

21

332/

3)(

)1(

1

0

321

0

2

1

0

3

2/

21

0

3

2/

1

0

2

dyyyydyy

yxx

dxdyydzdxdydVVyy

y

D

Ch9.10~9.17_106

Example 2

Change the order of integration in

Fig 9.126(a) Fig 9.126(b)

. to

),,(6

0

3/24

0

4/32/3

0

dydxdz

dzdydxzyxFx yx

Ch9.10~9.17_107

Example 2 (2)

Solution As in Fig 9.126(a), the region D is the solid in the first octant bounded by the three coordinates and the plane 2x + 3y + 4x = 12. Referring to Fig 9.126(b) and the table, we have

3

0

26

0

3/43/24

0

6

0

3/24

0

4/32/3

0

),,(

),,(

z zx

x yx

dydxdzzyxF

dzdydxzyxF

Ch9.10~9.17_108

Example 2 (3)

3 to0 26 to0 34

32

4 to0

6 to0 3/24 to0 4

32

3 to0

3rd 2nd 1st Order

zzx

dydxdz

xyx

dzdydx

Ch9.10~9.17_109

Cylindrical Coordinates

Refer to Fig 9.127.

Ch9.10~9.17_110

Conversion of Cylindrical Coordinates to Rectangular Coordinates

Thus between the cylindrical coordinates (r, , z) and rectangular coordinates (x, y, z), we have

x = r cos , y = r sin , z = z (3)

Ch9.10~9.17_111

Example 1

Convert (8, /3, 7) in cylindrical coordinates to rectangular coordinates.

SolutionFrom (3)

7

34)3/8sin(

,4)3/8cos(

z

y

x

Ch9.10~9.17_112

Conversion of Rectangular Coordinates to Cylindrical Coordinates

Also we have

(4) ,tan,222 zzxy

yxr

Ch9.10~9.17_113

Example 4

Solution

s.coordinate lcylindrica

toscoordinater rectangulain )1,2,2(Convert

.4/3 takewe

,0,0fact ith thetogether w,2 take weIf

11,2

2tan,4)2()2( 222

yxr

zr

Ch9.10~9.17_114

Fig 9.128

Ch9.10~9.17_115

Triple Integrals in Cylindrical Coordinates

See Fig 9.129.We have

)(

)(

),(

),(

),(

),(

2

1

2

1

2

1

),,(

),,(),,(

g

g

rf

rf

R

rf

rfD

rdzdrdzyxF

dAzrFdVzrF

Ch9.10~9.17_116

Fig 9.129

Ch9.10~9.17_117

Example 5

A solid in the first octant has the shape determined by the graph of the cone z = (x2 + y2)½ and the planes z = 1, x = 0 and y = 0. Find the center of the mass if the density is given by (r, , z) = r.

Solution

2

0

1

0

322

0

1

0

2

2

0

1

0

1

24)(

1

)(

drdrrdrdr

zr

rdzdrdrrdVmr

D

Ch9.10~9.17_118

Fig 9.130

Ch9.10~9.17_119

Example 5 (2)

Similarly, we have

2

0

1

0

422

0

1

0

22

2

0

1

0

1 2

30)(

211

2drdrrdrd

rr

z

dzdrdzrzrdVMr

Dxy

201

coscos

201

sinsin

2

0

1

0

1 32

2

0

1

0

1 32

rD

yz

rD

xz

dzdrdrdVrM

dzdrdrdVrM

Ch9.10~9.17_120

Example 5 (3)

8.024/30/

,38.024/20/1

,38.024/20/1

Hence

zyx

Ch9.10~9.17_121

Spherical Coordinates

See Fig 9.131.

Ch9.10~9.17_122

Conversion of Spherical Coordinates to Rectangular and Cylindrical Coordinates

We have

(6) cos,,sin

:),,( scoordinate lcylindrica to

),,( scoordinate spherical From

(5) cos,sinsin,cossin

:),,( scoordinater rectangula to

),,( scoordinate spherical From

zr

zr

zyx

zyx

Ch9.10~9.17_123

Example 6

Convert (6, /4, /3) in spherical coordinates to rectangular and cylindrical coordinates.

Solution

23cos ,3

,23sin

23cos2

63sinsin ,

223

cossin

)3/ ,4/ ,6() , ,(

zr

z

yx

Ch9.10~9.17_124

Inverse Conversion

(7) cos

tan ,

222

2222

zyx

zxy

zyx

Ch9.10~9.17_125

Triple Integrals in Spherical Coordinates

See Fig 9.132.

Ch9.10~9.17_126

We have

)(

)(

),(

),(

22

1

2

1sin),,(

),,(

g

g

f

f

D

dddF

dVF

Ch9.10~9.17_127

Example 7

Find the moment of inertia about the z-axis of the homogeneous solid bounded between the spheres

x2 + y2 + z2 = a2 and x2 + y2 + z2 = a2, a < bFig 9.133

Ch9.10~9.17_128

Example 7 (2)

Solution If (, , ) = k is the density, then

2

0 0

222

22222222

22

)sin(sinThen

. , are spheres theof equations The

,sin

have we(5), From .)(

b

az

Dz

dddkI

ba

zyxyx

kdVyxI

Ch9.10~9.17_129

Example 7 (3)

2

0 0

25

55

2

0 0

35

2

0 0

34

sin)cos1(5

)(5

sin5

sin

ddabk

dda

bk

dddkb

a

)(158

)(154

0cos

31

cos)(5

552

0

55

2

0

355

abk

dabk

dabk

Ch9.10~9.17_130

9.16 Divergence Theorem

Another Vector Form of Green’s TheoremLet F(x, y) = P(x, y)i + P(x, y)j be a vector field, and let T = (dx/ds)i + (dy/ds)j be a unit tangent to a simple closed plane curve C. If n = (dy/ds)i – (dx/ds)j is a unit normal to C, then

)(

RR

CC

dAyQ

xP

dAyQ

xP

QdxPdydsnF

Ch9.10~9.17_131

that is,

The result in (1) is a special case of the divergence or Gauss’ theorem.

(1) div)( R

CdAds FnF

Ch9.10~9.17_132

Let D be a closed and bounded region on 3-space with

a piecewise smooth boundary S that is oriented outward.

Let F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k be

a vector field for which P, Q, and R are continuous

and have continuous first partial derivatives in a region

of 3-space containing D. Then

(2)

THEOREM 9.15Divergence Theorem

( ) div S D

dS dV F n F

Ch9.10~9.17_133

Example 1

Let D be the region bounded by the hemisphere

SolutionThe closed region is shown in Fig 9.140.

.)1(

if theoremdivergence Verify the .1 plane the

and ,41,9)1( 222

kjiF

zyx

z

zzyx

Ch9.10~9.17_134

Fig 9.140

Ch9.10~9.17_135

Example 1 (2)

Triple Integral: Since F = xi + yj + zk, we see div F = 3. Hence

(10)

Surface Integral: We write S = S1 + S2, where S1 is the hemisphere and S2 is the plane z = 1. If S1 is a level surfaces of g(x, y) = x2 + y2 + (z – 1)2, then a unit outer normal is

543div DDD

dVdVdVF

Ch9.10~9.17_136

Example 1 (3)

RS

dAyx

ds

zyx

zyx

zyx

zyxgg

22

222

222

9

3)3( so and

33

)1(33

Now

31

33)1(

)1(

1

nF

nF

kjikji

n

2

0

3

0

2/12 54)9(9 rdrdr

Ch9.10~9.17_137

Example 1 (4)

54)( that see weHence,

.0)1( ,1 sinceBut

.1 that so ,On

2

2

S

S

dS

dSzz

zS

nF

nFkn

Ch9.10~9.17_138

Example 2

IF F = xyi + y2zj + z3k, evaluate S (F n)dS, where S is the unit cube defined by 0 x 1, 0 y 1, 0 z 1.

SolutionWe see div F = F = x + 2yz + 3z2. Then

1

0

1

0

1

0

2

2

)32(

)32()(

dxdydzzyzy

dVzyzydSD

SnF

Ch9.10~9.17_139

Example 2 (2)

20

1)

21

2()3

21

(

0

1)3

2(

)32(

321

0

2

1

0

222

1

0

1

0

2

zzz

dzzz

dzyzzyy

dydzzyzy

Ch9.10~9.17_140

9.17 Change of Variables in Multiple Integrals

Introduction If f is continuous on [a, b], x = g(u) and dx = g(u) du, then

where c = g(a), d = g(b).If we write J(u) = dx/du, then we have

(1) )('))(()( d

c

b

aduugugfdxxf

(2) )())(()( d

c

b

aduuJugfdxxf

Ch9.10~9.17_141

Double Integrals

If we have x= f(u, v), y = g(u, v)

(3)we expect that a change of variables would take the form

where S is the region in the uv-plane, and R is the region in the xy-plane.

(4) '),()),(),,((),( SR

dAvuJvugvufFdAyxF

Ch9.10~9.17_142

Example 1

Find the image of the region S shown in Fig 9.146(a) under the transformations x = u2 + v2, y = u2 − v2.

SolutionFig 9.146(a) Fig 9.146(b)

Ch9.10~9.17_143

Example 1 (2)

9.146(b). Fig See

)1,1( to)1,4(),()0,1( to),(),(

4 then ,1:

)1,4( to)4,4(),(),( to)0,2(),(

4 then ,4:

)4,4( to)1,1(),()0,2( to)0,1(),(

then ,,

then,0:

23

25

223

23

25

222

222222

1

yxvu

yvuS

yxvu

xvuS

yxvu

xyuvuyuvux

vS

Ch9.10~9.17_144

Some of the Assumptions

1. The functions f, g have continuous first partial derivatives on S.

2. The transformation is one-to-one.

3. Each of region R and S consists of a piecewise smooth simple closed curve and its interior.

4. The following determinant is not zero on S.

(7) uy

vx

vy

ux

vy

uy

vx

ux

Ch9.10~9.17_145

Equation (7) is called the Jacobian of the transformation T and is denoted by (x, y)/(u, v).Similarly, the inverse transformation of T is denoted by T-1. See Fig 9.147.

Ch9.10~9.17_146

If it is possible to solve (3) for u, v in terms of x, y, then we have

u = h(x,y), v = k(x,y)(8)The Jacobian of T-1 is

(10) 1),(),(

),(),(

and

(9) ),(),(

yxvu

vuyx

yv

xv

yu

xu

yxvu

Ch9.10~9.17_147

Example 2

The Jacobian of the transformation x = r cos , y = r sin

is

rr

ry

ry

xrx

ryx

cossin

sincos

),(),(

Ch9.10~9.17_148

If F is continuous on R, then

(11)

THEOREM 9.6Change of Variables in aDouble Integral

Advuyx

vugvufFAdyxFSR

),(

),()),(),,((),(

Ch9.10~9.17_149

Example 3

Evaluate

over the region R in Fig 9.148(a).

Fig 9.148(a) Fig 9.148(b)

R

dAyxyx )2cos()2sin(

Ch9.10~9.17_150

Example 3 (2)

Solution We start by letting u = x + 2y, v = x – 2y.

9.148(b). Fig See

)2,2( to)2,2(),()0,2( to),0(),(

2 then ,22:

)2,2( to)0,0(),(),0( to)0,0(),(

or 2,2 then ,0:

)0,0( to)2,2(),()0,0( to)0,2(),(

or , then ,0:

3

2

1

vuyx

uyxS

vuyx

vuyvyuxS

vuyx

vuxvxuyS

Ch9.10~9.17_151

Example 3 (3)

The Jacobian matrix is

41

21

41

21

21

),(),(

vy

uy

vx

ux

vuyx

Ch9.10~9.17_152

Example 3 (4)

Thus

2)2cos1(

41

sin21

sinsin41

cossin41

'41

cossin)2cos()2sin(

2

0

2

02

2

0

2

0

duuudu

duu

uvuvdvduu

dAvudAyxyx

u

u

SR

Ch9.10~9.17_153

Example 4

Evaluate over the region R in Fig 9.149(a).

Fig 9.149(a) Fig 9.149(b)

R

xydA

Ch9.10~9.17_154

Example 4 (2)

Solution The equations of the boundaries of R suggest

u = y/x2, v = xy (12)The four boundaries of the region R become u = 1, u = 4, v = 1, v = 5. See Fig 9.149(b).The Jacobian matrix is

uyx

vuyxvu

yx31

3),(),(

1),(),( 2

Ch9.10~9.17_155

Example 4 (3)

Hence

4ln41

4ln4

14

1

5

231

31

'31

4

1

4

1

2

4

1

5

1

uduu

duu

v

dvduuv

dAu

vxydASR

Ch9.10~9.17_156

Triple Integrals

Let x = f(u, v, w), y = g(u, v, w), z = h(u, v, w)be a one-to-one transformation T from a region E in the uvw-space to a region in D in xyz-space. If F is continuous in D, then

E

D

dVwvuizyx

wvuhwvugwvufF

dVzyxF

'),,(),,(

)),,(),,,(),,,((

),,(

Ch9.10~9.17_157

sin),,(),,(

then

(13) cos,sinsin,cossin

ifat verify thPlease

),,(),,(

where

2

wvuzyx

xzyx

wz

vz

uz

wy

vy

uy

wx

vx

ux

wvuzyx