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
