1. CHAPTER 9.10~9.17
Vector Calculus

2. Contents 9.10 Double Integrals
9.11 Double Integrals in Polar Coordinates
9.12 Green’s Theorem
9.13 Surface Integrals
9.14 Stokes’ Theorem
9.15 Triple Integrals
9.16 Divergence Theorem
9.17 Change of Variables in Multiple Integrals

3. 9.10 Double Integrals
Definition 9.10: Let f be a function of two variables defined on a closed region R. Then the double integral of f over R is given by
(1)
Integrability: If the limit in (1) exists, we say that f is integrable over R, and R is the region of integration.
Area: When f(x,y)=1 on R.
Volume: When f(x,y) 0 on R.

4. Properties of Double Integrals
Theorem 9.11: Let f and g be functions of two variables that are integrable over a region R. Then
(i) , where k is any constant
(ii)
(iii) where
and

5. Regions of Type I and II
Region of Type I See the region in Fig 9.71(a) R: a  x  b, g1(x)  y  g2(x)
Region of Type II See the region in Fig 9.71(b) R: c  y  d, h1(y)  x  h2(y)

6. Fig 9.71

7. Iterated Integral
For Type I: (4)
For Type II: (5)

8. Let f be continuous on a region R.
(i) For Type I: (6) (ii) For Type II: (7)
THEOREM 9.12
Evaluation of Double Integrals

9. Note:
Volume = where z = f(x, y) is the surface.

10. Example 1
Evaluate over the region bounded by y = 1, y = 2, y = x, y = −x + 5. See Fig 9.73.
Solution The region is Type II

11. Fig 9.73

12. Example 2
Evaluate over the region in the first quadrant bounded by y = x2, x = 0, y = 4.
Solution From Fig 9.75(a) , it is of Type I However, this integral can not be computed.

13. Fig 9.75(a) Fig 9.75(b)

14. Example 2 (2)
Trying Fig 9.75(b), it is of Type II

15. 9.11 Double Integrals in Polar Coordinates
Double Integral Refer to the figure The double integral is

16. Refer to the figure. The double integral is

17. 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

18. Example 2
Evaluate
Solution From the graph is shown in Fig Using x2 + y2 = r2, then 1/(5 + x2 + y2 ) = 1/(5 + r2)

19. Fig 9.84

20. Example 2 (2)
Thus the integral becomes

21. 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.
Solution Fig 9.85

22. Example 3 (2)
We find that and the equations become and r = sin . Now

23. Example 3 (3)

24. Area
If f(r, ) = 1, then the area is

25. 9.12 Green’s Theorem
Along Simple Closed Curves For different orientations of simple closed curves, please refer to Fig Fig 9.88(a) Fig 9.88(b) Fig 9.88(c)

26. Notations for Integrals Along Simple Closed Curves
We usually write them as the following forms where and represents in the positive and negative directions, respectively.

27. IF P, Q, P/y, Q/x are continuous on R, which is
THEOREM 9.13
IF P, Q, P/y, Q/x are continuous on R, which is
bounded by a simply closed curve C, then
Green’s Theorem in the Plane
Partial Proof For a region R is simultaneously of Type I and Type II,

28. Fig 9.89(a) Fig 9.89(b)

29. Partial Proof
Using Fig 9.89(a), we have

30. Partial Proof
Similarly, from Fig 9.89(b), From (2) + (3), we get (1).

31. 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

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

33. Example 2 (2)
Solution We have P(x, y) = x5 + 3y and then Hence Since the area of this circle is 4, we have

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

35. Example 3 (2)
Solution We have Hence from Green’s theorem In view of R, it is better handled in polar coordinates, since R:

36. Example 3 (3)

37. Example 4
The curve is shown in Fig Green’s Theorem is not applicable to the integral since P, Q, P/x, Q/y are not continuous at the region.

38. Fig 9.94

39. 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)

40. 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.

41. Example 5 Evaluate where C = C1 C2 is shown in Fig 9.96.
Solution Because

42. Example 5 (2)
are continuous on the region bounded by C, then

43. Fig 9.96

44. Conditions to Simplify 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

45. Fig 9.97

46. Example 6 Evaluate the line integral in Example 4.
Solution We 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.

47. Fig 9.98

48. Example 6 (2)
Moreover, we have

49. 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 origin in its interior.

50. 9.13 Surface Integrals
DEFINITION 9.11
Let f be a function with continuous first derivatives
fx, fy on a closed region. Then the area of the surface
z=f(x,y) over R is given by (2)
Surface Area

51. 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.
Solution If we define then
Thus where R is shown in Fig

52. Fig 9.103

53. Example 1 (2)
Change to polar coordinates:

54. Differential of Surface Area
The function is called the differential of surface area.

55. Let G be a function of three variables defined over a
DEFINITION 9.12
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)
Surface Integral

56. Method of Evaluation
(5) where we define z = f(x, y) to be the equation of S projecting into a region R of the xy-plane.

57. Projection of S Into Other Planes
If we define y = g(x, z) to be the equation of S projecting into a region R of the xz-plane, then (6)
Similarly, if x = h(y, z) is the equation of S projecting into a region R of the yz-plane, then (7)

58. 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

59. Example 3 (2)
Let y = g(x, z) = 2x Since gx(x, z) = 4x and gz(x, z) = 0, then

60. 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. e.g: S is defined by g(x, y, z) = 0, then n = g / ||g|| (9) where is the gradient.

61. Fig

62. Fig 9.107

63. 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

64. Fig 9.108

65. Computing Flux We have (10) See Fig 9.109. Flux of F through S:
The total volume of a
fluid passing through S
per unit time.

66. 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.
Solution Refer to the figure.

67. Example 5 (2)
We define g(x, y, z) = 3x + 2y + z – 6 = 0. Then a unit normal vector with a positive k component (it should be upward) is Thus With R the projection of the surface onto the xy-plane, we have

68. 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

69. Let S be a piecewise smooth orientable surface bounded
THEOREM 9.14
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 have
continuous 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.
Stokes’ Theorem

70. 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

71. Example 1 (2)
Solution See Fig Surface Integral: From F = xyi + yzj + xzk, we find

72. Example 1 (3)

73. Example 1 (4)

74. Example 1 (5)

75. 9.15 Triple Integrals
DEFINITION 9.13
Let F be a function of three variables defined over a
Closed region D of 3-space. Then the triple integral of F
over D is given by (1)
The Triple Integral

76. Evaluation by Iterated Integrals:
See Fig

77. Fig 9.123

78. 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)

79. 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

80. Cylindrical Coordinates
Refer to Fig

81. Conversion of Cylindrical Coordinates to Rectangular Coordinates
The relationship between the cylindrical coordinates (r, , z) and rectangular coordinates (x, y, z): x = r cos , y = r sin , z = z (3)

82. Example 1
Convert (8, /3, 7) in cylindrical coordinates to rectangular coordinates.
Solution From (3)

83. Conversion of Rectangular Coordinates to Cylindrical Coordinates
Also we have

84. Example 4
Solution

85. Fig 9.128

86. Triple Integrals in Cylindrical Coordinates
See Fig
We have

87. Fig 9.129

88. Spherical Coordinates
See Fig

89. Conversion of Spherical Coordinates to Rectangular and Cylindrical Coordinates

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

91. Inverse Conversion

92. Triple Integrals in Spherical Coordinates
See Fig

93. We have

94. 9.16 Divergence Theorem
Another Vector Form of Green’s Theorem Let F(x, y) = P(x, y)i + Q(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

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

96. Let D be a closed and bounded region in 3-space with
THEOREM 9.15
Let D be a closed and bounded region in 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)
Divergence Theorem

97. Example 1 Let D be the region bounded by the hemisphere
Solution The closed region is shown in Fig

98. Fig 9.140

99. Example 1 (2)
Triple Integral: Since F = xi + yj + (z-1)k, 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

100. Example 1 (3)

101. Example 1 (4)

102. 9.17 Change of Variables in Multiple Integrals
Introduction If f (x) is continuous on [a, b], then if x = g(u) and dx = g(u) du, we have where c = g(a), d = g(b). If we write J(u) = dx/du, then we have

103. 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. J(u,v) is some function obtained from the partial derivatives of the equation in (3).

104. Example 1
Find the image of the region S shown in Fig 9.146(a) under the transformations x = u2 + v2, y = u2 − v2.
Solution Fig 9.146(a) Fig 9.146(b)

105. Example 1 (2)

106. Some of the Assumptions
The functions f, g have continuous first partial derivatives on S.
The transformation is one-to-one.
Each of region R and S consists of a piecewise smooth simple closed curve and its interior.
The following determinant is not zero on S.

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

108. 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

109. Example 2
The Jacobian of the transformation x = r cos , y = r sin  is

110. If F is continuous on R, then (11)
THEOREM 9.6
If F is continuous on R, then (11)
Change of Variables in a
Double Integral

111. Example 3 Evaluate over the region R in Fig 9.148(a).
Fig 9.148(a) Fig 9.148(b)

112. Example 3 (2)
Solution We start by letting u = x + 2y, v = x – 2y.

113. Example 3 (3)
The Jacobian matrix is

114. Example 3 (4)
Thus

115. Example 4
Evaluate over the region R in Fig 9.149(a) Fig 9.149(a) Fig 9.149(b)

116. 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

117. Example 4 (3)
Hence

118. 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

120. Thank You !

121. Quiz (2)
1. (40%) (i) Find the level surface of passing through (1,1,1). (ii) Derive the equation of the tangent plane to the above level surface at (1,1,1). (iii) Find the unit vector such that the directional derivative of in at (1,1,1) achieves the minimum value among all possible .
2. (30%) Let the curve C be represented by i j,
, where and Let , please compute (i) , (ii) , and (iii)

122. 1.(10%)(i) F(1,1,1)= = -2
(15%)(ii) = 2xi - 8yj + 2zk
2(x-1)+(-8)(y-1)+2(z-1)=0
2x-2-8y+8+2z-2=0
The tangent plane equation is
2x-8y+2z+4=0

123. (iii)(15%)

124. 2. r(t) = x(t) i + y(t) j , x(t) = 2cost, y(t) = 2sint
dx = -2sint dt dy = 2cost dt
= -2[1-(-1)] = -4 (10%)
= 2[0-0] = 0 (10%)
(10%)