1. Vector Calculus CHAPTER 9.10~9.17

2. Ch9.10~9.17_2 Contents  9.10 Double Integrals 9.10 Double Integrals  9.11 Double Integrals in Polar Coordinates 9.11 Double Integrals in Polar Coordinates  9.12 Green’s Theorem 9.12 Green’s Theorem  9.13 Surface Integrals 9.13 Surface Integrals  9.14 Stokes’ Theorem 9.14 Stokes’ Theorem  9.15 Triple Integrals 9.15 Triple Integrals  9.16 Divergence Theorem 9.16 Divergence Theorem  9.17 Change of Variables in Multiple Integrals 9.17 Change of Variables in Multiple Integrals

3. 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, g 1 (y)  y  g 2 (y)  Region of Type II See the region in Fig 9.71(b) R: c  y  d, h 1 (x)  x  h 2 (x)

4. Ch9.10~9.17_4 Fig 9.71

5. Ch9.10~9.17_5 Iterated Integral  For Type I: (4)  For Type II: (5)

6. Ch9.10~9.17_6 Let f be continuous on a region R. (i) For Type I: (6) (ii) For Type II: (7) THEOREM 9.12 Evaluation of Double Integrals

7. Ch9.10~9.17_7 Note:  Volume = where z = f(x, y) is the surface.

8. 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. Solution The region is Type II

9. Ch9.10~9.17_9 Fig 9.73

10. Ch9.10~9.17_10 Example 2 Evaluate over the region in the first quadrant bounded by y = x 2, x = 0, y = 4. Solution From Fig 9.75(a), it is of Type I However, this integral can not be computed.

11. Ch9.10~9.17_11 Fig 9.75(a) Fig 9.75(b)

12. Ch9.10~9.17_12 Example 2 (2) Trying Fig 9.75(b), it is of Type II

13. 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.

14. 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. Solution See Fig 9.76.

15. Ch9.10~9.17_15 Example 3 (2)

16. Ch9.10~9.17_16 Example 3 (3)

17. Ch9.10~9.17_17 Example 3 (4)

18. Ch9.10~9.17_18 Example 3 (5) Hence

19. Ch9.10~9.17_19 Moments of Inertia  (12) are the moments of inertia about the x-axis and y-axis, respectively.

20. Ch9.10~9.17_20 Example 4 Refer to Fig 9.77. Find I y of the thin homogeneous disk of mass m. Fig 9.77

21. Ch9.10~9.17_21 Example 4 (2) Solution Since it is homogeneous, the density is the constant  (x, y) = m/  r 2.

22. Ch9.10~9.17_22 Example 4 (3)

23. Ch9.10~9.17_23 Radius of Gyration  Defined by (13) In Example 4,

24. Ch9.10~9.17_24 9.11 Double Integrals in Polar Coordinates  Double Integral Refer to the figure. The double integral is

25. Ch9.10~9.17_25 Refer to the figure. The double integral is

26. Ch9.10~9.17_26 Example 1 Refer to Fig 9.83. Find the center of mass where r = 2 sin 2  in the first quadrant and  is proportional to the distance from the pole. Fig 9.83

27. Ch9.10~9.17_27 Example 1 (2) Solution We have: 0     /2,  = kr, then

28. Ch9.10~9.17_28 Example 1 (3) Since x = r cos  and then

29. Ch9.10~9.17_29 Example 1 (4) Similarly, y = r sin , then

30. 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: x 2 + y 2 = r 2 and

31. Ch9.10~9.17_31 Example 2 Evaluate Solution From the graph is shown in Fig 9.84. Using x 2 + y 2 = r 2, then 1/(5 + x 2 + y 2 ) = 1/(5 + r 2 )

32. Ch9.10~9.17_32 Fig 9.84

33. Ch9.10~9.17_33 Example 2 (2) Thus the integral becomes

34. Ch9.10~9.17_34 Example 3 Find the volume of the solid that is under and above the region bounded by x 2 + y 2 – y = 0. See Fig 9.85. Solution Fig 9.85

35. Ch9.10~9.17_35 Example 3 (2) We find that and the equations become and r = sin . Now

36. Ch9.10~9.17_36 Example 3 (3)

37. Ch9.10~9.17_37 Area  If f(r,  ) = 1, then the area is

38. 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)

39. 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.

40. Ch9.10~9.17_40  Partial Proof For 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 9.13 Green’s Theorem in the Plane

41. Ch9.10~9.17_41 Fig 9.89(a) Fig 9.89(b)

42. Ch9.10~9.17_42 Partial Proof Using Fig 9.89(a), we have

43. Ch9.10~9.17_43 Partial Proof Similarly, from Fig 9.89(b), From (2) + (3), we get (1).

44. 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

45. Ch9.10~9.17_45 Example 1  Evaluate where C is shown in Fig 9.91.

46. Ch9.10~9.17_46 Example 1 (2) Solution If P(x, y) = x 2 – y 2, Q(x, y) = 2y – x, then and Thus

47. 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.

48. Ch9.10~9.17_48 Example 2 (2) Solution We have P(x, y) = x 5 + 3y and then Hence Since the area of this circle is 4 , we have

49. Ch9.10~9.17_49 Example 3  Find the work done by F = (– 16y + sin x 2 )i + (4e y + 3x 2 )j along C shown in Fig 9.93.

50. Ch9.10~9.17_50 Example 3 (2) Solution We have Hence from Green’s theorem In view of R, it is better handled in polar coordinates, since R:

51. Ch9.10~9.17_51 Example 3 (3)

52. 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.

53. Ch9.10~9.17_53 Fig 9.94

54. 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 C 1 and C 2. Now We introduce cross cuts as shown is Fig 9.95(b), R is divided into R 1 and R 2. By Green’s theorem: (4)

55. 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.

56. Ch9.10~9.17_56 Example 5 Evaluate where C = C 1  C 2 is shown in Fig 9.96. Solution Because

57. Ch9.10~9.17_57 Example 5 (2) are continuous on the region bounded by C, then

58. Ch9.10~9.17_58 Fig 9.96

59. Ch9.10~9.17_59 Conditions to Simply the Curves  As shown in Fig 9.97, C 1 and C 2 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 C 1 and C 2, then we have

60. Ch9.10~9.17_60 Fig 9.97

61. Ch9.10~9.17_61 Example 6 Evaluate the line integral in Example 4. Solution We find P = – y / (x 2 + y 2 ) and Q = x / (x 2 + y 2 ) have continuous first partial derivatives in the region bounded by C and C’. See Fig 9.98.

62. Ch9.10~9.17_62 Fig 9.98

63. Ch9.10~9.17_63 Example 6 (2) Moreover, we have

64. 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.

65. Ch9.10~9.17_65 9.13 Surface Integrals Let f be a function with continuous first derivatives f x, f y on a closed region. Then the area of the surface over R is given by (2) DEFINITION 9.11 Surface Area

66. Ch9.10~9.17_66 Example 1 Find the surface area of portion of x 2 + y 2 + z 2 = a 2 and is above the xy-plane and within x 2 + y 2 = b 2, where 0 < b < a. Solution If we define then Thus where R is shown in Fig 9.103.

67. Ch9.10~9.17_67 Fig 9.103

68. Ch9.10~9.17_68 Example 1 (2) Change to polar coordinates:

69. Ch9.10~9.17_69 Differential of Surface Area  The function is called the differential of surface area.

70. 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 9.12 Surface Integral

71. 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.

72. 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)

73. 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)

74. Ch9.10~9.17_74 Example 2 Find the mass of the surface of z = 1 + x 2 + y 2 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 + x 2 + y 2, then

75. Ch9.10~9.17_75 Fig 9.104

76. Ch9.10~9.17_76 Example 2 (2) Change to polar coordinates

77. Ch9.10~9.17_77 Example 3 Evaluate, where S is the portion of y = 2x 2 + 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.

78. Ch9.10~9.17_78 Example 3 (2) Let y = g(x, z) = 2x 2 + 1. Since g x (x, z) = 4x and g z (x, z) = 0, then

79. 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.

80. Ch9.10~9.17_80 Fig. 9.106

81. Ch9.10~9.17_81 Fig 9.107

82. Ch9.10~9.17_82 Example 4 Consider x 2 + y 2 + z 2 = a 2, a > 0. If we define g(x, y, z) = x 2 + y 2 + z 2 – a 2, then Thus the two orientations are where n defines outward orientation, n 1 = − n defines inward orientation. See Fig 9.108.

83. Ch9.10~9.17_83 Fig 9.108

84. Ch9.10~9.17_84 Computing Flux  We have (10) See Fig 9.109.

85. 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. Solution Refer to the figure.

86. 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

87. 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

88. 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 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. THEOREM 9.14 Stokes’ Theorem

89. Ch9.10~9.17_89 Example 1  Let S be the part of the cylinder z = 1 – x 2 for 0  x  1, −2  y  2. Verify Stokes’ theorem if F = xyi + yzj + xzk.  Fig 9.116

90. Ch9.10~9.17_90 Example 1 (2) Solution See Fig 9.116. Surface Integral: From F = xyi + yzj + xzk, we find

91. Ch9.10~9.17_91 Example 1 (3)

92. Ch9.10~9.17_92 Example 1 (4)

93. Ch9.10~9.17_93 Example 1 (5)

94. Ch9.10~9.17_94 Example 2  Evaluate where C is the trace of the cylinder x 2 + y 2 = 1 in the plane y + z = 2. Orient C counterclockwise as viewed from above. See Fig 9.117.

95. Ch9.10~9.17_95 Fig 9.117

96. Ch9.10~9.17_96 Example 2 (2) Solution The given orientation of C corresponding to an upward orientation of the surface S.

97. 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

98. 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 9.13 The Triple Integral

99. Ch9.10~9.17_99 Evaluation by Iterated Integrals: See Fig 9.123.

100. Ch9.10~9.17_100 Fig 9.123

101. Ch9.10~9.17_101 Applications

102. Ch9.10~9.17_102

103. Ch9.10~9.17_103

104. Ch9.10~9.17_104 Example 1  Find the volume of the solid in the first octant bounded by z = 1 – y 2, y = 2x and x = 3.  Fig 9.125(a) Fig 9.125(b)

105. 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 – y 2. From Fig 9.125(b), we see that the projection of D in the xy-plane is a region of Type II. Hence

106. Ch9.10~9.17_106 Example 2 Change the order of integration in Fig 9.126(a) Fig 9.126(b)

107. 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

108. Ch9.10~9.17_108 Example 2 (3)

109. Ch9.10~9.17_109 Cylindrical Coordinates  Refer to Fig 9.127.

110. 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)

111. Ch9.10~9.17_111 Example 1 Convert (8,  /3, 7) in cylindrical coordinates to rectangular coordinates. Solution From (3)

112. Ch9.10~9.17_112 Conversion of Rectangular Coordinates to Cylindrical Coordinates  Also we have

113. Ch9.10~9.17_113 Example 4 Solution

114. Ch9.10~9.17_114 Fig 9.128

115. Ch9.10~9.17_115 Triple Integrals in Cylindrical Coordinates  See Fig 9.129.  We have

116. Ch9.10~9.17_116 Fig 9.129

117. Ch9.10~9.17_117 Example 5 A solid in the first octant has the shape determined by the graph of the cone z = (x 2 + y 2 ) 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

118. Ch9.10~9.17_118 Fig 9.130

119. Ch9.10~9.17_119 Example 5 (2) Similarly, we have

120. Ch9.10~9.17_120 Example 5 (3)

121. Ch9.10~9.17_121 Spherical Coordinates  See Fig 9.131.

122. Ch9.10~9.17_122 Conversion of Spherical Coordinates to Rectangular and Cylindrical Coordinates  We have

123. Ch9.10~9.17_123 Example 6 Convert (6,  /4,  /3) in spherical coordinates to rectangular and cylindrical coordinates. Solution

124. Ch9.10~9.17_124 Inverse Conversion

125. Ch9.10~9.17_125 Triple Integrals in Spherical Coordinates  See Fig 9.132.

126. Ch9.10~9.17_126  We have

127. Ch9.10~9.17_127 Example 7 Find the moment of inertia about the z-axis of the homogeneous solid bounded between the spheres x 2 + y 2 + z 2 = a 2 and x 2 + y 2 + z 2 = a 2, a < b Fig 9.133

128. Ch9.10~9.17_128 Example 7 (2) Solution If  ( , ,  ) = k is the density, then

129. Ch9.10~9.17_129 Example 7 (3)

130. Ch9.10~9.17_130 9.16 Divergence Theorem  Another Vector Form of Green’s Theorem Let 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

131. Ch9.10~9.17_131  that is, The result in (1) is a special case of the divergence or Gauss’ theorem.

132. 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 9.15 Divergence Theorem

133. Ch9.10~9.17_133 Example 1 Let D be the region bounded by the hemisphere Solution The closed region is shown in Fig 9.140.

134. Ch9.10~9.17_134 Fig 9.140

135. 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 S 1 is the hemisphere and S 2 is the plane z = 1. If S 1 is a level surfaces of g(x, y) = x 2 + y 2 + (z – 1) 2, then a unit outer normal is

136. Ch9.10~9.17_136 Example 1 (3)

137. Ch9.10~9.17_137 Example 1 (4)

138. Ch9.10~9.17_138 Example 2 IF F = xyi + y 2 zj + z 3 k, evaluate  S (F  n)dS, where S is the unit cube defined by 0  x  1, 0  y  1, 0  z  1. Solution We see div F =   F = x + 2yz + 3z 2. Then

139. Ch9.10~9.17_139 Example 2 (2)

140. 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

141. 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.

142. Ch9.10~9.17_142 Example 1 Find the image of the region S shown in Fig 9.146(a) under the transformations x = u 2 + v 2, y = u 2 − v 2. Solution Fig 9.146(a) Fig 9.146(b)

143. Ch9.10~9.17_143 Example 1 (2)

144. 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.

145. 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.

146. 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

147. Ch9.10~9.17_147 Example 2 The Jacobian of the transformation x = r cos , y = r sin  is

148. Ch9.10~9.17_148 If F is continuous on R, then (11) THEOREM 9.6 Change of Variables in a Double Integral

149. 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)

150. Ch9.10~9.17_150 Example 3 (2) Solution We start by letting u = x + 2y, v = x – 2y.

151. Ch9.10~9.17_151 Example 3 (3) The Jacobian matrix is

152. Ch9.10~9.17_152 Example 3 (4) Thus

153. 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)

154. Ch9.10~9.17_154 Example 4 (2) Solution The equations of the boundaries of R suggest u = y/x 2, 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

155. Ch9.10~9.17_155 Example 4 (3) Hence

156. 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

157. Ch9.10~9.17_157