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Vector Calculus CHAPTER 9.10~9.17
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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
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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)
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Ch9.10~9.17_4 Fig 9.71
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Ch9.10~9.17_5 Iterated Integral For Type I: (4) For Type II: (5)
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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
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Ch9.10~9.17_7 Note: Volume = where z = f(x, y) is the surface.
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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
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Ch9.10~9.17_9 Fig 9.73
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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.
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Ch9.10~9.17_11 Fig 9.75(a) Fig 9.75(b)
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Ch9.10~9.17_12 Example 2 (2) Trying Fig 9.75(b), it is of Type II
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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.
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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.
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Ch9.10~9.17_15 Example 3 (2)
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Ch9.10~9.17_16 Example 3 (3)
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Ch9.10~9.17_17 Example 3 (4)
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Ch9.10~9.17_18 Example 3 (5) Hence
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Ch9.10~9.17_19 Moments of Inertia (12) are the moments of inertia about the x-axis and y-axis, respectively.
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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
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Ch9.10~9.17_21 Example 4 (2) Solution Since it is homogeneous, the density is the constant (x, y) = m/ r 2.
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Ch9.10~9.17_22 Example 4 (3)
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Ch9.10~9.17_23 Radius of Gyration Defined by (13) In Example 4,
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Ch9.10~9.17_24 9.11 Double Integrals in Polar Coordinates Double Integral Refer to the figure. The double integral is
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Ch9.10~9.17_25 Refer to the figure. The double integral is
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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
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Ch9.10~9.17_27 Example 1 (2) Solution We have: 0 /2, = kr, then
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Ch9.10~9.17_28 Example 1 (3) Since x = r cos and then
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Ch9.10~9.17_29 Example 1 (4) Similarly, y = r sin , then
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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
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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 )
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Ch9.10~9.17_32 Fig 9.84
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Ch9.10~9.17_33 Example 2 (2) Thus the integral becomes
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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
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Ch9.10~9.17_35 Example 3 (2) We find that and the equations become and r = sin . Now
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Ch9.10~9.17_36 Example 3 (3)
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Ch9.10~9.17_37 Area If f(r, ) = 1, then the area is
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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)
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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.
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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
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Ch9.10~9.17_41 Fig 9.89(a) Fig 9.89(b)
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Ch9.10~9.17_42 Partial Proof Using Fig 9.89(a), we have
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Ch9.10~9.17_43 Partial Proof Similarly, from Fig 9.89(b), From (2) + (3), we get (1).
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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
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Ch9.10~9.17_45 Example 1 Evaluate where C is shown in Fig 9.91.
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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
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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.
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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
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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.
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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:
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Ch9.10~9.17_51 Example 3 (3)
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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.
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Ch9.10~9.17_53 Fig 9.94
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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)
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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.
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Ch9.10~9.17_56 Example 5 Evaluate where C = C 1 C 2 is shown in Fig 9.96. Solution Because
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Ch9.10~9.17_57 Example 5 (2) are continuous on the region bounded by C, then
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Ch9.10~9.17_58 Fig 9.96
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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
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Ch9.10~9.17_60 Fig 9.97
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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.
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Ch9.10~9.17_62 Fig 9.98
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Ch9.10~9.17_63 Example 6 (2) Moreover, we have
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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.
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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
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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.
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Ch9.10~9.17_67 Fig 9.103
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Ch9.10~9.17_68 Example 1 (2) Change to polar coordinates:
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Ch9.10~9.17_69 Differential of Surface Area The function is called the differential of surface area.
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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
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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.
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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)
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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)
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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
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Ch9.10~9.17_75 Fig 9.104
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Ch9.10~9.17_76 Example 2 (2) Change to polar coordinates
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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.
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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
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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.
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Ch9.10~9.17_80 Fig. 9.106
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Ch9.10~9.17_81 Fig 9.107
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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.
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Ch9.10~9.17_83 Fig 9.108
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Ch9.10~9.17_84 Computing Flux We have (10) See Fig 9.109.
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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.
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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
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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
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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
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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
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Ch9.10~9.17_90 Example 1 (2) Solution See Fig 9.116. Surface Integral: From F = xyi + yzj + xzk, we find
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Ch9.10~9.17_91 Example 1 (3)
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Ch9.10~9.17_92 Example 1 (4)
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Ch9.10~9.17_93 Example 1 (5)
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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.
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Ch9.10~9.17_95 Fig 9.117
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Ch9.10~9.17_96 Example 2 (2) Solution The given orientation of C corresponding to an upward orientation of the surface S.
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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
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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
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Ch9.10~9.17_99 Evaluation by Iterated Integrals: See Fig 9.123.
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Ch9.10~9.17_100 Fig 9.123
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Ch9.10~9.17_101 Applications
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Ch9.10~9.17_102
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Ch9.10~9.17_103
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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)
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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
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Ch9.10~9.17_106 Example 2 Change the order of integration in Fig 9.126(a) Fig 9.126(b)
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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
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Ch9.10~9.17_108 Example 2 (3)
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Ch9.10~9.17_109 Cylindrical Coordinates Refer to Fig 9.127.
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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)
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Ch9.10~9.17_111 Example 1 Convert (8, /3, 7) in cylindrical coordinates to rectangular coordinates. Solution From (3)
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Ch9.10~9.17_112 Conversion of Rectangular Coordinates to Cylindrical Coordinates Also we have
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Ch9.10~9.17_113 Example 4 Solution
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Ch9.10~9.17_114 Fig 9.128
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Ch9.10~9.17_115 Triple Integrals in Cylindrical Coordinates See Fig 9.129. We have
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Ch9.10~9.17_116 Fig 9.129
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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
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Ch9.10~9.17_118 Fig 9.130
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Ch9.10~9.17_119 Example 5 (2) Similarly, we have
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Ch9.10~9.17_120 Example 5 (3)
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Ch9.10~9.17_121 Spherical Coordinates See Fig 9.131.
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Ch9.10~9.17_122 Conversion of Spherical Coordinates to Rectangular and Cylindrical Coordinates We have
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Ch9.10~9.17_123 Example 6 Convert (6, /4, /3) in spherical coordinates to rectangular and cylindrical coordinates. Solution
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Ch9.10~9.17_124 Inverse Conversion
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Ch9.10~9.17_125 Triple Integrals in Spherical Coordinates See Fig 9.132.
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Ch9.10~9.17_126 We have
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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
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Ch9.10~9.17_128 Example 7 (2) Solution If ( , , ) = k is the density, then
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Ch9.10~9.17_129 Example 7 (3)
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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
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Ch9.10~9.17_131 that is, The result in (1) is a special case of the divergence or Gauss’ theorem.
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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
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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.
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Ch9.10~9.17_134 Fig 9.140
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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
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Ch9.10~9.17_136 Example 1 (3)
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Ch9.10~9.17_137 Example 1 (4)
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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
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Ch9.10~9.17_139 Example 2 (2)
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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
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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.
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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)
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Ch9.10~9.17_143 Example 1 (2)
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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.
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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.
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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
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Ch9.10~9.17_147 Example 2 The Jacobian of the transformation x = r cos , y = r sin is
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Ch9.10~9.17_148 If F is continuous on R, then (11) THEOREM 9.6 Change of Variables in a Double Integral
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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)
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Ch9.10~9.17_150 Example 3 (2) Solution We start by letting u = x + 2y, v = x – 2y.
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Ch9.10~9.17_151 Example 3 (3) The Jacobian matrix is
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Ch9.10~9.17_152 Example 3 (4) Thus
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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)
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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
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Ch9.10~9.17_155 Example 4 (3) Hence
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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
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Ch9.10~9.17_157
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