CHAPTER 9 109 17 Vector Calculus Contents v
- Slides: 124
CHAPTER 9. 10~9. 17 Vector Calculus
Contents v 9. 10 Double Integrals v 9. 11 Double Integrals in Polar Coordinates v 9. 12 Green’s Theorem v 9. 13 Surface Integrals v 9. 14 Stokes’ Theorem v 9. 15 Triple Integrals v 9. 16 Divergence Theorem v 9. 17 Change of Variables in Multiple Integrals 2
9. 10 Double Integrals v 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. 3
Properties of Double Integrals v 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 4
Regions of Type I and II § Region of Type I See the region in Fig 9. 71(a) R: a x b, g 1(x) y g 2(x) § Region of Type II See the region in Fig 9. 71(b) R: c y d, h 1(y) x h 2(y) 5
Fig 9. 71 6
Iterated Integral v For Type I: (4) v For Type II: (5) 7
THEOREM 9. 12 Evaluation of Double Integrals Let f be continuous on a region R. (i) For Type I: (6) (ii) For Type II: (7) 8
Note: v Volume = where z = f(x, y) is the surface. 9
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 10
Fig 9. 73 11
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. 12
Fig 9. 75(a) Fig 9. 75(b) 13
Example 2 (2) Trying Fig 9. 75(b), it is of Type II 14
9. 11 Double Integrals in Polar Coordinates v Double Integral Refer to the figure. The double integral is 15
Refer to the figure. The double integral is 16
Change of Variables v 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 17
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) 18
Fig 9. 84 19
Example 2 (2) Thus the integral becomes 20
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 21
Example 3 (2) We find that and the equations become and r = sin . Now 22
Example 3 (3) 23
Area v If f(r, ) = 1, then the area is 24
9. 12 Green’s Theorem v Along Simple Closed Curves For different orientations of simple closed curves, please refer to Fig 9. 88(a) Fig 9. 88(b) Fig 9. 88(c) 25
Notations for Integrals Along Simple Closed Curves v We usually write them as the following forms where and represents in the positive and negative directions, respectively. 26
THEOREM 9. 13 Green’s Theorem in the Plane IF P, Q, P/ y, Q/ x are continuous on R, which is bounded by a simply closed curve C, then v Partial Proof For a region R is simultaneously of Type I and Type II, 27
Fig 9. 89(a) Fig 9. 89(b) 28
Partial Proof Using Fig 9. 89(a), we have 29
Partial Proof Similarly, from Fig 9. 89(b), From (2) + (3), we get (1). 30
Note: v 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. v Fig 9. 90 31
Example 2 Evaluate where C is the circle (x – 1)2 + (y – 5)2 = 4 shown in Fig 9. 92. 32
Example 2 (2) Solution We have P(x, y) = x 5 + 3 y and then Hence Since the area of this circle is 4 , we have 33
Example 3 v Find the work done by F = (– 16 y + sin x 2)i + (4 ey + 3 x 2)j along C shown in Fig 9. 93. 34
Example 3 (2) Solution We have Hence from Green’s theorem In view of R, it is better handled in polar coordinates, since R: 35
Example 3 (3) 36
Example 4 v The curve is shown in Fig 9. 94. Green’s Theorem is not applicable to the integral since P, Q, P/ x, Q/ y are not continuous at the region. 37
Fig 9. 94 38
Region with Holes v 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) 39
Fig 9. 95(a) Fig 9. 95(b) v The last result follows from that fact that the line integrals on the crosscuts cancel each other. 40
Example 5 Evaluate where C = C 1 C 2 is shown in Fig 9. 96. Solution Because 41
Example 5 (2) are continuous on the region bounded by C, then 42
Fig 9. 96 43
Conditions to Simplify the Curves v 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 44
Fig 9. 97 45
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. 46
Fig 9. 98 47
Example 6 (2) Moreover, we have 48
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. 49
9. 13 Surface Integrals DEFINITION 9. 11 Surface Area 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) 50
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. 51
Fig 9. 103 52
Example 1 (2) Change to polar coordinates: 53
Differential of Surface Area v The function is called the differential of surface area. 54
DEFINITION 9. 12 Surface Integral 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) 55
Method of Evaluation v (5) where we define z = f(x, y) to be the equation of S projecting into a region R of the xy-plane. 56
Projection of S Into Other Planes v If we define y = g(x, z) to be the equation of S projecting into a region R of the xz-plane, then (6) v Similarly, if x = h(y, z) is the equation of S projecting into a region R of the yz-plane, then (7) 57
Example 3 Evaluate , where S is the portion of y = 2 x 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. 58
Example 3 (2) Let y = g(x, z) = 2 x 2 + 1. Since gx(x, z) = 4 x and gz(x, z) = 0, then 59
Orientable Surface v 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. 60
Fig. 9. 106 61
Fig 9. 107 62
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. 63
Fig 9. 108 64
Computing Flux v We have (10) See Fig 9. 109. v Flux of F through S: The total volume of a fluid passing through S per unit time. 65
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 – 3 x – 2 y in the first octant oriented upward. Solution Refer to the figure. 66
Example 5 (2) We define g(x, y, z) = 3 x + 2 y + 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 67
9. 14 Stokes’ Theorem v 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 68
THEOREM 9. 14 Stokes’ Theorem 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. 69
Example 1 v 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. v Fig 9. 116 70
Example 1 (2) Solution See Fig 9. 116. Surface Integral: From F = xyi + yzj + xzk, we find 71
Example 1 (3) 72
Example 1 (4) 73
Example 1 (5) 74
9. 15 Triple Integrals DEFINITION 9. 13 The Triple Integral 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) 75
Evaluation by Iterated Integrals: See Fig 9. 123. 76
Fig 9. 123 77
Example 1 v Find the volume of the solid in the first octant bounded by z = 1 – y 2, y = 2 x and x = 3. v Fig 9. 125(a) Fig 9. 125(b) 78
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 79
Cylindrical Coordinates v Refer to Fig 9. 127. 80
Conversion of Cylindrical Coordinates to Rectangular Coordinates v The relationship between the cylindrical coordinates (r, , z) and rectangular coordinates (x, y, z): x = r cos , y = r sin , z = z (3) 81
Example 1 Convert (8, /3, 7) in cylindrical coordinates to rectangular coordinates. Solution From (3) 82
Conversion of Rectangular Coordinates to Cylindrical Coordinates v Also we have 83
Example 4 Solution 84
Fig 9. 128 85
Triple Integrals in Cylindrical Coordinates v See Fig 9. 129. v We have 86
Fig 9. 129 87
Spherical Coordinates v See Fig 9. 131. 88
Conversion of Spherical Coordinates to Rectangular and Cylindrical Coordinates 89
Example 6 Convert (6, /4, /3) in spherical coordinates to rectangular and cylindrical coordinates. Solution 90
Inverse Conversion 91
Triple Integrals in Spherical Coordinates v See Fig 9. 132. 92
v We have 93
9. 16 Divergence Theorem v 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 94
v that is, The result in (1) is a special case of the divergence or Gauss’ theorem. 95
THEOREM 9. 15 Divergence Theorem 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) 96
Example 1 Let D be the region bounded by the hemisphere Solution The closed region is shown in Fig 9. 140. 97
Fig 9. 140 98
Example 1 (2) Triple Integral: Since F = xi + yj + (z-1)k, we see div F = 3. Hence (10) Surface Integral: We write S = S 1 + S 2, 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 99
Example 1 (3) 100
Example 1 (4) 101
9. 17 Change of Variables in Multiple Integrals v 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 102
Double Integrals v 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). 103
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) 104
Example 1 (2) 105
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. 106
v 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 9. 147. 107
v 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 108
Example 2 The Jacobian of the transformation x = r cos , y = r sin is 109
THEOREM 9. 6 Change of Variables in a Double Integral If F is continuous on R, then (11) 110
Example 3 Evaluate over the region R in Fig 9. 148(a) Fig 9. 148(b) 111
Example 3 (2) Solution We start by letting u = x + 2 y, v = x – 2 y. 112
Example 3 (3) The Jacobian matrix is 113
Example 3 (4) Thus 114
Example 4 Evaluate Fig 9. 149(a) over the region R in Fig 9. 149(a). Fig 9. 149(b) 115
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 116
Example 4 (3) Hence 117
Triple Integrals v 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 118
119
Quiz (2) v 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) 121
1. (10%)(i) F(1, 1, 1)=1 - 4 + 1= -2 (15%)(ii) = 2 xi - 8 yj + 2 zk 2(x-1)+(-8)(y-1)+2(z-1)=0 2 x-2 -8 y+8+2 z-2=0 The tangent plane equation is 2 x-8 y+2 z+4=0 122
(iii)(15%) 123
2. r(t) = x(t) i + y(t) j , x(t) = 2 cost, y(t) = 2 sint dx = -2 sint dt dy = 2 cost dt = -2[1 -(-1)] = -4 (10%) = 2[0 -0] = 0 (10%) 124
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