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vec_int_calc_sol

# vec_int_calc_sol - MIT OpenCourseWare http/ocw.mit.edu...

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MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable Calculus Fall 200 7 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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6. Vector Integral Calculus in Space 6A. Vector Fields in Space 6A-1 a) the vectors are all unit vectors, pointing radially outward. b) the vector at P has its head on the y-axis, and is perpendicular to it 6A-4 A vector field F = M i + N j + P k is parallel to the plane 32 - 4y + z = 2 if it is perpendicular to the normal vector to the plane, 3 i - 4 j + k : the condition on M, N, P therefore is 3M - 4N + P = 0, or P = 4N - 3M. The most general such field is therefore F = M i + N j + (4N - 3M) k , where M and N are functions of x, y, z. 6B. Surface Integrals and Flux x i + y j + z k 6B-1 We have n = ; therefore F . n = a. a I P P Flux through S = F . n dS = a(area of S) = 47r a3 Is 6B-2 Since k is parallel to the surface, the field is everywhere tangent to the cylinder, hence the flux is 0. ll, i + j + k 1 6B-3 is a normal vector to the plane, so F . n = -. l.6 l.6 area of region - 4 (base) (height) - ( 4 )( 4 ) - - 1 Therefore, flux = - - - l.6 l.6 l.6 2' x i + y j + z k y2 6B-4 /k n = l" l" ; F . n = -. Calculating in spherical coordinates, a a flux = L 2 dS = a4 sin3 4 sin2 0 dm do = a3 ln 6" sin3 4 sin2 0 dmde. a Inner integral: sin2 O(- cos 4 + \$ cos3 4) \$ 1 1: = \$ sin2 8; " Outer integral: %a3 (\$8 - sin 28) J, = 2i7ra3.
2 E. 18.02 EXERCISES F . n - z i + j + k . 6B-5 n = a ' a. z dxdy 1 dx dy 1-Y Innerintegral: = x - ~ x 1 2 -xy ]I-. = \$ ( 1 - ~ ) ~ . 0 1 1 Outerintegral: = 2 3 6B-6 z = f (x, y) = x2 + y2 (a paraboloid). By (13) in Notes V9, dS = (-2xi - 2yj + k ) dxdy. (This points generally "up", since the k component is positive:) Since F = x i + y j + z k , where R is the interior of the unit circle in the xy-plane, i.e., the projection of S onto the xy-plane). Since z = x2 + y2, the above integral The answer is negative since the positive direction for flux is that of n, which here points into the inside of the paraboloidal cup, whereas the flow x i + y j + z k is generally from the inside toward the outside of the cup, i.e., in the opposite direction. x i + y j F . n = - . y2 6B-8 On the cylindrical surface, n = a a In cylindrical lri2 coordinates, Jdk since y = a sin 8, this gives us F . dS = F . n dS = a2sin28 dz dB. Flux = a2sin28 dz dB = a2h sin28d8=a2h ( - - - ~ i ; 2 8 ) ~ / ~ = -a2h. s -r/2 -r/2 2 6B-12 Since the distance from a point (x, y, 0) up to the hemispherical surface is z, JJs dS average distance = - JJs dS ' In spherical coordinates, / L z d S = Jd2rJdri2acos) .a2sin)d)d8. Inner: = a3Jd r/2 :/2 : . Jd 2 r sin)cos)d)=a3(- s~F] - - Outer: = d8 = sa3. s a 3 a Finally, dS = area of hemisphere = 2sa2, so average distance = - - - 2sa2 2'

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6. VECTOR INTEGRAL CALCULUS IN SPACE 6C. Divergence Theorem 6C-2 Using the product and chain rules for the first, symmetry for the others, x2 + y2 + z 2 adding these three, we get div F = npn-l + 3pn = pn(n+3). P Therefore, div F = 0 # n = -3. 6C-3 Evaluating the /k triple integral first, we have div F = 3, therefore 2 div F dV = 3(vol.of D) = 3 -7ra3 = 27ra3.
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vec_int_calc_sol - MIT OpenCourseWare http/ocw.mit.edu...

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