vec_int_calc_sol

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MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable Calculus Fall 2007 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 = Mi + N j + Pk is parallel to the plane 32 - 4y + z = 2 if it is perpendicular to the normal vector to the plane, 3 i - 4j + k : the condition on M, N, P therefore is 3M - 4N + P = 0, or P = - 3M. The most general such field is therefore F = + N j + (4N - 3M) k , where M and N are functions of x, y, z. 6B. Surface Integrals and Flux xi +yj +zk 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 area of region - 4 (base) (height) - (4) - - 1 Therefore, flux = - - - 2' 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" sin34 sin2 0 dmde. a Inner integral: sin2 O(- cos 4 + $ cos3 4) $ 1 1: = $ sin2 8; " Outer integral: %a3 ($8 - sin 28) J, = 2i7ra3.
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2 E. 18.02 EXERCISES F.n- z i+j+k. 6B-5 n= a ' a. z dxdy 1 dx dy 1-Y Innerintegral: =x-~x 1 2 -xy ]I-. =$(1-~)~. 0 11 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 = + 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 xi + y j + z k is generally from the inside toward the outside of the cup, i.e., in the opposite direction. xi+yj 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 . = F . n dS = a2sin28 dz dB. Flux = a2sin28 dz dB = a2h sin28d8=a2h ( - - - ~i;28)~/~ = -a2h. s -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 = - ' In spherical coordinates, /LzdS = Jd2rJdri2acos) .a2sin)d)d8. Inner: = a3Jd r/2 :/2 :. Jd 2r sin)cos)d)=a3(- s~F] - - Outer: = d8 = sa3. sa3 a Finally, = 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 +z2 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. 3 To evaluate the double integral over the closed surface S = S1 + S2,the respective normal vectors are: xi +yj +zk nl = (hemisphere S1), n2 = - k (disc 5'2); a using these, the surface integral for the flux through S is since x2 + y2 + z2 = P2 = a2 on S1, and z = 0 on S2. SO the value of the surface integral is a(area of S1)= a(27ra2) = 27ra3, which agrees with the triple integral above.
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vec_int_calc_sol - MIT OpenCourseWare http:/ocw.mit.edu...

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