Exam_solutions_3_

Exam_solutions_3_ - MATH 241, 3rd Examination Solutions and...

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Solutions and Grading Key Prof. Jonathan Rosenberg Monday, November 28, 2005 1. (40 points, 10 per part) Suppose that an object occupies the solid ball D bounded by the sphere x 2 + y 2 + ( z - 2) 2 = 4 , and has mass density f ( x, y, z ) = z . Set up, but do not evaluate , explicit iterated integrals (with the integrand and all limits of integration written out in the appropriate coordinate system) for the mass of the object RRR D f dV in: (a) rectangular (Cartesian) coordinates; (b) cylindrical coordinates; (c) spherical coordinates. (d) Now compute one of the integrals you obtained in (a), (b), or (c). (You will only get credit for evaluating one of the three, though you are free to compute more than one as a way to check your answer. Correct evaluation of an integral that is very far from being correct, in case you made a mistake in (a), (b), or (c), will be worth something, but is not necessarily worth full credit on this part.) Solution. (a) For this part, 1 point each for the outer 4 limits of integration, 3 points each for the inner 2 limits of integration. Remember they do not need to evaluate; I’ve just put the evaluation here in case they do it for (d). The equation of the sphere is ( z - 2) 2 = 4 - x 2 - y 2 or z = 2 ± p 4 - x 2 - y 2 . Its projection to the x - y plane is the circular disk x 2 + y 2 4 . With the z on the inside we get Z 2 - 2 Z 4 - x 2 - 4 - x 2 Z 2+ 4 - x 2 - y 2 2 - 4 - x 2 - y 2 z dz dy dx = 1 2 Z 2 - 2 Z 4 - x 2 - 4 - x 2 ± ² 2 + p 4 - x 2 - y 2 ³ 2 - ² 2 - p 4 - x 2 - y 2 ³ 2 ´ dy dx = Z 2 - 2 Z 4 - x 2 - 4 - x 2 4 p 4 - x 2 - y 2 dy dx. Without the 4 , this would be the volume of half a spherical ball, so the integral works out to 4 2 π 3 (2) 3 = 64 π 3 . Also,
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Exam_solutions_3_ - MATH 241, 3rd Examination Solutions and...

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