Calc 3 Exam 3 Sin - 27 . So the center of mass is at ( M y...

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CALCULUS AND ANALYTIC GEOMETRY III SOLUTIONS TO SAMPLE TEST 3 Please let me know if you find errors or typos. For most problems, your solution should include pictures; I have omitted them but you should not. 1. The surface is the part of the graph of f ( x,y ) = x 2 + y 2 which lies over the disk D given by x 2 + y 2 16. Applying to formula for surface area, using f x = 2 x , f y - 2 y , and then converting to polar coordinates we have SA = Z D p 1 + 4 x 2 + 4 y 2 dA = Z 2 π 0 Z 4 0 p 1 + 4 r 2 r dr dθ = π 6 (65 3 2 - 1) . 2. The plate is bounded below by y = x 2 and above by y = x for 0 x 1. We have m = Z 1 0 Z x x 2 ρ ( x,y ) dy dx = Z 1 0 Z x x 2 2 xy 2 dy x = Z 1 0 2 x 4 - 2 x 7 3 dx = 1 20 . Similarly, we compute the moments about the axes: M x = Z 1 0 Z x x 2 ( x,y ) dy dx = Z 1 0 Z x x 2 2 xy 3 dy x = Z 1 0 x 5 - x 9 2 dx = 1 30 and M y = Z 1 0 Z x x 2 ( x,y ) dy dx = Z 1 0 Z x x 2 2 x 2 y 2 dy x = Z 1 0 2 x 5 - 2 x 8 3 dx = 1
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Unformatted text preview: 27 . So the center of mass is at ( M y /m,M x /m ) = (27 / 20 , 3 / 2). 3. By drawing a picture, one sees that the region in question is the triangle with vertices (0 , 0), (0 , 1) and (1 , 1). So if we interchange the order of integration, the integral becomes Z 1 Z y (1-y 2 ) 1 3 dxdy = Z 1 y (1-y 2 ) 1 3 dy = 3 8 . 4. Converting to spherical coordinates, the integral becomes: Z π 2 Z π 2 Z 2 1 e-ρ 2 ρ ρ 2 sin φdρdφdθ = Z π 2 Z π 2 ±-1 2 e-ρ 2 ² 2 1 sin φdφdθ = π 4 ( e-1-e-4 ) ....
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This note was uploaded on 10/20/2011 for the course MAC 2313 taught by Professor Keeran during the Fall '08 term at University of Florida.

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