Quiz7_sol

# Quiz7_sol - Z 2 Z 3-x x 2 x y dydx = Z 2 xy y 2 2 | 3-x x 2...

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NAME: Spring 2011, MAC 2313, Quiz 7 UFID: Section: 3124 1. Evaluate RR D e - x 2 - y 2 dA by changing to polar coordinates, where D is the region bounded by the semicircle x = p 4 - y 2 and the y -axis. (3 points) Solution . Changing to polar coordinates we have Z 2 0 Z π/ 2 - π/ 2 e - r 2 rdrdθ = π Z 2 0 e - r 2 rdr = - π 2 e - r 2 | 2 0 = π 2 (1 - e - 4 ) . (1) 2. Find the mass of the lamina that occupies the region D where D is the triangular region with vertices (0 , 0), (2 , 1) and (0 , 3), and density function ρ ( x, y ) = x + y . (3 points) Solution . According to the vertices, the triangle is bounded by lines y = x/ 2 and y = 3 - x and x = 0. Therefore, D = { ( x, y ) | 0 x 2 , x/ 2 y 3 - x } , and hence the mass is m = ZZ D
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Unformatted text preview: Z 2 Z 3-x x/ 2 ( x + y ) dydx = Z 2 ( xy + y 2 2 ) | 3-x x/ 2 dx = Z 2 ( 9 2-9 x 2 8 ) dx = ( 9 x 2-3 x 3 8 ) | 2 = 6 . (2) 3. Evaluate the triple integral RRR E 2 xdV where E = { ( x,y,z ) | ≤ y ≤ 2 , ≤ x ≤ p 4-y 2 , ≤ z ≤ y } . (3 points) Solution . According to the bounds of E , we know the triple integral is ZZZ E 2 xdV = Z 2 Z √ 4-y 2 Z y 2 xdzdxdy = Z 2 ( Z √ 4-y 2 2 xdx )( Z y dz ) dy = Z 2 (4-y 2 ) ydy = (2 y 2-y 4 4 ) | 2 = 4 . (3)...
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