Math 293 practice prelim 1 solution

# Math 293 practice prelim 1 solution - Math 293 Answers to...

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Math 293 Answers to Practice Prelim #1 Spring 2000 x y y = 4 - x 2 4 2 1 . R 2 0 R 4 - x 2 0 xe 2 y 4 - y dy dx . This is awkward to in- tegrate in the given order, so switch the order to get R 4 0 R 4 - y 0 xe 2 y 4 - y dx dy . Then after the first inte- gration we have R 4 0 e 2 y 2 dy , which works out to be e 8 4 - 1 4 . 2 . R 1 0 R 1 - x 2 0 e - ( x 2 + y 2 ) dy dx . The region of integration is the portion of the unit disk lying in the first quadrant, so in polar coordinates this integral becomes R π/ 2 0 R 1 0 e - r 2 r dr dθ , which computes out to be ( - e - 1 +1 2 )( π 2 ) . 3 . In spherical coordinates the cone is given by the equation φ = π 4 , so the region below the cone and above the xy -plane will have π 4 φ π 2 . The region between the two spheres will have 3 ρ 4. Also, x 2 + y 2 = ρ 2 - z 2 . So we get the integral R 2 π 0 R π/ 2 π/ 4 R 4 3 ( ρ 2 - z 2 ) ρ 2 sin φ dρ dφ dθ . Substituting z = ρ cos φ , this becomes R 2 π 0 R π/ 2 π/ 4 R 4 3 ( ρ 2 - ρ 2 cos 2 φ ) ρ 2 sin φ dρ dφ dθ which works out to be ( 2 π )( 4 5 - 3 5 5 )( 1 2 - 1 3 · 2 2 ) . 4 . The vector field F ( x, y ) = ( y + e x
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