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Math 53 Worksheet Solutions - Integration and Change of Coordinates RRR 1. Evaluate xy dV , where E is bounded by the parabolic cylinders y = x2 and x = y 2 and the planes z = 0 and z = x + y. Solution. ZZZ 1 Z √ Z x Z x+y xy dz dy dx = xy dV = x2 0 0 3 . 28 2. Use a triple integral to find the volume of the solid bounded by the cylinder y = x2 and the planes z = 0, z = 4, and y = 9. Solution. Z 3 Z 9 4 Z dz dy dz = 144. V = x2 −3 0 RRR z 3. Use cylindrical coordinates to evaluate e dV , where E is enclosed by the E paraboloid z = 1 + x2 + y 2 , the cylinder x2 + y 2 = 5, and the xy-plane. √ Solution. Bounded by z = 1 + r2 and r = 5, cylindrically symmetric. ZZZ Z 2π Z √5 Z 1+r2 ez r dz dr dθ = π(e6 − e − 5). ez dV = E 0 0 0 4. Evaluate the integral ZZ ex+y dA, R where R is given by the inequality |x| + |y| ≤ 1. Solution. Set u = x + y and v = x − y, inspired by drawing the region and writing the equations for each of the lines forming the four parts of the boundary. Then we have x = (u + v)/2, y = (u − v)/2, and 1 ∂(x, y) 12 21 = 1 1 = − , −2 ∂(u, v) 2 2 so ZZ e R x+y Z 1 dA = −1 1 1 e du dv = e − e−1 . 2 −1 Z 1 u 5. Evaluate the integral Z 2 Z √4−y2 Z − −2 √ 4−y 2 2 √ xz dz dx dy, x2 +y 2 by changing to cylindrical coordinates. Solution. Z 2 −2 Z √4−y2 Z − √ 4−y 2 2 √ Z 2π Z 2 Z 2 xz dz dx dy = x2 +y 2 0 0 r2 cos θ dz dr dθ = 0. r Can also see answer immediately by a symmetry argument. 6. Let f be continuous on [0, 1] and let R be the triangular region with vertices (0, 0), (1, 0), and (0, 1). Show that Z 1 ZZ uf (u) du. f (x + y) dA = 0 R Solution. Set u = x + y and v = y. Then ∂(x, y) 1 1 = ∂(u, v) 0 1 and so ZZ Z 1 Z f (x + y) dA = R = 1, u Z f (u) dv du = 0 0 2 1 uf (u) du. 0