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Math 522 Homework 9 Solutions 1. To solve this integral we will use a change of variables to n-dimensional spherical coordinates and consider the limit as the “radius” variable goes to infinity. The variables in our new coordinate system will be r, φ1 , φ2 , . . . , φn−1 where 1 ≤ r < ∞, φ1 , φ2 , . . . , φn−2 ∈ [0, π], and φn−1 ∈ [0, 2π). We introduce the new variables by defining f by the following. f : x1 = r cos(φ1 ) f : x2 = r sin(φ1 ) cos(φ2 ) f : x3 = r sin(φ1 ) sin(φ2 ) cos(φ3 ) .. . f : xn−1 = r sin(φ1 ) sin(φ2 ) · · · sin(φn−2 ) cos(φn−1 ) f : xn = r sin(φ1 ) sin(φ2 ) · · · sin(φn−1 ). We then see that |x| = r and det ∇f = rn−1 sinn−2 (φ1 ) sinn−3 sin(φ2 ) · · · sin(φn−2 ) so Z F 1 dx = lim t→∞ |x|k Z 2π Z π Z π Z π Z t ··· φn−1 =0 φn−2 =0 Z t n−1−k = cn lim t→∞ r φn−3 =0 φ1 =0 r=1 1 | det ∇f |drdφ1 · · · dφn−1 . rk dr 1 n−k cn t 1 − t→∞ n − k n−k = lim where cn is a constant. This converges when n − k < 0 so the given integral is finite when k > n. Page 368 6. Graph of F in the x1 , x2 -plane: Graph of F in the u1 , u2 -plane: Solving for each variable we find that x21 q + = u21 + u22 , s q 1 2 2 x1 = u1 + u2 + u1 , and 2 s q 1 2 2 u1 + u2 − u1 . x2 = 2 x22 Now f −1 : u1 = x21 − x22 , u2 = 2x1 x2 so det ∇f −1 = 4x21 + 4x22 . Therefore det ∇f = 4x21 1 . + 4x22 From the change of variables we have Z x21 + x22 dV2 (x) Z 4 Z 4 Z 2 = 2 F Z 1 2 = 2 1 p u2 + u22 p1 du1 du2 4 u21 + u22 1 du1 du2 4 1 = 2 9. The given inequalities tell us that ρ ∈ [0, 1], ϕ ∈ [0, 2π) and θ ∈ [0, π4 ]. Finding the Jacobian of the given change of variables we get ρ2 sin θ. Thus, Z Z 2π Z π/4 Z x3 dV3 (x) = 0 F π 2 π = 8 Z = 0 π/4 1 ρ3 sin θ cos θ dρ dθ dϕ 0 sin θ cos θ dθ 0