Math 293 Answers to Practice Prelim #1Spring 2000xyy = 4 - x2421.R20R4-x20xe2y4-ydy dx.This is awkward to in-tegrate in the given order, so switch the order togetR40R√4-y0xe2y4-ydx dy. Then after the first inte-gration we haveR40e2y2dy, which works out to bee84-14.2.R10R√1-x20e-(x2+y2)dy dx. The region of integration is the portion ofthe unit disk lying in the first quadrant, so in polar coordinates this integralbecomesRπ/20R10e-r2r dr dθ, which computes out to be(-e-1+12)(π2).3. In spherical coordinates the cone is given by the equationφ=π4, so theregion below the cone and above thexy-plane will haveπ4≤φ≤π2. Theregion between the two spheres will have 3≤ρ≤4. Also,x2+y2=ρ2-z2.So we get the integralR2π0Rπ/2π/4R43(ρ2-z2)ρ2sinφ dρ dφ dθ.Substitutingz=ρcosφ, this becomesR2π0Rπ/2π/4R43(ρ2-ρ2cos2φ)ρ2sinφ dρ dφ dθwhichworks out to be(2π)(45-355)(1√2-13·2√2).4. The vector fieldF(x, y) = (y+ex
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