# Answers to Even Recommended problems from Ch. 12 Review

## Mathematics 267 with Ni at Arizona State University - Tempe *

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- Answers to Even Recommended problems from Ch. 12 Review

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Calculus: Early Transcendentals (Stewart's Calculus Series)
ANSWERS TO EVEN ASSIGNED PROBLEMS CHAPTER 12 REVIEW (pages 723-724) 26. x y z Using the projection on the yz plane we can evaluate the integral as follows: integraldisplayintegraldisplayintegraldisplay E zdV = integraldisplay 1 0 integraldisplay √1−y2 0 integraldisplay 2−y 0 z dx dz dy = integraldisplay 1 0 integraldisplay √1−y2 0 (2−y)z dz dy = integraldisplay 1 0 1 2(2−y)(1−y 2)dy = integraldisplay 1 0 1 2(2−y−2y 2 + y3)dy = 13 14 28. Using spherical coordinates we have that 0 ≤ ρ ≤ 1(since the solid is bounded by the sphere of radius 1), 0 ≤ φ ≤ pi2 since the solid lies above the xy plane) and 0 ≤ θ ≤ 2pi. Thus integraldisplayintegraldisplayintegraldisplay H z3 radicalbig x2 + y2 + z2 dV = integraldisplay 2pi 0 integraldisplay pi/2 0 integraldisplay 1 0 (ρ3 cos3 φ)ρ(ρ2 sinφ) dρ dφ dθ = integraldisplay 2pi 0 dθ integraldisplay pi/2 0 cos3 φsinφ dφ integraldisplay 1 0 ρ6dρ = 2pi bracketleftbigg −14 cos4 φ bracketrightbiggpi/2 0 parenleftbigg1 7 parenrightbigg = pi14 32. x y z The xy projection of the solid is the disk of radius 2 centered at the origin. In cylindrical coordinates the plane y + z = 3 has equation z = 3−rcosθ. Thus the volume is given by V = integraldisplayintegraldisplayintegraldisplay E dV = integraldisplay 2pi 0 integraldisplay 2 0 integraldisplay 3−rsinθ 0 r dz dr dθ = integraldisplay 2pi 0 integraldisplay 2 0 (3r−r2 sinθ) dr dθ = integraldisplay 2pi 0 [6− 83 sinθ] dθ = 6θ]2pi0 + 0 = 12pi 34. x y z The paraboloid and the half-cone intersect when x2 + y2 =radicalbig x2 + y2, that is when x2 + y2 = 1. Thus the projection of the solid on the xy plane is a circle of radius 1. In cylindrical coordi- nates, the equation of the paraboloid is z = r2 and the equation of the cone is z = r and the volume is given by V = integraldisplayintegraldisplayintegraldisplay E dV = integraldisplay 2pi 0 integraldisplay 1 0 integraldisplay r r2 r dz dr dθ = integraldisplay 2pi 0 integraldisplay 1 0 (r2 −r3) dr dθ = integraldisplay 2pi 0 (13 − 14) dθ = 112(2pi) = pi6 42. (a) The surface is a vertical plane at an angle of pi4 with the positive x-axis. In cartesian coordinates, the plane has equation y = x. (b) The surface is a cone at an angle of pi4 radians with the positive z-axis. In cartesian coordinates the cone has equation z = radicalbigx2 + y2. 1 answers_ch12_review.dvi

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