Math 234, Practice Test #3Show your work in all the problems.1. Find the volume of the region bounded above by the paraboloid z = 9−x2−y2, below by the xy-plane and lying outside the cylinder x2+y2= 1.2. Evaluate the integral by changing t o polar coordinatesZ1−1Z√1−y2−√1−y2ln(x2+ y2+ 1) dx dy3. Describe the region of integration. Convert the integral to sphericalcoordinates and evaluate itZ+1−1Z√1−x2−√1−x2Z1√x2+y2dz dy dx4. Sketch the region of integration, and write an integral with the orderof integration reversed. Do not evaluate the integral.Z40Z(y−4)/2−√4−ydx dy5. Find the centroid of the triangular regio n cut from the first quadrantby the line x + y = 3.1Solutions1. The most convenient coordinates for this problem are cylindrical coordi-nates x = r cos θ, y = r sin θ. We have to figure out when the paraboloidintersects t he xy-plane. This is the case for 0 = z = 9 − x2− y2, i.e.x2+ y2= 9 or r = 3. The integral is then given byZ2π0Z31Z9−r20r dz dr dθ =Z2π0Z31(9 − r2)r dr dθ=Z2π09r22−r4431dθ= 32π2. The equations x = ±√1 − y2are equivalent to x2+ y2= 1 which de-scribes a circle with radius 1 centered at the origin. In polar coor dina tesx = r cos θ, y = r sin θ this is given by r = 1. HenceZ1−1Z√1−y2−√1−y2ln(x2+ y2+ 1) dx dy =Z2π0Z10ln(r2+ 1) r dr d θ(substitute u = r2+ 1, du = 2rdr)=12Z2π0Z21ln(u) d u=12Z2π0(u ln(u) − u)21dθ= π ( 2 ln(2) − 1)Recall from Calculus II that the integral of the logarithm function iscalculated as follows:Zln(x)dx =Z1 · ln (x)dx(integration by parts d u = 1 dx, v = ln(x),dv = dx/x, u = x)= x ln(x) −Zdx= x ln(x) − x + C3. The equation z =√x2+ y2describes a cone in upper half space z ≥ 0with tip at the origin and opening angle of π/4. The region of integra-tion is bounded below by the cone and a bove by the horizontal plane2z = 1. In spherical coordinates the plane z = 1 corresponds to1 = z = ρ cos φ i.e. ρ = sec φ.The converted integral is thenZ2π0Zπ/40Zsec φ0ρ2sin φ dρ dφ dθ =13Z2π0Zπ/40sec3φ sin φ dφ dθ=13Z2π0Zπ/40sec φ(sec φ tan φ)dφ dθ=2π3sec2φ2π/40=2π31 −12=π34. The regio n of integration is the region enclosed by the parabola y =4−x2and the line y = 4 +2x. Reversing the order of int egr ation yieldsZ0−2Z4−x24+2xdy dx =Z0−2(−x2− 2x)dx= − x33+ x2!0−2=435. The area of the triangle equals 9/2. We compute the first momentsMx=Z30Z3−x0y dy dx=12Z30(3 − x)2dx=12Z30(9 − 6x + x2)dx=12 9x − 3x2+x33!30=923andMy=Z30Z3−x0x dy dx=Z30(3x − x2) dx= 3x22−x33!30=92The center of mass is then given by(¯x, ¯y) = (My/M, Mx/M) = (1,
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