Math 251-copyright Joe Kahlig, 15C Page 11. (a) True.(b) False. This would be true for ex+y.(c) False. To be true the inside integral needs to start at z = 0 not z = 1.(d) True(e) True. When using the values of θ, the only requirement is that the starting value andending value can not have a difference of more than 2π.2. (a) x2+ y2= 6y this is a circle.(b) ρ = 4 sec(φ)ρ cos φ = 4z = 4 this is a plane3. Note: Since f (x, y) is not a constant function, the volume of each of the 5 solids formed withthe base being the leaf and the height being f (x, y) will not be equal.Note 2: you can not integrate from 0 to 2π since on parts of this interval the radius will benegative.I have set up the integral for the leaf that is centered on the positive x-axis.Zπ/10−π/10Zcos(5θ)05r3cos2θ drdθ4. Change the limits of the integral to get:Z30Z2y0ey2dxdy.integrating this gives e9− 15. The correct region has been shaded in the graph.x=1y=xx+y=8Z41Z8−xxp32+ 52+ 1 dydx = ... = 9√356. The surface is given by z =p25 − y2so we get zx= 0 and zy=−yp25 − y2Z30Z√9−x20sy225 − y2+ 1 dydx =Z30Z√9−x20r2525 − y2dydxorZπ/20Z30rr2525 − r2sin2θdrdθMath 251-copyright Joe Kahlig, 15C Page 27. First find the intersection of the sphere and the paraboloid.The paraboloid gives z+5 = x2+y2substituting this into the sphere equation gives z+5+z2= 25.Solving gives z = 4 and z = −5. Since this is the top of the sphere z = 4 is what we want.The intersection is x2+ y2+ (4)2= 25 or x2+ y2= 9Also note that y ≥ 0 means that 0 ≤ θ ≤ π.Zπ0Z30Z√25−r2r2−5r dzdrdθ8.Z40Z16−4y0Z(16−2y−z)/2yf(x, y, z) dxdzdy9. (a)Zπ0Z30Z√18−r2rzr2cos θ dzdrdθ(b)Zπ0Zπ/40Z√180ρ4sin2φ cos φ cos θ dρdφdθ10.Z2π0Z63Zr sin θ+401r2sin θ dzdrdθ11.Z√3−√3Z15−y24y2k((x − 20)2+ y2)
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