Math 251-copyright Joe Kahlig, 15C Page 11. (a) False. This would be true for ex+y.(b) True.(c) True(d) 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.(e) False. To be true the inside integral needs to start at z = 0 not z = 1.2. (a) x2+ y2= 4y this is a circle.(b) ρ = 2 sec(φ)ρ cos φ = 2z = 2 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)03r3sin2θ drdθ4. Change the limits of the integral to get:Z20Z2y0ey2dxdy.integrating this gives e4− 15. The correct region has been shaded in the graph.x=3y=xx+y=10Z53Z10−xxp42+ 22+ 1 dydx = ... = 4√216. The surface is given by z =p9 − y2so we get zx= 0 and zy=−yp9 − y2Z20Z√4−x20sy29 − y2+ 1 dydx =Z20Z√4−x20r99 − y2dydxorZπ/20Z20rr99 − r2sin2θdrdθMath 251-copyright Joe Kahlig, 15C Page 27. Step 1 find the intersection of the sphere and the paraboloid.The paraboloid gives z+4 = x2+y2substituting this into the sphere equation gives z+4+z2= 6.Solving gives z = 1 and z = −2. Since this is the top of the sphere z = 1 is what we want.The intersection is x2+ y2+ (1)2= 6 or x2+ y2= 5Also note that y ≥ 0 means that 0 ≤ θ ≤ π.Zπ0Z√50Z√6−r2r2−4r dzdrdθ8.Z30Z12−4y0Z(12−2y−z)/2yf(x, y, z) dxdzdy9. (a)Zπ/2−π/2Z50Z√50−r2rzr2sin θ dzdrdθ(b)Zπ/2−π/2Zπ/40Z√500ρ4sin2φ cos φ sin θ dρdφdθ10.Z2π0Z63Zr sin θ+401r2cos θ dzdrdθ11.Z2−2Z8−y2y2k((x − 14)2+ y2)
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