Question

Express the volume of the solid that the cylinder r = 4cos(theta) cuts out of the sphere of radius 4 centered at the origin with a triple integral in cylindrical coordinates. I have already found the intervals, but I cannot solve it. The bounds are -sqrt(16-r^2) < z < sqrt(16-r^2), 0 < r < 4cos(theta), -pi/2 < theta < pi/2. Thank you.

Answer 1

Using symmetry, you can just take 4 times the volume in the 1st octant:

4∫[0,π/2]∫[0,4cosθ]∫[0,√(16-r^2)] r dz dr dθ
= 4∫[0,π/2]∫[0,4cosθ] r√(16-r^2) dr dθ
= -4/3∫[0,π/2] (16-r^2)^(3/2) dθ [0,4cosθ]
= -4/3∫[0,π/2] (64sin^3θ)-(64) dθ
= -256/3∫[0,π/2] sin^3θ-1 dθ
= 256/3 ∫[0,π/2] 1- sinθ(1-cos^2θ) dθ
= 256/3 (θ + cosθ - 1/3 cos^3θ) [0,π/2]
= 256/9 (3θ + 3cosθ - cos^3θ) [0,π/2]
= 256/9 ((3π/2)-(3-1))
= 128/9 (3π-4)