Question

Find volume of solid bounded by x^2+y^2+z^2=16 and below by x^2+y^2=4

Answer 1

∫2π0∫6cosθ0∫r0rdzdrdθ

Since the xy plane was a bound, I assumed you needed the top part of the cone so z=x2+y2−−−−−−√ or z=r for the top z bound.

Since x2+y2=6x, polar conversion equations give r2=6rcosθ or r2−6rcosθ=0 so r=0 and r=6cosθ from that.

x2+y2=6x is a full circle so I let θ=0 to θ=2π bounds. There is a shift in the circle but I don't believe it affects the θ bounds.