Find the volume laying inside x^2 + y^2 + z^2 =2z and inside z^2 = x^2 + y^2. This is a problem my professor made, so I have no way of checking my answer. What I did first was completed the square for the sphere and got x^2 + y^2 + (z-1)^2 = 1, which is a sphere of radius one shifted above the z axis by one. and z^2 = x^2 + y^2 is two cones, but I am only worried about the one above the z axis. I changed x^2 + y^2 + (z-1)^2 = 1 into spherical coordinates: (ρ^2)sin^2(Φ) + (ρ^2)cos^2(Φ) - 2ρcosΦ = 0 and got ρ = 2cosΦ. So the limits for θ are from 0 to 2pi, ρ is from 0 to 2cosΦ, and since I am going from the top of the sphere to the bottom of the cone above the z axis I think my limits for Φ are from 0 to pi/2. So my integral is ∫(0,2pi)∫(0,pi/2)∫(0,2cosΦ) (ρ^2)sinΦ dρdΦdθ which is a pretty simple integral. Does this look right? I'm not 100% sure on my ρ and Φ limits.