Find the volume of the solid that lies within the sphere $𝑥^2+𝑦^2+𝑧^2=81$, above the 𝑥𝑦 plane, and outside the cone $𝑧=5\sqrt{𝑥^2+𝑦^2})$.
My difficulties are that having it be outside the cone, you must also have a way to find the volume of the tip of the sphere over the cone. I tried to represent the tip by looking at the xz plane shown in the picture below. I see no way to express this spherically or cylindrically. My best bet is to calculate the hemisphere volume by $\frac43\pi r^3$ and then subtract the cone volume from that which is found rather easily cylindrically.
Thank you.
Best Answer
There are three ways to do this with one integral directly (as opposed to an indirect symmetry argument, but I don't recommend those because it is fair question to ask to set up and do the integrals over the same region but with an integrand that does not share that symmetry).
Cylindrical ($dr$ first):
$$\int_0^{2\pi} \int_0^{\frac{45}{\sqrt{26}}} \int_{\frac{z}{5}}^{\sqrt{81-z^2}} rdrdzd\theta$$
Spherical ($d\rho$ first):
$$\int_0^{2\pi} \int_{\cot^{-1}(5)}^{\frac{\pi}{2}} \int_0^9 \rho^2\sin\phi d\rho d\phi d\theta$$
Spherical ($d\phi$ first):
$$\int_0^{2\pi} \int_0^9 \int_{\cot^{-1}(5)}^{\frac{\pi}{2}} \rho^2\sin\phi d\phi d\rho d\theta$$
which follows from Fubini's theorem in hindsight, but it is important to consider the last integration order since most people don't when it could in certain situations make life easier.