Volume above cone $z = a\sqrt{x^2+y^2}$ and inside sphere $x^2+y^2+z^2=b^2$

Question

Find the volume of the region: $$\iiint(x^2+y^2+z^2)dV $$ where $R$ is the region above the cone $z = a\sqrt{x^2+y^2}$ and inside the sphere $x^2+y^2+z^2=b^2$.

Answer 1

Just find the intersection point. Write $x=\rho\sin\phi\cos\theta,$ $y=\rho \sin\phi \sin\theta$, and $z=\rho \cos \phi$. At the place of intersection (see a picture), $\rho= b$. Moreover, by symmetry, we can find the angle $\phi$ for any $\theta$.

We have $z^2=a^2(x^2+y^2)$ so plugging this into the second one, we have \begin{align} x^2+y^2+a^2(x^2+y^2)=b^2\implies x^2+y^2=\frac{b^2}{a^2+1}. \end{align} On the cone, this means $z=a\sqrt{x^2+y^2}=\frac{ab}{\sqrt{a^2+1}} = b\cos\phi$. Therefore, $$ \phi=\cos^{-1}(\frac{a}{\sqrt{a^2+1}}):=\alpha. $$ Using $dV=\rho^2\sin\phi\,d\rho d\phi d\theta$the setup in spherical coordinates is then $$ \int_0^{2\pi}\int_0^\alpha\int_0^b \rho^4\sin\phi\, d\rho d\phi d\theta $$ The angle $\alpha$ looks ugly, but when you plug it in, it ends up inverting with the cosine function.

Answer 2

In cylindrical coordinates, the bottom and top surfaces are $z=ar$ and $z=\sqrt{b^2-r^2}$, respectively. Then, the enclosed volume is

$$\int_0^{2\pi}\int_0^R \left(\sqrt{b^2-r^2}-ar\right) rdrd\theta = \frac{2\pi b^3}3\left(1-\frac a{\sqrt{1+a^2}} \right)$$

where the radial upper limit is the intersection of the two curves, i.e. $R=\frac{b}{\sqrt{1+a^2}}$.