TITLE: Integrating volume of a sphere with a cylinder "drilled" out of it
QUESTION [6 upvotes]: Unfortunately, I am stuck again on another integration problem.
Famous last words, this should be simple.
$$
\text{A cylindrical drill with radius 5 is used to bore a hole through}\\\text{the center of a sphere of radius 7. Find the volume of the ring shaped solid that remains.}
$$
So we can setup our problem by first defining our change in $\theta$ as the first region, because we can do some simple multiplication to fill the rest of the sphere symmetrically. We should just be able to calculate the change in $r$ inside of that.
$$
\begin{align}
&=8 \int_{0}^{\frac{\pi}{4}} \int_{5}^{7} r\:dr\:d\theta\\
&=\frac{8}{2}\int_{0}^{\frac{\pi}{4}}49-25\:d\theta\\
&=24 \times \frac{8}{2} \times \frac{\pi}{4}\\
&=24\pi
\end{align}
$$
Edit: I attempted to re-evaluate my process, but the problem was still not correct.
I attempted to set my integrand to the arc length of the sphere - the arc length of the cylinder, but the integrand $\frac{r^3}{2}-5r$ was not the correct integrand to use.

REPLY [2 votes]: This is the napkin ring problem.
This illustration hints at using Cavalieri's principle on it.
This section gives a complete solution.