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.
Best Answer
Why not to use simple geometry, In general
The volume of the sphere, having a radius $R$, thoroughly drilled along its diameter by a cylinder having a radius $r$, is $$=\text{(volume of the original sphere)}-\text{(volume of of cylindrical hole with ends as spherical caps)}$$ $$V_{\text{drilled sphere}}=\frac{4\pi}{3}R^3-\frac{4\pi}{3}(R^3-(R^2-r^2)^{3/2})$$ $$\bbox[4pt, border:1px solid blue;]{\color{blue}{V_{\text{drilled sphere}}=\frac{4\pi}{3}(R^2-r^2)^{3/2}}} $$ where $\color{red}{0\leq r<R}$.
Now, substituting the given values, radius of the sphere, $R=7$ & the radius of the cylindrical drill $r=5$, we get the volume of ring shaped solid (i.e. drilled sphere) $$V_{\text{drilled sphere}}=\frac{4\pi}{3}(7^2-5^2)^{3/2}$$ $$=\frac{4\pi}{3}(24)^{3/2}$$ $$=\frac{4\pi}{3}(8)(6)^{3/2}$$ $$=\color{blue}{\frac{32\pi 6^{3/2}}{3}\approx 492.4991348\ldots \text{ cubic units}}$$