I want to evaluate the lost volume of a sphere of radius $R$ after the cylinder of radius $r$ is punctured through it's center using integration. What the the value of $r$ such that the sphere maintains exactly the half of it's volume?
Assume that we don't know any formula from the Napkin ring problem - how to evaluate the lost volume of a sphere using integration?
On
Another approach in spherical coordinates:
The solid is defined by $$ E=\{(\rho,\theta,\phi)\;|\; 0\le \theta \le 2 \pi, \phi_1 \le \phi \le \phi_2, \frac{r}{\sin \phi} \le \rho \le R \} $$ where $\phi_1$ and $\phi_2$ are the angles where the sphere $\rho = R$ and the cylinder $\rho =\frac{r}{\sin \phi}$ intersect, namely $$ \phi_1 = \arcsin \frac{r}{R}\quad \mbox{and}\quad \phi_2 = \pi-\arcsin \frac{r}{R} $$ It follows that $$ V(r)=\int_0^{2\pi}\int_{\phi_1}^{\phi_2}\int_{\frac{r}{\sin \phi}}^{R}\rho^2\sin\phi\; d\rho d\phi d \theta = \frac{4\pi}{3}(R^2-r^2)^{3/2} $$
Assume that boring is parallel to the $z$-axis. The napkin ring is a rotational body whose volume $V$ can be computed using the "shell method". The shells have height $2\sqrt{R^2-x^2}$, radius $x$ (hence circumference $2\pi x$), and thickness $dx$. Therefore $$\eqalign{V&=\int_r^R2\sqrt{R^2-x^2}\>2\pi x\>dx=2\pi\int_{r^2}^{R^2}\sqrt{R^2-u}\>du\cr &=-{4\pi\over3}\bigl(R^2-u\bigr)^{3/2}\Biggr|_{u=r^2}^{u=R^2}={4\pi\over3}(R^2-r^2)^{3/2}\ .\cr}$$ I leave it to you to compute the $r$ that produces $V={1\over2}V_{\rm sphere}$.