Find the volume of the solid that remains after a circular hole of radius a is bored through the center of a solid sphere of radius r > a.

Question

Find the volume of the solid that remains after a circular hole of radius a is bored through the center of a solid sphere of radius r > a. So in the picture it looks like a circle with a cylinder cut out of the middle. I am not even sure where to start with this. I know this has to do with integrals but I am not sure how to set this up to even get an integral. Any help would be greatly appreciated.

Answer 1

We can decompose the sphere as as a stack of discs with a hole in the middle: $$ dV(z) = A(z)\, dz $$ where the height $z$ over a full sphere with center at the origin of the coordinate system would range from $-r$ to $r$.

The radius $R(z)$ of a disc at height $z$ is $$ R(z) = \sqrt{r^2 - z^2} $$ The area of a solid disc is $$ A(z) = \pi R(z)^2 $$ From this we subtract the citcular area of the hole and get: $$ V = \int\limits_{-h}^h \pi \left(\left(r^2 - z^2\right) - a^2\right) dz $$ The above equation features a height parameter $h \le r$. We should choose it such that the disc is not smaller than its hole. $$ 0 = r^2 - h^2 - a^2 \iff h = \sqrt{r^2 - a^2} $$ This gives $$ V = \pi\int\limits_{-\sqrt{r^2-a^2}}^{\sqrt{r^2-a^2}} \left( r^2-a^2 - z^2 \right) dz $$ This 1D integral should be easy to solve. For $a=0$ the volume of a sphere with radius $r$ should result.

Answer 2

Another way to set this up is to take the region under the semicircle $y=\sqrt{r^2-x^2}$ for $a\le x\le r$ and revolve this around the y-axis to get the top half of the solid.

Using the shell method (and symmetry),

this gives $\displaystyle V=2\int_a^r2\pi x\sqrt{r^2-x^2}\;dx$.