# Thread: Volume of sphere with hole

1. ## Volume of sphere with hole

A hole of diameter d is drilled through a sphere of radius r in such a way that the axis of the hole passes through the centre of the sphere. Find the volume of the solid that remains.

I drew a picture of a sphere centred on the cartesian plane, with a hole going through its centre (in the shape of a cylinder).

I already knew that the volume of the sphere was $\frac {4}{3}\pi r^{3}$. I then worked out the volume of the hole (cylinder):
$
\int_{-r}^{r} \frac{1}{4}\pi d^{2}dy$

$= \frac{1}{2}\pi d^{2} r
$

Volume of solid remaining = $\frac {4}{3}\pi r^{3} - \frac{1}{2}\pi d^{2} r$

but the correct answer was $\frac{1}{6}\pi (4r^{2} - d^{2})^{\frac{3}{2}}$

I cant see where I went wrong cause surely the objective of this problem is to minus the hole (cylinder) from a sphere to get the remaining volume?

2. The volume subtracted out by drilling the hole is not precisely a cylinder, because a cylinder usually has flat ends. The material removed has curved ends (matches up with the sphere). So what do you get when you change this?

3. Very intresting problem. It sounds like Ackbeet is right in the fact that the figure removed is not
Spoiler:
only
a cylinder.

Good Luck!

4. Rather than subtracting the round-ended cylinder from the volume of the sphere, I think it's probably easier to set up the integral for the volume. You can use cylindrical shell elements of width $dx$, circumference $2 \pi x$, and height $2 \sqrt {R^2 - x^2}$, and integrate from x = d/2 to x = R.

5. I got the integral $V=-2\pi\int_{r^{2}-\frac{d^{2}}{4}}^{0} \sqrt{u} du$ after making the substitution of $u=r^{2} - x^{2}$ then my final answer was:

$\frac{4}{3}\pi(r^{2}-\frac{d^{2}}{4})^{\frac{3}{2}}$. Did I do something wrong or is there any way to simplify it to $\frac{1}{6}\pi(4r^{2} -d^{2)^\frac{3}{2}$ ?

6. They are the same answer. Proof:

$\displaystyle{\frac{4}{3}\pi(r^{2}-\frac{d^{2}}{4})^{\frac{3}{2}}}$

$\displaystyle{=\frac{4}{3}\pi(4r^{2}-d^{2})^{\frac{3}{2}}\left(\frac{1}{4}\right)^{\fra c{3}{2}}}$

$\displaystyle{=\frac{4}{3}\pi(4r^{2}-d^{2})^{\frac{3}{2}}\,\frac{1}{4^{3/2}}}$

$\displaystyle{=\frac{1}{8}\,\frac{4}{3}\pi(4r^{2}-d^{2})^{\frac{3}{2}}}$

$\displaystyle{=\frac{1}{6}\pi(4r^{2}-d^{2})^{\frac{3}{2}}.}$