# Volume of sphere with hole

• Aug 30th 2010, 09:18 AM
SyNtHeSiS
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 $\displaystyle \frac {4}{3}\pi r^{3}$. I then worked out the volume of the hole (cylinder):
$\displaystyle \int_{-r}^{r} \frac{1}{4}\pi d^{2}dy$
$\displaystyle = \frac{1}{2}\pi d^{2} r$

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

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

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?
• Aug 30th 2010, 09:23 AM
Ackbeet
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?
• Aug 30th 2010, 01:29 PM
Topher
Very intresting problem. It sounds like Ackbeet is right in the fact that the figure removed is not
Spoiler:
only
a cylinder.

(Itwasntme)

Good Luck!
• Aug 30th 2010, 01:46 PM
ebaines
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 $\displaystyle dx$, circumference $\displaystyle 2 \pi x$, and height $\displaystyle 2 \sqrt {R^2 - x^2}$, and integrate from x = d/2 to x = R.
• Aug 31st 2010, 06:28 AM
SyNtHeSiS
I got the integral $\displaystyle V=-2\pi\int_{r^{2}-\frac{d^{2}}{4}}^{0} \sqrt{u} du$ after making the substitution of $\displaystyle u=r^{2} - x^{2}$ then my final answer was:

$\displaystyle \frac{4}{3}\pi(r^{2}-\frac{d^{2}}{4})^{\frac{3}{2}}$. Did I do something wrong or is there any way to simplify it to $\displaystyle \frac{1}{6}\pi(4r^{2} -d^{2)^\frac{3}{2}$ ?
• Aug 31st 2010, 06:37 AM
Ackbeet
They are the same answer. Proof:

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

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

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

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

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