Volume of sphere with hole

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August 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 $\frac {4}{3}\pi r^{3}$ . I then worked out the volume of the hole (cylinder):
$<br /> \int_{-r}^{r} \frac{1}{4}\pi d^{2}dy$
$= \frac{1}{2}\pi d^{2} r<br />$

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?
August 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?
August 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!
August 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 $dx$ , circumference $2 \pi x$ , and height $2 \sqrt {R^2 - x^2}$ , and integrate from x = d/2 to x = R.
August 31st 2010, 06:28 AM
SyNtHeSiS

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 $\frac{1}{6}\pi(4r^{2} -d^{2)^\frac{3}{2}$ ?
August 31st 2010, 06:37 AM
Ackbeet

They are the same answer. Proof:

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

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

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

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

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