Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We're learning about triple integrals and such in class.

Here's one of the problems I'm working on:

A cylindrical drill with radius 3 is used to bore a hole through the center of a sphere of radius 5. Find the volume of the ring shaped solid that remains.

Now here's what I'm thinking:

Triple integrate this: $r \,dr\, d\theta\, dz$

Bounds for $r$: $3$ to $5$

Bounds for $\theta$: $0$ to $2\pi$

Bounds for $z$ : ...$0$ to $5$? (times $2$? I mean, it's $-5$ to $5$...)

However, maybe the bounds for $r$ should depend on $z$. That would make sense... right?

I wish I knew when to have the bounds depend on another variable and such. I think this one would though, because the "radius" would shrink as you increased (or decreased) z. By how much though? When $z$ is $0$, $r$ is $5$. When $z$ is $1$, $r$ is $5$, so the new distance from the $z$-axis would be $\sqrt{1 + 5^2}$. Right...?

I'd sure appreciate some pointers! I'll respond as quickly as possible.

share|cite|improve this question
What you have is the surface of revolution of a circular segment, no? – J. M. Nov 18 '11 at 3:18
$3^2 + 4^2 = 5^2$ – The Chaz 2.0 Nov 18 '11 at 3:20
Whoops! \sqrt{5^2 - 1^2} And thanks for the link, I'm digesting the material now. And thanks for the edit Zev! – user13327 Nov 18 '11 at 3:24

1 Answer 1

up vote 1 down vote accepted

Your thinking looks quite right to me.

Cylindrical coordinates seem like the way to go, so that's good. A nice check is to try to figure out what region your bounds define; in this case, if $r$ goes from 3 to 5, $\theta$ from $0$ to $2\pi$, and $z$ from $0$ to $5$, we see that we're defining a cylinder with a hole bored through it instead of the sphere with a hole that we want.

We want to integrate over the whole sphere, so certainly $\theta$ goes from $0$ to $2\pi$. However, as you noted, one of $r$ or $z$ must depend on the other (and the other should have the same bounds you already had, since the first will determine the sphere). It seems like it would be a little hard to put $r$ in terms of $z$, since $r$ has weirder bounds. Also consider the fact that the equation of a sphere (and so the outer limit of integration) is $z^2 + r^2 = 5^2$. You can use this to find the bounds for $z$ (and it's an excellent idea to integrate the top half and just multiply by two).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.