# Math Help - Volume for a cone in cylindrical coordinates.

1. ## Volume for a cone in cylindrical coordinates.

Hi there. I haven't used iterated integrals for a while, and I'm studying some mechanics, the inertia tensor, etc. so I need to use some calculus. And I'm having some trouble with it.

I was trying to find the volume of a cone, and then I've found lots of trouble with such a simple problem.

So I thought of using cylindrical coordinates this way:
$\begin{Bmatrix}{ x=r\cos\theta} \\y=r\sin\theta \\z=r\end{matrix}$

And then I've stated the integral this way:

$\displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^ {r}\displaystyle\int_{r}^{h}rdzdrd\theta=\displays tyle\int_{0}^{2\pi}\displaystyle\int_{0}^{r}r(h-r)drd\theta=\displaystyle\int_{0}^{2\pi}\displayst yle\frac{r^2h}{2}-\displaystyle\frac{r^3}{3}=\pi r^2h-\displaystyle\frac{2\pi\r^3}{3}=\pi r^2(h-\displaystyle\frac{2}{3}r)$

But I should get: $V_{cone}=\displaystyle\frac{\pi r^2 h}{3}$

I think I'm giving wrong limits for the integration.

Help pls

2. I think your limits are ok. You have a notational error here:

$\displaystyle\int_{0}^{2\pi}\left(\frac{r^{2}h}{2}-\frac{r^{3}}{3}\right)d\theta=\pi r^{2}h-\frac{2\pi r^{3}}{3}.$

But as that gets fixed later on, that won't explain what's going on. I think part of the issue is that you've determined that $z=r.$ There's no a priori reason to assume that. I would probably have done $z=hr/a,$ where $a$ is the radius of the base, to account for differently "squashed" cones. Be that as it may, I think your formula is actually correct, because in your case, since $z=r,$ you have that $r=h,$ and hence

$\displaystyle \pi r^{2}h-\frac{2\pi r^{3}}{3}=\frac{\pi r^{3}}{3}.$

This formula is correct if $r=h.$

Try going through the integration again with $z=hr/a,$ and see what that gets you.

Something else I would try: don't use $r$ both for the variable of integration and for the radius of the base of the cone. It's confusing.

3. Thanks for your answer Ackbeet. I thought of z=r because I thought of the cone as $z=(x^2+y^2)^{1/2}$, thats the cone I have integrated, and as you said, that means that r=h, because the cone in the xz plane goes through the line x=z. The relation I was missing I think that its the one that you gave.

Thank you again.

4. You're welcome. Have a good one!