Volume for a cone in cylindrical coordinates.

[Telemachus]

Homework Statement

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=\displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^{r}r(h-r)drd\theta=\displaystyle\int_{0}^{2\pi}\displaystyle\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 :)

 

[tiny-tim]

Hi Telemachus!

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

No, your integration is fine.

(I'd have used spherical coordinates, but your way does work)

But you're doing it for the 45° cone instead of a general cone (so r = h, which makes your formula the same as the given answer).

Somehow your variable of integration r has managed to survive into the afterlife under a new persona.

 

[Telemachus]

Thanks Tim :)

Haha sorry for the notation, I should used another name for the variable :P