# Thread: Using cylindrical coordinates to solve triple integrals

1. ## Using cylindrical coordinates to solve triple integrals

Hi,

I'm trying to solve a triple integrals problem. I need to find the volume of the solid delimited by the following functions:

$z=\sqrt{x^2+y^2}\rightarrow z=r$
$z=1-2 \sqrt{x^2+y^2}\rightarrow z=1-2r$

Can I do that?
$r=1-2r$ so that I find out $r=z=\frac{1}{3}$

So that I can pose the following triple integral to find the volume of the solid delimited by the 2 functions:
$\int^{2\pi}_{0} \int^{\frac{1}{3}}_{0} \int^{1-2r}_{r}r^2 \mathrm{d}z \mathrm{d}r \mathrm{d}\theta = \frac{\pi }{162}$

Thank you so much for checking out,

Bazinga

2. Hi,

The solid $z=r$ is a cone starting from the origin (0;0;0) and going up with a slope of 1

The solid $z=1-2r$ is a cone starting from the point (0;0;1) and going down with a slope of 2

The 2 cones intersect at a circle center (0;0;1/3), radius 1/3 and parallel to the xOy plane

Can you see all of this ?

3. ## Follow-up

Thanks for taking the time to answer.

Yes, I see the circle on the Oxy plan and know that z=r is indeed the bottom one as the other is on top. Am I posing the integral the right way? I thought of it as a circle (0 to 2pi) of 0 to 1/3 as radius delimited by both cones from bottom to top in z.

Regards,

Bazinga

4. The "elementary" volume is $r \mathrm{d}\theta \mathrm{d}z \mathrm{d}r$ and not $r^2 \mathrm{d}\theta \mathrm{d}z \mathrm{d}r$

The volume of the first cone (at the bottom) is
$\int^{2\pi}_{0} \int^{\frac{1}{3}}_{0} \int^{z}_{0}r \mathrm{d}\theta \mathrm{d}z \mathrm{d}r$

Sorry if my English vocabualry is not the appropriated one ...

5. Je suis confus. Comment dois-je poser l'intégrale qui résout le problème? Je croyais que le volume du cône était r et l'élément de volume r dr dz d theta. Vous posez une intégrale de 0 à z en d theta dont je ne saisis pas le sens. Merci beaucoup de vos réponses constructives, elles m'aident à comprendre le cheminement de résolution de tels problèmes.

I am confused. Would you please pose the correct integral so that I understand my mistake. Why is there a 0 to z integral. Thanks for your previous answers, they were much helpful.

6. Je pense qu'on ne peut pas continuer en Français sur ce forum ou alors éventuellement en MP

In cylindrical coordinates the elementary volume is $r dr d\theta dz$
The volume of the cone is $\int \int \int r dr d\theta dz$

Concerning the bounds, $\theta$ varies from 0 to $2 \pi$, z varies from 0 to 1/3 and r varies from 0 to a bound which is z-dependent. If you consider a given value for z, you can see that r varies from 0 to z (I am talking about the lower cone)

Therefore $V = \int_{\theta=0}^{\theta=2\pi} \int_{z=0}^{z=\frac{1}{3}} \int_{r=0}^{r=z}r dr d\theta dz$