# Write down this integral as a triple integral with cylindrical coordinates

I have the integral: $${\iiint} x^2 dx dy dz$$ which is bounded from above by the elliptic paraboloid $$z=2-x^2 - y^2$$ and from below by the upper part of the cone $$z^2 = x^2 + y^2$$

I want to write this integral as a triple integral with cylindrical coordinates in the order $$dz dr d\theta$$

I know $$r^2 = x^2 + y^2$$, so by using this I can say $$z^2 = x^2 + y^2$$ becomes $$z^2 = r^2$$ (and so $$z=r$$) and $$z=2-x^2 - y^2$$ becomes $$z=2-r^2$$

I also set $$r=2-r^2$$ (since they both equal $$z$$) and found $$r=1$$ (since we want a positive $$r$$)

So I have what I believe is the correct set-up for the upper and lower bounds of my triple integration: $${\int}_0 ^{2\pi} {\int}_0 ^1 {\int}_r ^{2-r^2} *\text{something}* dz dr d\theta$$

What I'm not sure of is what $$*\text{something}*$$ is, it was $$x^2$$ but I'm not sure how it changes now that my triple integral has cylindrical coordinates.

If someone could show me (or give me a hint to do it myself) what happens to $$x^2$$ when we convert to cylindrical coordinates it would really help.

I know what to do after I find out what $$*\text{something}*$$ is so there's no need to do the triple integral as well (unless you want to).

• Why do you say elliptic paraboloid? It is not. All your working is based on $x = r \cos\theta, y = r\sin\theta$. That gives you value of $x^2$. Jul 30 '21 at 14:13
In cylindrical coordinates, $$x^2$$ becomes $$r^2\cos^2\theta$$. Besides, you have to multiply everything by the volume element, which is $$r$$. So, compute$$\int_0^{2\pi}\int_0^1\int_r^{2-r^2}r^3\cos^2(\theta)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta.$$
• Ok, I have 2 short questions. 1) When you say $x^2$ becomes $r^2 cos^2 (\theta)$, did you work this out or is it just known that this is the case? 2) Do we multiply everything by $r$ every time regardless of what we have in the middle? Jul 30 '21 at 13:42
• 1) It is universal. Working with cylindrical coordinates means that you are doing $x=r\cos(\theta)$ and $y=r\sin(\theta)$. 2) Yes. Jul 30 '21 at 16:26