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).

Thanks in advance

  • $\begingroup$ 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 $x^2$. $\endgroup$
    – Math Lover
    Jul 30, 2021 at 14:13
  • $\begingroup$ In my first line I just typed the question I was given, in the question they say it's an elliptic paraboloid so I just assumed it to be true. To be honest I don't think it affects the question much so it's not too important. $\endgroup$
    – Charlie P
    Jul 30, 2021 at 15:39

1 Answer 1


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$$\int_0^{2\pi}\int_0^1\int_r^{2-r^2}r^3\cos^2(\theta)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta.$$

  • $\begingroup$ 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? $\endgroup$
    – Charlie P
    Jul 30, 2021 at 13:42
  • $\begingroup$ 1) It is universal. Working with cylindrical coordinates means that you are doing $x=r\cos(\theta)$ and $y=r\sin(\theta)$. 2) Yes. $\endgroup$ Jul 30, 2021 at 16:26
  • $\begingroup$ Any more questions? $\endgroup$ Jul 31, 2021 at 9:16
  • $\begingroup$ No that's perfect thanks. Sorry, I forgot to reply and upvote. Thanks for the help $\endgroup$
    – Charlie P
    Aug 2, 2021 at 12:23

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