# Evaluating $\iiint x\,dx\,dy\,dz$ limited by paraboloid of equation $x=4 y^2+4z^2$ & for plane $x=4$

Hi we have the following problem:

$\iiint x\,dx\,dy\,dz$ limited by the paraboloid of equation $x=4 y^2+4z^2$ and for the plane $x=4$.

We are having difficulty on finding the limits of each integral and how to turn to polar coordinates.

Could you offer any tips? I can provide more details on the comments on what we tried. Thank you.

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Normally, polar coordinates are given as $x=r\cos \theta$ and $y=r\sin \theta$, but that does not necessarily have to be. In this case, I think it is wise to use $y=r\cos \theta$ and $z=r\sin \theta$. Then, your limits of integration would be from the paraboloid to the plane (in $dx$), then from zero to the radius of the bounding circle in the $y$-$z$ plane, then from zero to $2\pi$ in $\theta$ to complete the full revolution around the circle.

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Thank you we are trying this right now! –  Edoardo Sep 7 '12 at 3:03
If you use y=rcos(θ) and z=rsin(θ), what do you use for x since the integral is ∫∫∫xdxdydz? –  Edoardo Sep 7 '12 at 3:11
You don't need to substitute anything for x... the change in variables result in a differential of the form: $r dxdrd\theta$. –  Paul Sep 7 '12 at 13:42

An idea: make a substitution change $\,x\leftrightarrow z\,$, so that you have the paraboloid $\,z=4x^2+4y^2\,$ and the plane$\,z=4\,$, and now use cylindrical coordinates and the symmetry of the paraboloid around the $\,z-\,$ axis:

$$\iiint z\,dx\,dy\,dz=4\int_0^1dr\int_0^{\pi/2}d\theta\int_{4r^2}^4 4r^3\,dz$$

Please do note that $\,4r^3=4r^2\cdot r=z|J|$, where $|J|$ is the Jacobian of the transformation into cylindrical coordinates.

Added: We can also do the following:

\begin{align} \iiint z\,dx\,dy\,dz &= 4\int_0^1dr\int_0^{\pi/2}d\theta\int_{4r^2}^4 zr\,dz\\ & = 2\pi\int_0^1r\left[\frac{1}{2}z^2\right]\Bigg|_{4r^2}^4\,dr \\ & = \pi\int_0^1(16r-16r^5)\,dr \\ & = 16\pi\left(\frac{1}{2}-\frac{1}{6}\right) \\ & = \frac{16\pi}{3} \end{align}

which is the right answer according to the book. I still am not sure what went wrong with the first method which I leave here for others to check and comment.

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Sorry the answer doesn't matches with the calculator :/ –  Edoardo Sep 7 '12 at 3:15
What's *the answer" and by whom ( the book's?)? And what did you get above? Just like that it's hard to say what could have happened... –  DonAntonio Sep 7 '12 at 3:16
The answer according to my text book is 16pi/3 and what I got from above was 64pi/15 –  Edoardo Sep 7 '12 at 3:18
Well, the result of the integral in my answer is $\,\frac{8\pi}{3}\,$ , so it is off by a simple factor of 2 which I can't see right now from where it comes. Perhaps later I'll add something to my answer. Anyways, your calculation seesm to be wrong –  DonAntonio Sep 7 '12 at 3:28
@Edoardo , what book is that? –  DonAntonio Sep 7 '12 at 3:50