# Volume between two paraboloids

Find the volume of the solid enclosed by the paraboloids $z=9(x^2+y^2)$ and $z=32−9(x^2+y^2)$

I'm not sure how to even find the volume enclosed to begin with. I know that the paraboloids intersect when

$$9(r^2) = 32−9(r^2) \implies r = \frac43 \implies z = 16$$ If this is the plane where the two intersect, then the bounds are $16 \leq z\leq 32−9(x^2+y^2)$

Am I correct in this?

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Try curve tracing to visualize the problem in 3D. –  Anurag Kalia Jan 17 '13 at 8:45

Yes. you are correct. I think if we use the cylindrical coordinates, the volume could find better. As follows: $$4\int_0^{\pi/2}\int_{r=0}^{r=4/3}\int_{z=9r^2}^{32-9r^2}rdzdrd{\theta}$$. I added the following picture for you to consider the volume better.

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Thank you very much –  Cactus BAMF Jan 17 '13 at 9:05
Thank you so much, I've been trying to visualize this problem for hours. It finally is clear! –  Cactus BAMF Jan 17 '13 at 9:08
@CactusBAMF: I made it by Maple 16. It was easy. Welcome. ;-) –  Babak S. Jan 17 '13 at 9:09
I'm always big on nice pictures! + 1 –  amWhy Feb 15 '13 at 0:04

Hint: Write the volume as an integral.

The region of space you're interested in is $$R = \{(x,y,z)\in\mathbf R^3 : 9(x^2+y^2) \leqslant z \leqslant 32-9(x^2+y^2) \},$$ so the volume can be written as $$V = \iiint_R \mathrm dx\,\mathrm dy\,\mathrm dz.$$ Of course, given the presence of $x^2+y^2$ everywhere, you probably want to apply a change of variables to use cylindrical coordinates. The new domain is then $R'=\{(r,\theta,z):9r^2\leqslant z\leqslant 32-9r^2\}$. You can simplify again just a bit using what you have already noticed: there is space between the paraboloids only when $0\leqslant r\leqslant 4/3$, so the new domain can be written $$R'=\{(r,\theta,z):0\leqslant r\leqslant 4/3,~9r^2\leqslant z\leqslant 32-9r^2\}.$$ Now you have to actually do the change of variables, and compute an integral.

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@BabakSorouh Woops! :) –  jathd Jan 17 '13 at 9:13