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?
 A: 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.

A: 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.
