Triple Integral to Find Volume Question: Use a triple integral to find the volume of the solid enclosed by the parabaloids $y=x^2+z^2$ and $y=8-x^2-z^2$.
My attempt: The best I can figure, this object looks kind of like a football oriented along the $y$-axis from $y=0$ to $y=8$ and is symmetric about the $y$-axis and the plane $y=4$.
It seems best to integrate first with respect to $y$, and $x^2+z^2 \le y \le 8-(x^2+z^2)$.
The widest part of the football is at $y=4$; substitute that into both of the equations above to find that the projection onto the $xz$-plane is $x^2+z^2=4$, or a circle of radius 2, so my bounds for $z$ are $-\sqrt{4-x^2} \le z \le \sqrt{4-x^2}$ and my bounds for $x$ are $-2 \le x \le 2$.
But I believe I can make this easier by integrating from $0\le z \le \sqrt{4-x^2}$ and multiplying by 2 and integrating from $0 \le x \le 2$ and multiplying by another 2. (I actually think I can integrate $y$ from $4 \le y \le 8-(x^2+z^2)$ and multiply by another 2, but that doesn't seem to simplify anything.)
Since I'm finding the volume, the function I integrate is one. I come up with this
$$4\int_0^2 \int_0^{\sqrt{4-x^2}} \int_{x^2+z^2}^{8-(x^2+z^2)} 1\ dy\ dz\ dx.$$
(Is this right so far?)
If nothing's wrong yet, I still can't finish this integral
$$4\int_0^2 \int_0^{\sqrt{4-x^2}} \int_{x^2+z^2}^{8-\left(x^2+z^2\right)} 1\ dy\ dz\ dx \\ 
4\int_0^2 \int_0^{\sqrt{4-x^2}} y \Big|_{x^2+z^2}^{8-\left(x^2+z^2\right)} dz\ dx \\ 
4\int_0^2 \int_0^{\sqrt{4-x^2}} \left(8-\left(x^2+z^2\right)\right)-\left(x^2+z^2\right) dz\ dx \\ 
4\int_0^2 \int_0^{\sqrt{4-x^2}} \left(8-2x^2-2z^2\right)\ dz\ dx \\ 
4\int_0^2 \left[\left(8-2x^2\right)z-\frac 23 z^3\right]_0^{\sqrt{4-x^2}} \ dx \\
4\int_0^2 \left[ \left(8-2x^2\right)\sqrt{4-x^2} - \frac 23 \sqrt{4-x^2}^3 \right]\ dx \\
-\frac{16}3\int_0^2 \left[ \left(x^2 -4\right)\sqrt{4-x^2} \right]\ dx \\
\vdots \\ ??$$
 A: The volume will be represented by the triple integral:
$\int \int \int _{\Omega} dV = \Gamma$ where $ \Omega$ is the solid enclosed by the two paraboloids. The paraboloids intersect in $y = 4$ So:
$\Gamma = \int \int_{\Omega'} \int_{x^{2}+z^{2}}^{8-(x^{2}+z^{2})}dV$. Where $\Omega'= $$\{(x,z) \lvert  x^{2} + z^{2} \leq 4 \}$.
So:
$\Gamma = \int \int_{\Omega´}8-2(x^{2}+z^{2})dA$
Using the substitution:
$x = r cos(\theta)$ and $z = r sin(\theta)$, $0 \leq r \leq 2$ and $0\leq \theta \leq 2\pi$.
$\Gamma = \int_{0}^{2\pi} \int_{0}^{2}r(8-2r^{2})drd\theta$
So $\Gamma = 2\pi(4(2^2) - (2)^3) = 16\pi$ 
A: HINT:For a start, it is convenient to put $ x^2+z^2 = r^2$, so that you have solids of revolution:
$$ y_1=r^2 ,\ y_2=8-r^2 $$.
A: The integral in your last line can be simplified to 
$$
\frac{16}3\int_0^2 (4-x^2)^{3/2}\,dx
$$
To do this, first use the substitution $x=2\sin\theta$, obtaining
$$
\frac{16}3\int_0^{\pi/2}(4-4\sin^2\theta)^{3/2}\cos\theta\,d\theta=
8\cdot\frac{16}3\int_0^{\pi/2}\cos^4\theta\,d\theta
$$
To complete the integral, use the identity $\cos^2\phi = \frac{1+\cos(2\phi)}2$ to write
$$
\cos^4\theta = \left(\frac{1+\cos(2\theta)}{2}\right)^2=\frac12+\cos(2\theta)+\frac14\cos^22\theta=\frac12+\cos(2\theta)+\frac{1}{4}\left(\frac{1+\cos(4\theta)}{2}\right)
$$
