Answer
$16 \pi$
Work Step by Step
Here, we have $V=\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{x^2+z^2}^{8-x^2-z^2} dy dz dx $
This implies that $V=\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} [y]_{x^2+z^2}^{8-x^2-z^2} dz dx $
In polar coordinate system, we have $x=r \cos \theta \\ y=r \sin \theta \\z=z$ and $r^2=x^2+y^2 \\ \tan \theta=\dfrac{y}{x} \\z=z$
and $\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} [8-2(x^2+z^2)] dz dx =\int_{0}^{2\pi} \int_0^2 (8-2r^2) r dr d\theta$
Now, we have $\int_0^2 (8r-2r^3) dr \int_{0}^{2\pi} d \theta= (2 \pi) \times [4r^2-(\dfrac{1}{2}) r^4]_0^2$
or, $V= (2 \pi) [4(2-0)^2-(\dfrac{1}{2}) (2-0)^4]= (2 \pi) (16-8)= 16 \pi$
