Answer
$16 \pi$
Work Step by Step
Here, $V=\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{x^2+z^2}^{8-x^2-z^2} dy dz dx$
$= \int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} [y]_{x^2+z^2}^{8-x^2-z^2} dz dx $
$= \int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (8-x^2-z^2-x^2-z^2) dz dx $
$=\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} [8-2(x^2+z^2)] dz dx $
$=\int_{0}^{2\pi} \int_0^2 (8-2r^2) r dr d\theta$
$=\int_0^2 (8r-2r^3) dr \cdot \int_{0}^{2\pi} d \theta$
$= (2 \pi) [4r^2-\dfrac{r^4}{2}]_0^2$
$= (2 \pi) [4(2)^2-\dfrac{(2)^4}{2}$
$= 16 \pi$
