Answer
$ 16 \pi$
Work Step by Step
Consider $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 $
or, $=\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} [y]_{x^2+z^2}^{8-x^2-z^2} dz dx $
or, $=\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} [8-2(x^2+z^2)] dz dx $
or, $=\int_{0}^{2\pi} \int_0^2 (8-2r^2) r dr d\theta$
or, $=\int_0^2 (8r-2r^3) dr \times \int_{0}^{2\pi} d \theta$
or, $= (2 \pi) [4r^2-(1/2) r^4]_0^2$
or, $= (2 \pi) (16-8)$
or, $= 16 \pi$
