Answer
$20 \pi$
Work Step by Step
Let us consider that $V=\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (5-z) dz dx $
and $\int_{-2}^2 (5z-z^2)_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dx =\int_{-2}^{2} \sqrt{4-x^2} dx$
Consider $x=2 \sin \theta$ and $dx=2 \cos \theta d \theta$
Thus, we have $\int_{-2}^{2} \sqrt{4-(2 \sin \theta)^2} (2 \cos \theta d \theta)=(10) \int_{-\pi/2}^{\pi/2}(2 \cos \theta) (2\cos \theta) d \theta$
Hence, $V= 10\int_{-\pi/2}^{\pi/2}(2 \cos \theta)^2 d \theta= (20) \int_{-\pi/2}^{pi/2} (2\cos \theta+1)=20 \pi$
