Answer
$20 \pi$
Work Step by Step
Here, we have $V=\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (5-z) dz dx $
$=\int_{-2}^2 [5z-z^2]_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dx $
$=\int_{-2}^2 [5[\sqrt{4-x^2})+\sqrt{4-x^2}]-[\sqrt{4-x^2})+\sqrt{4-x^2}]^2]_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dx $
$=\int_{-2}^{2}(5) \sqrt{4-x^2} dx$
Suppose $x=2 \sin \theta$ and $ dx=2 \cos \theta d \theta$
$=\int_{-2}^{2} \sqrt{4-(2 \sin \theta)^2} (2 \cos \theta d \theta)$
$=10 \int_{-\pi/2}^{\pi/2}(2 \cos \theta)^2 d \theta$
$= 20 \int_{-\pi/2}^{\pi/2} (2\cos \theta+1)$
$=20 \pi$
