Answer
$20 \pi$
Work Step by Step
Consider $V=\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (5-z) dz dx $
or, $=\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (5-z) dz dx $
or, $=\int_{-2}^2 (5z-z^2)_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dx $
or, $=\int_{-2}^{2} \sqrt{4-x^2} dx$
Plug $x=2 \sin \theta \implies dx=2 \cos \theta d \theta$
or, $=(10) \int_{-\pi/2}^{pi/2}(2 \cos \theta) (2\cos \theta) d \theta$
or, $= (10)\int_{-\pi/2}^{pi/2}(2 \cos \theta)^2 d \theta$
or, $= (20) \int_{-\pi/2}^{pi/2} (2\cos \theta+1)$
or, $=20 \pi$
