University Calculus: Early Transcendentals (3rd Edition)

$$8\pi -\dfrac{32}{3}$$
Our aim is to integrate the triple integral as follows: $$V=\int^4_0 \int^{\sqrt{16-y^2}/2}_0 \int^{4-y}_0 \space dx \space dz \space dy \\=\int^4_0 \int^{\sqrt{16-y^2}/2}_0(4-y) \space dz \space dy\\=\int^4_0 \dfrac{\sqrt{16-y^2}}{2}(4-y) \space dy \\= \int^4_0 2\sqrt{16-y^2} dy -\dfrac{1}{2} \times \int^4_0 y\sqrt{16-y^2} \space dy \\=[y\sqrt{16-y^2}+16 \sin^{-1}(\dfrac{y}{4})]^4_0+[\dfrac{1}{6}(16-y^2)^{3/2}]^4_0 \\=8\pi -\dfrac{32}{3}$$