## Calculus (3rd Edition)

$\dfrac{4}{3} \pi r^3$
The volume of a region can be calculated as: $V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$ where, $R_{outside}=x^3+2-(-2)=x^3+4$ and $R_{inside}=4-x^2-(-2)=6-x^2$ $V=\pi \int_0^r 2y \sqrt {r^2-y^2} \ dy$ Let us apply the substitution method such that: $a=r^2-y^2 \implies dx=\dfrac{-da}{2y}$ Now, $V = - 2\pi \int_0^r 2y \sqrt a \times \dfrac{1}{2y} \ da \\= - 2\pi \int_0^r \dfrac{2a^{3/2}}{3} \ da \\=-2\pi [\dfrac{2(r^2-y^2)^{3/2}}{3}]_0^r \\=2 \pi [\dfrac{2}{3} r^3]\\=\dfrac{4}{3} \pi r^3$