## Calculus (3rd Edition)

$\dfrac{32 \pi}{105}$
The volume of a region can be calculated as: Now, $V=\pi \int_{-1}^{1} [(1-x^{2/3})^{3/2}]^2 \ dx \\ =\pi \int_{-1}^1 (1-x^{2/3})^3 \ dx \\=\pi \int_{-1}^1 (-x^2+3x^{4/3}-3x^{2/3}+1) \ dx \\=\pi [-\dfrac{x^3}{3}+\dfrac{9x^{7/3}}{7}- \dfrac{9x^{5/3}}{5}+x]_{-1}^1 \\=\dfrac{32 \pi}{105}$