Answer
$V=\frac{16\pi}{15} $
Work Step by Step
If y=0, then x=1, x=-1, so that a=1, b=-1. The volume is
$V=\int^{1}_{-1}\pi(1-x^{2})^{2}dx = \int^{1}_{-1}\pi(1-2x^{2}+x^{4})dx = \pi(x-\frac{2x^{3}}{3}+\frac{x^{5}}{5})|^{1}_{-1} =\frac{16\pi}{15} $