## Calculus with Applications (10th Edition)

$\approx 6\pi$
If y=0, then $x=\sqrt 2, x=-\sqrt 2$, so that $a=\sqrt 2, b=-\sqrt 2$. The volume is $V=\int^{\sqrt 2}_{-\sqrt 2}\pi(2-x^{2})^{2}dx = \int^{\sqrt 2}_{-\sqrt 2}\pi(4-4x^{2}+x^{4})dx = \pi(4x-\frac{4x^{3}}{3}+\frac{x^{5}}{5})|^{\sqrt 2}_{-\sqrt 2} \approx 6\pi$