Chapter 7: Transcendental Functions - Section 7.6 - Inverse Trigonometric Functions - Exercises 7.6 - Page 423: 121

$\frac{\pi^{2}}{2}$

$V$ = $\int_{{\,\frac{-\sqrt 3}{3}}}^{{\,\sqrt 3}}$$\pi(\frac{1}{\sqrt{1+x^{2}}})^{2}$ $dx$ $V$ = $\int_{{\,\frac{-\sqrt 3}{3}}}^{{\,\sqrt 3}}$$\frac{\pi}{{1+x^{2}}}$ $dx$ $V$ = $\pi[tan^{-1}(x)]$ $|_{{\,\frac{-\sqrt 3}{3}}}^{{\,\sqrt 3}}$ $V$ = $\pi[tan^{-1}(\sqrt 3))-tan^{-1}(\frac{-\sqrt 3}{3})]$ $V$ = $\pi(\frac{\pi}{3}-(\frac{-\pi}{6}))$ $V$ = $\pi(\frac{\pi}{3}+\frac{\pi}{6})$ $V$ = $\frac{\pi^{2}}{2}$