Chapter 7 - Section 7.3 - Trigonometric Substitution - 7.3 Exercises - Page 506: 43

$V = \frac{{3{\pi ^2}}}{8} + \frac{{3\pi }}{4}$

$$\eqalign{ & y = \frac{9}{{{x^2} + 9}},{\text{ }}y = 0,{\text{ }}x = 0,{\text{ and }}x = 3 \cr & {\text{The graph of the enclosed region is shown below}} \cr & {\text{Use the disk method about the }}x{\text{ - axis}} \cr & V = {\int_a^b {\pi \left[ {f\left( x \right)} \right]} ^2}dx \cr & {\text{Then}} \cr & V = \pi \int_0^3 {{{\left( {\frac{9}{{{x^2} + 9}}} \right)}^2}} dx \cr & V = 81\pi \int_0^3 {\frac{1}{{{{\left( {{x^2} + 9} \right)}^2}}}} dx \cr & {\text{Let }}x = 3\tan \theta ,{\text{ }}dx = 3{\sec ^2}\theta d\theta \cr & {x^2} + 9 = 9{\sec ^2}\theta \cr & \theta = {\tan ^{ - 1}}\left( {\frac{x}{3}} \right) \cr & {\text{The new limits of integration are:}} \cr & x = 3 \to \theta = {\tan ^{ - 1}}\left( {\frac{3}{3}} \right) = \frac{\pi }{4} \cr & x = 0 \to \theta = {\tan ^{ - 1}}\left( {\frac{0}{3}} \right) = 0 \cr & {\text{Substituting}}{\text{, we obtain}} \cr & V = 81\pi \int_0^{\pi /4} {\frac{1}{{{{\left( {9{{\sec }^2}\pi } \right)}^2}}}} \left( {3{{\sec }^2}\theta } \right)d\theta \cr & V = 3\pi \int_0^{\pi /4} {\frac{{{{\sec }^2}\theta }}{{{{\sec }^4}\theta }}} d\theta \cr & V = 3\pi \int_0^{\pi /4} {{{\cos }^2}\theta } d\theta \cr & V = 3\pi \int_0^{\pi /4} {\left( {\frac{{1 + \cos 2\theta }}{2}} \right)} d\theta \cr & V = \frac{{3\pi }}{2}\int_0^{\pi /4} {\left( {1 + \cos 2\theta } \right)} d\theta \cr & {\text{Integrating}} \cr & V = \frac{{3\pi }}{2}\left[ {\theta + \frac{1}{2}\sin 2\theta } \right]_0^{\pi /4} \cr & V = \frac{{3\pi }}{2}\left[ {\frac{\pi }{4} + \frac{1}{2}\sin \left( {\frac{\pi }{2}} \right)} \right] - \frac{{3\pi }}{2}\left[ {0 + \frac{1}{2}\sin \left( 0 \right)} \right] \cr & V = \frac{{3\pi }}{2}\left[ {\frac{\pi }{4} + \frac{1}{2}} \right] - \frac{{3\pi }}{2}\left[ 0 \right] \cr & V = \frac{{3{\pi ^2}}}{8} + \frac{{3\pi }}{4} - 0 \cr & V = \frac{{3{\pi ^2}}}{8} + \frac{{3\pi }}{4} \cr} $$