## Calculus: Early Transcendentals (2nd Edition)

$V = 6\pi + 2\pi \ln \,\left( {\frac{2}{5}} \right)$
$\begin{gathered} We\,\,have\,\,to\,\,find\,\,the\,\,volume\,\,of\,\,the\,\,solid\,\,obtained\,\,by \hfill \\ revolving\,\,the\,\,region\,\,bounded\,\,by \hfill \\ \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = \frac{1}{{x + 2}}\,\,\,,\,\,\,\,y = 0\,\,\,\,,\,\,x = 0\,\,and\,\,x = 3 \hfill \\ \hfill \\ is\,\,revolved\,\,about\,\,x = - 1 \hfill \\ \hfill \\ V = 2\pi \int_0^3 {\,\left( {x + 1} \right)ydx} \hfill \\ \hfill \\ substitute\,\,\,\,\,\,\,\,y = \frac{1}{{x + 2}} \hfill \\ \hfill \\ V = 2\pi \int_0^3 {\,\left( {\frac{{x + 1}}{{x + 2}}} \right)dx} \hfill \\ \hfill \\ by\,\,the\,\,long\,\,division\,\,\frac{{x + 1}}{{x + 2}} = 1 - \frac{1}{{x + 2}} \hfill \\ \hfill \\ V = 2\pi \int_0^3 {\,\left( {1 - \frac{1}{{x + 2}}} \right)dx} \hfill \\ \hfill \\ integrate\,\,and\,\,evaluate \hfill \\ \hfill \\ V = 2\pi \,\,\left[ {x - \ln \,\left( {x + 2} \right)} \right]_0^3 \hfill \\ \hfill \\ V = 2\pi \,\,\left[ {3 - \ln 5 + \ln 2} \right] \hfill \\ \hfill \\ solution \hfill \\ \hfill \\ V = 6\pi + 2\pi \ln \,\left( {\frac{2}{5}} \right) \hfill \\ \end{gathered}$