Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.5 Area Of A Surface Of Revolution - Exercises Set 5.5 - Page 380: 12

$$S = \frac{{\left( {65\sqrt {65} - 5\sqrt 5 } \right)\pi }}{6}$$

$$\eqalign{ & x = \sqrt {16 - y} ,\,\,\,\,0 \leqslant y \leqslant 15 \cr & {\text{Calculate }}\frac{{dx}}{{dy}} \cr & \frac{{dx}}{{dy}} = - \frac{1}{{2\sqrt {16 - y} }} \cr & {\text{Use }}S = \int_a^b {2\pi x\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} } dy \cr & S = \int_0^{15} {2\pi \left( {\sqrt {16 - y} } \right)\sqrt {1 + {{\left( { - \frac{1}{{2\sqrt {16 - y} }}} \right)}^2}} } dy \cr & S = \int_0^{15} {2\pi \left( {\sqrt {16 - y} } \right)\sqrt {1 + \frac{1}{{4\left( {16 - y} \right)}}} } dy \cr & S = \int_0^{15} {2\pi \left( {\sqrt {16 - y} } \right)\sqrt {\frac{{4\left( {16 - y} \right) + 1}}{{4\left( {16 - y} \right)}}} } dy \cr & S = \int_0^{15} {2\pi \left( {\frac{{\sqrt {16 - y} }}{{2\sqrt {16 - y} }}} \right)\sqrt {64 - 4y + 1} } dy \cr & S = \pi \int_0^{15} {\sqrt {65 - 4y} } dy \cr & {\text{Integrate}} \cr & S = - \frac{\pi }{4}\left[ {\frac{{2{{\left( {65 - 4y} \right)}^{3/2}}}}{3}} \right]_0^{15} \cr & S = - \frac{\pi }{4}\left[ {\frac{{2{{\left( {65 - 4\left( {15} \right)} \right)}^{3/2}}}}{3} - \frac{{2{{\left( {65 - 4\left( 0 \right)} \right)}^{3/2}}}}{3}} \right] \cr & S = - \frac{\pi }{4}\left[ {\frac{{10\sqrt 5 - 130\sqrt {65} }}{3}} \right] \cr & S = \frac{{\left( {130\sqrt {65} - 10\sqrt 5 } \right)\pi }}{{12}} \cr & S = \frac{{\left( {65\sqrt {65} - 5\sqrt 5 } \right)\pi }}{6} \cr} $$