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: 7

$$S = 24\pi $$

$$\eqalign{ & x = \sqrt {9 - {y^2}} ,\,\,\,\, - 2 \leqslant y \leqslant 2 \cr & {\text{Calculate }}\frac{{dx}}{{dy}} \cr & \frac{{dx}}{{dy}} = \frac{{ - 2y}}{{2\sqrt {9 - {y^2}} }} = \frac{{ - y}}{{\sqrt {9 - {y^2}} }} \cr & {\text{Use }}S = \int_a^b {2\pi x\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} } dy \cr & S = \int_{ - 2}^2 {2\pi \left( {\sqrt {9 - {y^2}} } \right)\sqrt {1 + {{\left( {\frac{{ - y}}{{\sqrt {9 - {y^2}} }}} \right)}^2}} } dy \cr & S = 2\pi \int_{ - 2}^2 {\sqrt {9 - {y^2}} \sqrt {1 + \frac{{{y^2}}}{{9 - {y^2}}}} } dy \cr & S = 2\pi \int_{ - 2}^2 {\sqrt {9 - {y^2}} \sqrt {\frac{{9 - {y^2} + {y^2}}}{{9 - {y^2}}}} } dy \cr & S = 2\pi \int_{ - 2}^2 3 dy \cr & {\text{Integrate}} \cr & S = 6\pi \left[ y \right]_{ - 2}^2 \cr & S = 6\pi \left[ {2 + 2} \right] \cr & S = 24\pi \cr} $$