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: 3

$$S = 8\pi $$

$$\eqalign{ & y = \sqrt {4 - {x^2}} ,\,\,\,\, - 1 \leqslant x \leqslant 1 \cr & {\text{Calculate }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = \frac{{ - 2x}}{{2\sqrt {4 - {x^2}} }} = - \frac{x}{{\sqrt {4 - {x^2}} }} \cr & {\text{Use }}S = \int_a^b {2\pi y\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx \cr & S = \int_{ - 1}^1 {2\pi \left( {\sqrt {4 - {x^2}} } \right)\sqrt {1 + {{\left( { - \frac{x}{{\sqrt {4 - {x^2}} }}} \right)}^2}} } dx \cr & S = \int_{ - 1}^1 {2\pi \left( {\sqrt {4 - {x^2}} } \right)\sqrt {1 + \frac{{{x^2}}}{{4 - {x^2}}}} } dx \cr & S = \int_{ - 1}^1 {2\pi \left( {\sqrt {4 - {x^2}} } \right)\sqrt {\frac{{4 - {x^2} + {x^2}}}{{4 - {x^2}}}} } dx \cr & S = \int_{ - 1}^1 {2\pi \left( {\frac{{\sqrt {4(4 - {x^2)}} }}{{\sqrt {4 - {x^2}} }}} \right)\sqrt 1 } dx \cr & S = 2\pi \int_{ - 1}^12 {dx} \cr & {\text{Integrate}} \cr & S = 2\pi \left[ 2x \right]_{ - 1}^1 \cr & S = 4\pi \left[ {\left( 1 \right) - \left( { - 1} \right)} \right] \cr & S = 4\pi \left( 2 \right) \cr & S = 8\pi \cr} $$