Answer
$$S = 8\pi $$
Work Step by Step
$$\eqalign{
& y = \sqrt {4 - {x^2}} ,{\text{ }} - 1 \leqslant x \leqslant 1 \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\sqrt {4 - {x^2}} } \right] \cr
& \frac{{dy}}{{dx}} = - \frac{x}{{\sqrt {4 - {x^2}} }} \cr
& {\text{Formula for surface area}} \cr
& S = 2\pi \int_a^b {r\left( x \right)\sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} } dx,{\text{ }}r\left( x \right) = f\left( x \right) \cr
& {\text{Then,}} \cr
& S = 2\pi \int_{ - 1}^1 {\sqrt {4 - {x^2}} \sqrt {1 + {{\left[ { - \frac{x}{{\sqrt {4 - {x^2}} }}} \right]}^2}} } dx \cr
& S = 2\pi \int_{ - 1}^1 {\sqrt {4 - {x^2}} \sqrt {1 + \frac{{{x^2}}}{{4 - {x^2}}}} } dx \cr
& S = 2\pi \int_{ - 1}^1 {\sqrt {4 - {x^2}} \sqrt {\frac{{4 - {x^2} + {x^2}}}{{4 - {x^2}}}} } dx \cr
& S = 2\pi \int_{ - 1}^1 {\frac{{\sqrt {4 - {x^2}} }}{{\sqrt {4 - {x^2}} }}\sqrt 4 } dx \cr
& S = 4\pi \int_{ - 1}^1 {dx} \cr
& {\text{Integrate}} \cr
& S = 4\pi \left[ x \right]_{ - 1}^1 \cr
& S = 4\pi \left( {1 - \left( { - 1} \right)} \right) \cr
& S = 8\pi \cr} $$
