Answer
$$S = \frac{\pi }{3}\left( {16\sqrt 2 - 8} \right)$$
Work Step by Step
$$\eqalign{
& y = 1 - \frac{{{x^2}}}{4},{\text{ }}0 \leqslant x \leqslant 2 \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {1 - \frac{{{x^2}}}{4}} \right] \cr
& \frac{{dy}}{{dx}} = - \frac{x}{2} \cr
& S = 2\pi \int_a^b {r\left( x \right)\sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} } dx,{\text{ }}r\left( x \right) = x \cr
& {\text{Then,}} \cr
& S = 2\pi \int_0^2 {x\sqrt {1 + {{\left( { - \frac{x}{2}} \right)}^2}} } dx \cr
& S = 2\pi \int_0^2 {x\sqrt {1 + \frac{{{x^2}}}{4}} } dx \cr
& S = 2\pi \int_0^2 {\frac{x}{2}\sqrt {4 + {x^2}} } dx \cr
& S = \pi \int_0^2 {x\sqrt {4 + {x^2}} } dx \cr
& S = \frac{\pi }{2}\int_0^2 {\left( {2x} \right)\sqrt {4 + {x^2}} } dx \cr
& {\text{Integrate}} \cr
& S = \frac{\pi }{2}\left[ {\frac{{{{\left( {4 + {x^2}} \right)}^{3/2}}}}{{3/2}}} \right]_0^2 \cr
& S = \frac{\pi }{3}\left[ {{{\left( {4 + {x^2}} \right)}^{3/2}}} \right]_0^3 \cr
& S = \frac{\pi }{3}\left[ {{{\left( 8 \right)}^{3/2}} - {{\left( 4 \right)}^{3/2}}} \right] \cr
& S = \frac{\pi }{3}\left( {16\sqrt 2 - 8} \right) \approx 15.31779 \cr} $$
