Answer
$$S = \frac{\pi }{6}\left( {37\sqrt {37} - 1} \right)$$
Work Step by Step
$$\eqalign{
& y = 9 - {x^2},{\text{ }}0 \leqslant x \leqslant 3 \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {9 - {x^2}} \right] \cr
& \frac{{dy}}{{dx}} = - 2x \cr
& S = 2\pi \int_a^b {r\left( x \right)\sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} } dx,{\text{ }}r\left( x \right) = x \cr
& {\text{Then,}} \cr
& S = 2\pi \int_0^3 {x\sqrt {1 + {{\left( { - 2x} \right)}^2}} } dx \cr
& S = 2\pi \int_0^3 {x\sqrt {1 + 4{x^2}} } dx \cr
& S = \frac{{2\pi }}{8}\int_0^3 {8x\sqrt {1 + 4{x^2}} } dx \cr
& S = \frac{\pi }{4}\int_0^3 {\left( {8x} \right)\sqrt {1 + 4{x^2}} } dx \cr
& {\text{Integrate}} \cr
& S = \frac{\pi }{4}\left[ {\frac{{{{\left( {1 + 4{x^2}} \right)}^{3/2}}}}{{3/2}}} \right]_0^3 \cr
& S = \frac{\pi }{6}\left[ {{{\left( {1 + 4{x^2}} \right)}^{3/2}}} \right]_0^3 \cr
& S = \frac{\pi }{6}\left[ {{{\left( {37} \right)}^{3/2}} - 1} \right] \cr
& S = \frac{\pi }{6}\left( {37\sqrt {37} - 1} \right) \approx 117.3187 \cr} $$
