Answer
$$S = \frac{{8\left( {3\sqrt 3 - 2\sqrt 2 } \right)}}{3}\pi $$
Work Step by Step
$$\eqalign{
& x = 2\sqrt {1 - y} ,\,\,\,\, - 1 \leqslant y \leqslant 0 \cr
& {\text{Calculate }}\frac{{dx}}{{dy}} \cr
& \frac{{dx}}{{dy}} = 2\left( {\frac{{ - 1}}{{2\sqrt {1 - y} }}} \right) = - \frac{1}{{\sqrt {1 - y} }} \cr
& {\text{Use }}S = \int_a^b {2\pi x\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} } dy \cr
& S = \int_{ - 1}^0 {2\pi \left( {2\sqrt {1 - y} } \right)\sqrt {1 + {{\left( { - \frac{1}{{\sqrt {1 - y} }}} \right)}^2}} } dy \cr
& S = 4\pi \int_{ - 1}^0 {\left( {\sqrt {1 - y} } \right)\sqrt {1 + \frac{1}{{1 - y}}} } dy \cr
& S = 4\pi \int_{ - 1}^0 {\left( {\sqrt {1 - y} } \right)\sqrt {\frac{{1 - y + 1}}{{1 - y}}} } dy \cr
& S = 4\pi \int_{ - 1}^0 {\sqrt {2 - y} } dy \cr
& {\text{Integrate}} \cr
& S = - 4\pi \left[ {\frac{{2{{\left( {2 - y} \right)}^{3/2}}}}{3}} \right]_{ - 1}^0 \cr
& S = - \frac{8}{3}\pi \left[ {{{\left( {2 - 0} \right)}^{3/2}} - {{\left( {2 + 1} \right)}^{3/2}}} \right] \cr
& S = \frac{{8\left( {3\sqrt 3 - 2\sqrt 2 } \right)}}{3}\pi \cr} $$
