Answer
$$S = 2\pi \left[ {\sqrt 2 + \ln \left( {1 + \sqrt 2 } \right)} \right]$$
Work Step by Step
$$\eqalign{
& y = \sin x,{\text{ 0}} \leqslant x \leqslant \pi \cr
& {\text{The area of surface is given by the formula}} \cr
& S = \int_a^b {2\pi f\left( x \right)} \sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} dx \cr
& {\text{,then}} \cr
& S = 2\pi \int_0^\pi {\left( {\sin x} \right)} \sqrt {1 + {{\cos }^2}x} dx \cr
& {\text{Let }}u = \cos x,{\text{ }}du = - \sin xdx, \cr
& {\text{The new limits of integration are:}} \cr
& x = 0 \Rightarrow u = 1 \cr
& x = \pi \Rightarrow u = - 1 \cr
& S = - 2\pi \int_1^{ - 1} {\sqrt {1 + {u^2}} } du \cr
& S = 2\pi \int_{ - 1}^1 {\sqrt {1 + {u^2}} } du \cr
& {\text{Integrate by the formula }} \cr
& \int {\sqrt {{u^2} + {a^2}} du = \frac{u}{2}\sqrt {{u^2} + {a^2}} + \frac{{{a^2}}}{2}\ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C} \cr
& S = 2\pi \left[ {\frac{u}{2}\sqrt {1 + {u^2}} + \frac{1}{2}\ln \left( {u + \sqrt {1 + {u^2}} } \right)} \right]_{ - 1}^1 \cr
& S = 2\pi \left[ {\frac{1}{2}\sqrt 2 + \frac{1}{2}\ln \left( {1 + \sqrt 2 } \right)} \right] - 2\pi \left[ { - \frac{1}{2}\sqrt 2 + \frac{1}{2}\ln \left( {1 + \sqrt 2 } \right)} \right] \cr
& {\text{Simplifying}} \cr
& S = 2\pi \left[ {\sqrt 2 + \ln \left( {1 + \sqrt 2 } \right)} \right] \cr} $$
