The sum of the circumferences of infinitely many, infinitesimally thin (dx), circles should give us the surface area of this solid. As always, when we integrate a nth-dimensional quantity (in this case, the 1-d circumference), we get an n+1st answer (in this case, the 2-d area):
SA = ∫0π 2πsinxdx = -2πcosx]0π = 4π (Note: the height of the function represents the radius of these circles.)