The surface area of the circular disk is then \pi \left(r\frac{(h-x)}{h}\right)^2 = \pi r^2\frac{(h-x)^2}{h^2}.
Integrating gives us \frac{\pi r^2}{h^2}\left(\frac{h^3}{3}\right) = \frac{1}{3}\pi r^2 h.