Quote Originally Posted by Walagaster View Post
In spherical coordinates, if the radius of the sphere is $\rho = r$, the element of surface area is $dS = r^2\sin\phi~d\phi d\theta$. So for the surface area of the cap try$$
S = \int_0^{2\pi}\int_0^{\arccos\frac {r - h}r}r^2\sin\phi~d\phi d\theta$$
Quote Originally Posted by Vinod View Post
The equation of the right semicircle $x=\sqrt{r^2-y^2},\frac{dx}{dy}=\frac{-y}{\sqrt{r^2-y^2}}=\frac{-y}{x}$

$ds=\sqrt{1+(\frac{dx}{dy})^2}dy =\sqrt{1+(\frac{-y}{x})^2}dy,= \frac{\sqrt{x^2+y^2}}{x}dy, =\frac{r}{x}dy$

Then $S_y=2\pi \int_c^d xds = 2\pi \int_{r-h}^{r} xds= 2\pi \int_{r-h}^{r}x*\frac{r}{x}dy=2\pi r|y|_{r-h}^{r}$

$S_y=2\pi r[r-(r-h)]=2\pi rh$
Right. Now how about trying the formula in spherical coordinates? You might like how it works out without all the square roots and derivatives. It is actually the natural way to set up this problem.