## Re: Volume and Area of a spherical cap

Originally Posted by Walagaster
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$$S = \int_0^{2\pi}\int_0^{\arccos\frac {r - h}r}r^2\sin\phi~d\phi d\theta$$
Originally Posted by Vinod
Hello,
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.