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 $S_y=2\pi \int_c^d xds = 2\pi \int_{r-h}^{r} xds= 2\pi \int_{r-h}^{r}x*\frac{r}{x}dy=2\pi r|y|_{r-h}^{r}$
$S_y=2\pi r[r-(r-h)]=2\pi rh$