surface area of a slice of a hemisphere

Question

Imagine a hemisphere on it's base in the horizontal plane (center at the origin). Imagine another plane P which passes through the center (origin) and inclined at alpha to the horizontal (the base of the hemisphere). Now, if the plane P slices the hemisphere what would be the surface areas of the two pieces, in terms of the total surface area of the hemisphere?

Answer 1

The total surface area $ A$ of a hemisphere of a sphere with the radius $ R $ is given by $ A=3\pi R^2 $. Equating ratios of angles $\alpha $ to $\pi$ and external surface area of the first slice to the external surface area of a hemisphere, we get that the total surface area $ A_1 $ of a first slice is then given by $ A_1=(2\alpha +\pi) R^2$ and the total surface area $ A_2 $ of a second slice is given by $ A_2=(3\pi - 2\alpha) R^2$. Given these conditions, we can deduce that $ A_1=\frac {A}{3}(\frac {2\alpha }{\pi}+1) $ and $ A_2=A (1-\frac {2\alpha}{3}) $.