# Dividing hemisphere in equally-sized patches

I want to divide a sphere in equally-sized patches, meaning, the surface of all the patches needs to be the same. The way I am doing it is to divide the radius in $$n$$ parts, and the surfaces contained between the planes dividing the radius should be the same. I will post an example:

My question is: in the picture, is it true that the surface of the red sector is equal to the green sector? (the two sectors cover 360 degrees). This should apply if I divide the radius by $$n$$ parts.

My calculus

My answer is yes, but I want to be sure it is correct. When r = 0.5*R, $$\theta$$ $$\approx$$ 0.52 rad
When r = R, $$\theta$$ $$\approx$$ 1.57 rad
$$\int_0^{0,52} 2 \pi R^2 cos\theta d\theta$$ $$\int_{0,52}^{1,57} 2 \pi R^2 cos\theta d\theta$$
Solving these two integrals, we can prove they are equal. This means that, if I divide the sectors vertically in $$m$$ parts equally distantiated, all the patches would have the same surface.

Note that $${\rm Area} \ \{ x^2+y^2=1,\ t_1 for $$-1
• If the polar angle $\theta$ would be subdivided into a number of symmetric intervals (with respect to the plane $\theta=0$), is it possible to choose such azimuthal angles $\phi$, so that the resulting points on the sphere will be have 90 degree symmetry? Basically make all 8 quadrants have the same distribution of points on them. Commented Aug 31, 2019 at 15:09