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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:
Hemisphere

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.

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Note that $${\rm Area} \ \{ x^2+y^2=1,\ t_1<z<t_2\} = {\rm Area}\ \{ x^2+y^2+z^2=1,\ t_1<z<t_2\}$$ for $-1<t_1<t_2<2$

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  • $\begingroup$ 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. $\endgroup$
    – Alexander Cska
    Commented Aug 31, 2019 at 15:09

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