<title> Let </title> Let (r, <font face = "symbol"> l </font>, <font face = "symbol"> q </font>) is a spherical coordinate system, here r > 0, 0 < <font face = "symbol"> l </font> < 2 <font face = "symbol"> p </font>, 0 < <font face = "symbol"> q </font> < <font face = "symbol"> p </font>. The square (or a rectangle) on the surface (<font face = "symbol"> l </font> ₀, <font face = "symbol"> l </font> _N) x (<font face = "symbol"> q </font> ₀, <font face = "symbol"> q </font> _M) is given. How to get an area of the square? Thanks.

For constant r (i.e., a spherical surface), the area is given by integation over the polar angle, theta, and the azimuthal angle, phi, of

dA = r*sin(theta)d(theta)*d(phi)

which results, for your case, in

A = r * delta(phi) * delta(-cos(theta))

where delta(f) is the difference fN-f0.

Thanks you. I.e. we receive A=r(<font face="symbol">l</font> _N<font face="symbol">-</font> <font face="symbol">l</font> ₀)(cos<font face="symbol">q</font> ₀ <font face="symbol">-</font> cos<font face="symbol">q</font> _M). Simply I haven't found this formula in the handbook.

Erratum probably: r replace by r^2