# Area of square on shpere

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 February 5, 2005, 03:44 Area of square on shpere #1 Jim K Guest   Posts: n/a Let Let (r, l , q ) is a spherical coordinate system, here r > 0, 0 < l < 2 p , 0 < q < p . The square (or a rectangle) on the surface ( l 0 , l N ) x ( q 0 , q M ) is given. How to get an area of the square? Thanks.

 February 6, 2005, 04:49 Re: Area of square on shpere #2 Rami Guest   Posts: n/a 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.

 February 6, 2005, 08:26 Re: Area of square on shpere #3 Jim K Guest   Posts: n/a Thanks you. I.e. we receive A=r(l N- l 0)(cosq 0 - cosq M). Simply I haven't found this formula in the handbook.

 February 7, 2005, 00:08 Re: Area of square on shpere #4 Jim K Guest   Posts: n/a Erratum probably: r replace by r^2

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