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Area of square on shpere

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Old   February 5, 2005, 03:44
Default Area of square on shpere
  #1
Jim K
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<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> <sub> 0 </sub>, <font face = "symbol"> l </font> <sub> N </sub>) x (<font face = "symbol"> q </font> <sub> 0 </sub>, <font face = "symbol"> q </font> <sub> M </sub>) is given. How to get an area of the square? Thanks.
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Old   February 6, 2005, 04:49
Default Re: Area of square on shpere
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Rami
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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.
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Old   February 6, 2005, 08:26
Default Re: Area of square on shpere
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Jim K
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Thanks you. I.e. we receive A=r(<font face="symbol">l</font> <sub>N</sub><font face="symbol">-</font> <font face="symbol">l</font> <sub>0</sub>)(cos<font face="symbol">q</font> <sub>0</sub> <font face="symbol">-</font> cos<font face="symbol">q</font> <sub>M</sub>). Simply I haven't found this formula in the handbook.

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Old   February 7, 2005, 00:08
Default Re: Area of square on shpere
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Jim K
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Erratum probably: r replace by r^2
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