Surface area in spherical co-ordinates

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SUMMARY

The discussion focuses on calculating the surface area in spherical coordinates to determine the fraction of total radiation emitted by a point source within specified elevation angles (phi) and azimuth angles (theta). The solution involves performing a double integration over phi and theta, utilizing the function dxdydz = r² dr du cos(v) dv, where r represents the radial direction, u is the longitude, and v is the latitude. The user seeks assistance in adapting their knowledge of Cartesian coordinates to solve this problem in spherical coordinates.

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  • Understanding of spherical coordinates and their applications
  • Familiarity with double integration techniques
  • Knowledge of radiant intensity and radiation emission concepts
  • Proficiency in calculus, specifically surface integrals
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  • Study the conversion between Cartesian and spherical coordinates
  • Learn about surface integrals in spherical coordinates
  • Research the application of double integrals in physics
  • Explore examples of radiant intensity calculations in spherical systems
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Physicists, engineers, and students involved in radiation analysis, particularly those working with spherical coordinate systems and surface integrals.

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I have a generalised function in spherical polar co-ordinates that describes the radiant intensity in any given direction of a point source emitter. I need to calculate the fraction of the total radiation that is emitted into a region of space specified by some min to max range of elevation angles phi and a min to max range of azimuth angles theta.

I know that the solution to the problem is essentially finding the surface area of the function by a double integration over phi and theta. However although I know how to solve a generalised surface integral in cartesian co-ordinates I'm not getting anywhere trying to solve this is spherical co-ordinates. Any help would be appreciated.
 
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dxdydz=r2drducosvdv, where r is the radial direction, u is "longitude" (full circle), and v is "latitude" (-π/2 < v < π/2). You could replace cosvdv by sinwdw for 0 < w < π.
 

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