Mathematica How do I calculate an area in Mathematica?

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To calculate the area of a region defined by inequalities in Mathematica, particularly the intersection of a hollow sphere and a solid toroid, one suggested approach is to randomly generate points on the sphere's surface. By checking how many of these points fall within the toroidal surface, a fraction can be determined, which can then be multiplied by the sphere's total area to estimate the area of the intersection region. It is crucial to ensure that the points are uniformly distributed across the sphere to avoid inaccuracies; simply selecting random latitudes and longitudes can lead to a denser point distribution near the poles, skewing the results. This method, involving approximately one million points, provides a practical approximation for the area of the specified region.
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Hi!

Say I have a region described by any number of inequalities. This region is a surface in 3D space. How can I ask Mathematica to calculate the region's area?

If it helps, my particular region is the intersection of a hollow sphere and a solid (i.e. filled-out) toroid-like surface. I'm using Mathematica 9, but I think I can find a Mathematica 8 .dmg file and my old license key if necessary.

Thanks!
 
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Since no one has offered any idea and perhaps you might still see this...

Suppose you randomly generate a point on the surface of your sphere.
Then check to see whether that point is inside the toroid like surface.
Repeat that perhaps 10^6 times. See what fraction of those are inside.
Does that fraction tell you something interesting or useful?

Note: If you did not have a hollow sphere to start with then this might
be more complicated. You might think about why that might be.
 
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Thanks for the suggestion. With a million points, that would make a very good approximation. (Obviously, the area of my region would be the sphere's area times the fraction of points inside the torus.)

Cheers!
 
Wonderful. Just be very careful with that random generation so that the points are uniformly distributed across the sphere. Doing something like randomly choosing a latitude and then randomly choosing a longitude would not make the points uniformly distributed and would lead to an inaccurate estimate of the area.
 
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Ah, thanks for the advice. It's perfectly reasonable now that you've mentioned it, yet I would have probably gone and done just that --randomly select a latitude and a longitude-- without realising that the point distribution would then be denser near the poles. I'll be more careful than that with my random-point selection.

Cheers!
 
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