How do I calculate an area in Mathematica?

  • Context: Mathematica 
  • Thread starter Thread starter IridescentRain
  • Start date Start date
  • Tags Tags
    Area Mathematica
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 3K views
IridescentRain
Messages
16
Reaction score
0
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!
 
Physics news on Phys.org
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.
 
Last edited:
  • Like
Likes   Reactions: 1 person
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.
 
  • Like
Likes   Reactions: 1 person
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!