If you take the derivative of the volume of a sphere you get the surface area. f(r)=(4/3)Pi*r^3 f'(r)=4Pi*r^2 Does this also apply to a spheroid/ellipsoid?
you have to take partial of ellipsoid since it depends on more than one variable
I don't understand what you mean with "take partial of ellipsoid"
Partial derivatives are used when there are more than one variable in a function
Well, before you even get there, what's the formula for the volume of an ellipsoid in terms of r?
is there r in ellipsoid ?
precisely.
yep,depends on the choice of coordinates, spherical coordinates would employ, r, and \[\phi\]
\[\rho\]
Well i didn't say that there was a 'r', the sphere was just an example.
What I want to point out to you is the r in the equation for a sphere corresponds to the physical distance r, which is the the r of spherical coordinates. But for a ellipsoid, the expression for volume does not include r, or at least it doesn't in its canonical form. So unfortunately this method doesn't work. It would be nice if it did, because the surface area of an ellipsoid would be quite simple, like that of a sphere. But if you try and calculate the surface area of an ellipsoid directly, you'll find it's slightly complex and involving integrating functions that don't have anti-derivatives in terms of elementary functions.
Hmm. This is going to be more difficult than I though it would. My friend and I want to find the surface of a triangle on a spheroid, the points are randomly chosen. Do you have any kind of tips?
ah, that's a different thing, and quite doable ... http://mathworld.wolfram.com/SphericalTriangle.html
Thank you
Join our real-time social learning platform and learn together with your friends!