Finding the Volume of x^4+y^4+z^4=1: A Challenge!

[hypermonkey2]

Homework Statement

What would be the most efficient way to find the volume of the solid x^4+y^4+z^4=1?

Homework Equations

The Attempt at a Solution

Cylindrical and spherical coordinates end up messy with integrals that cannot be computed by hand. I am at a loss to find something that will work in the long run! Thanks!

 

[Avodyne]

Spherical coords. You have to integrate powers of sines and cosines; look these up somewhere. Or integrate by parts. Or use e.g. cos(x)=(1/2)(e^ix + e^-ix) and expand.

 

[hypermonkey2]

So i just let x=rcos(o)sin(phi) etc etc? i wouldn't need to try something like x^2=rcos(o)sin(phi)?
Because i don't see how to find limits for r in the first case...

 

[hypermonkey2]

in either case, the jacobian gives me an expression that i can only integrate with mathematica using error functions or elliptical integrals... I am starting to think that this is impossible!

 

[hypermonkey2]

but there must be a way to do this... no one has any ideas?

 

[Avodyne]

Oops, sorry, it's harder than I thought. Cylindrical coords look like your best bet. Do the z integral first (easy), then the rho integral (Mathematica will do it), and finally the phi integral (ditto).

I got (pi^2 Gamma[1/4])/(3 Gamma[3/4]^3) = 6.48, which is a reasonable number (between a cube of edge length 2 and a sphere of diameter 2).

EDIT: I also got the same answer with rectangular coords, which is probably even easier.