How Do You Simplify the Triple Integral of |xyz| Over an Ellipsoid?

[MattL]

I'm having trouble with evaluating

[Triple Integral] |xyz| dx dy dz

over the region (x/a)^2 + (y/b)^2 + (z/c)^2 <= 1

Do I need to use some sort of parametrisation for the region, and is there some way of dealing with the absolute value function without integrating over the eight octants?

Whilst I've separated the integral into the product of three integrals, I'm not sure if this actually helps?

 

[dextercioby]

Well,the function is even and the domain of integration is symmetric wrt the origin,so that would give u a hint upon the limits of integration.The symmetry of the ellipsoid is really useful.

As for the parametrization,i'm sure u'll find the normal one

x=a\cos\varphi\sin\vartheta

y=b\sin\varphi\sin\vartheta

z=c\cos\vartheta

pretty useful.

Daniel.

 

[saltydog]

This integral looks like it's zero with all the symmetry: you know, four positives and four negatives for the integrand. Not sure though as I can't evaluate it. Would like to know though.

 

[arildno]

This integral looks like it's zero with all the symmetry: you know, four positives and four negatives for the integrand. Not sure though as I can't evaluate it. Would like to know though.

I think you missed the absolute value sign on the integrand..

 

[saltydog]

I think you missed the absolute value sign on the integrand..

Well . . . no, that's the reason I used for the symmetry but again, I qualify my statements by the fact I can't prove it. For example in the first octant:

|xyz|=xyz

That's a positive one.

However, in the octant with x<0, y>0 and z>0 we have:

|xyz|=-xyz

And so forth in the 8 octants leaving 4 positive and 4 negative ones integrated symmetrically (I think).

 

[arildno]

The integrand is positive almost everywhere; hence, the integral is strictly positive:
Let:
x=ar\sin\phi\cos\theta,y=br\sin\phi\sin\theta,z=cr\cos\phi
0\leq{r}\leq{1},0\leq\theta\leq{2}\pi,0\leq\phi\leq\pi
Thus, we may find:
dV=dxdydz=abcr^{2}\sin\phi{dr}d\phi{d}\theta
|xyz|=\frac{abcr^{3}}{2}\sin^{2}\phi|\cos\phi\sin(2\theta)|
Doing the r-integrations yield the double integral:
I=\frac{(abc)^{2}}{12}\int_{0}^{2\pi}\int_{0}^{\pi}\sin^{3}\phi|\cos\phi\sin(2\theta)|d\phi{d}\theta
We have symmetry about \phi=\frac{\pi}{2}; thus we gain:
I=\frac{(abc)^{2}}{24}\int_{0}^{2\pi}|\sin(2\theta)|d\theta
We have four equal parts here, and using the part 0\leq\theta\leq\frac{\pi}{2} yields:
I=\frac{(abc)^{2}}{12}

 

[saltydog]

Thanks Arildno. MattL, hope I didn't get in your way. I'll go through it to make sure I understand it.

 

[MattL]

Thanks Arildno. MattL, hope I didn't get in your way. I'll go through it to make sure I understand it.

No problem.

Think I should be able to give this question a fair go now.

 

[Ray Vickson]

Change variables to x = a*x1, y = b*x2, z = c*x3. The integration region is the unit ball x1^2 + x2^2 + x3^2 <= 1, the integrand is abc*|x1 x2 x3|, and dV = abc * dx1 dx2 dx3. Because of the absolute value and symmetry, the whole integral, I, equals 8 times the integral over the {x1,x2,x3 >= 0} portion of the ball. This gives I = 8abc*int_{x3=0..1} f(x3) dx3, where f(x3) = x3*int_{x1^2 + x2^2 <= 1-x3^2} x1 x2 dx1 dx2. Using polar coordinates (or first integrating over x2 for fixed x1, then integrating over x1) we can easily evaluate f(x3), then integrate it over x3 = 0-->1. The final result is I = abc/6.

R.G. Vickson