# Evaluating Triple integral

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

saltydog said:
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