# Thread: [SOLVED] triple integral issue

1. ## [SOLVED] triple integral issue

I need some help with a triple integral problem that I keep getting wrong in the last step somehow.

the problem is:

Evaluate $\int\int\int$ on E xyz dV where E lies between the spheres $\rho$=2 and $\rho$=4 and above the cone $\phi$= $\pi$/3.

There work I've done so far is set up the integral integrated it and got:

1/4 sin^4 $\phi$ from 0 to $\pi$ * 1/2sin^2 $\theta$ from 0 to 2 $\pi$ * 1/6 $\rho$^6 from 2 to 4.

I get that the middle term should be zero, but for some reason my prof gives the answer to be
(1562 $\pi$)/15

2. That is a large value your professor got. I am not so sure he/she is correct.

3. That's what I was thinking as I am pretty sure the answer should be zero, but I thought that it wouldn't hurt to check.

4. Consider symmetry. The region of integration is rotationally symmetric about the z axis, and so if we rotate the function being integrated about the z axis, the integral should be unchanged. Now, let us rotate the integrand by 90 degrees about the z axis: we transform x and y to x' and y' via x'=-y and y'=x. Then the integrand is f(x',y',z)=xyz=(y')(-x')z=-x'y'z.
Thus, we have $\iiint_{E}xyz\,dx\,dy\,dz=\iiint_{E'}-x'y'z\,dx'\,dy'\,dz$. But, if we rename the variables x' and y' as x and y, (and as E'=E), we have
$\iiint_{E}xyz\,dx\,dy\,dz=-\iiint_{E}xyz\,dx\,dy\,dz$
which means $\iiint_{E}xyz\,dx\,dy\,dz=0$.
Thus, due to the symmetry, the integral is zero.
[We can also look at this as the fact that xyz's dependence on the azimuthal angle $\theta$ is of the form $\sin(2\theta)$, and the region has no dependence on the azimuthal angle (the rotational symmetry), so the integral over $\theta$ will be 0]

--Kevin C.

5. Thanks everyone. I got an e-mail back from my professor and he said that it was a typo. Thanks again!