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I am unsure when to use symmetry with triple integrals.

Can I use symmetry for this centre of mass question?

$E$ is bounded by the parabolic cylinder $z=1-y^2$ and the planes $x+6z=6, x=0$, and $z=0 ; \quad \rho(x, y, z)=8$

My set up at the moment wo using symmetry is as follows:

$M_{y z}=\int_{y=-1}^1 \int_{z=0}^{1-y^2} \int_{x=0}^{6-6 z} 8 x d x d z d y$

$M_{x z}=\int_{y=-1}^1 \int_{z=0}^{1-y^2} \int_{x=0}^{6-6 z} 8 y d x d z d y$

$M_{x y}=\int_{y=-1}^1 \int_{z=0}^{1-y^2} \int_{x=0}^{6-6 z} 8 z d x d z d y$

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    $\begingroup$ The second integral has odd symmetry. You can safely say it cancels. $\endgroup$
    – Ninad Munshi
    Commented May 27 at 12:14
  • $\begingroup$ Ok thanks. When solving volume or density functions are there times when not to using consider symmetry. $\endgroup$
    – user41592
    Commented May 27 at 13:24
  • $\begingroup$ You need to examine the symmetry of both the region and the density function. If the region and density are both symmetric about an axis, then the center of mass will lie on that axis. (You can deduce this from usual properties of the integral of an odd function over an interval $[-a,a]$ in one-variable calculus.) $\endgroup$
    – Ted Shifrin
    Commented May 27 at 18:26

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