Why are these three integrals all $0$? I have the following three triple integrals:
$$\iiint_S x \sigma \mathrm{d} V$$
$$\iiint_S y \sigma \mathrm{d} V$$
$$\iiint_S z \sigma \mathrm{d} V$$
where $\sigma = k \left(1 - (\rho / a)^3 \right)$, $\rho$ is the distance from the origin in spherical coordinates, and $S$ represents a sphere at the origin with radius $a$.
I can easily compute all three integrals to be $0$. However, I'm asked to make an argument based on the symmetry of the sphere, rather than actually calculating the integrals. I could, for example, substitute $x = \rho \cos \theta \sin \phi$ to calculate the first integral as 0, but what specifically about the symmetry of the sphere is causing this to be true?
 A: Consider a small piece of volume $\Delta V$ at position $(x,y,z)$. There exist a mirror piece at $(-x,y,z)$ with equal volume. The contribution of these two mass elements to the integral $\int_V x\sigma{\rm d}V$ is (see picture below)
$$x \sigma(
\rho) \Delta V + (-x)\sigma(
\rho) \Delta V = 0$$
since both of these elements have the same distance $\rho$ from the origin. Thus by symmetry the integral has to be zero. This is the same observations as made by stochasticboy321 in the comment above.
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This observation can be generalized: whenever the integration region is symmetric about the origin and the integrand is an odd function of either of the coordinates then $\int_V f(x,y,z){\rm d}V = 0$.
A: You could think of $\sigma$ as density of the sphere, and since the only variable in $\sigma = k \left(1 - (\rho / a)^3 \right)$ is $\rho$, the distance from a point to the origin, you can see that for any spherical shell inside the sphere the density is uniform(only for a spherical shell).
From the definition of centre of mass, your integrals represent the product of mass and the x,y,z coordinates of the centre of mass, respectively. From symmetry due to uniform density of spherical shells, we argue that the centre of mass is $(0,0,0)$ and hence all three integrals are zero.
