I'm having trouble proving the following proof:
If a volume in 3d space occupied by the solid body "V" has reflection symmetry with the xy-plane, and the spread of mass in the solid body makes it so that $\delta(x,y,z) = \delta(x,y,-z).$ Prove the center of mass must be on the plane $z = 0$. Quick note: when a region of space has reflection symmetry with th xy-plane, it means that if point P(x,y,z) is in the region, so is P*(x,y,-z). Also,$\delta(x,y,z)$ is the density (mass per unit volume) of an object occupying a region D in space.
Some things I have done: when the problem says the z=0 plane, that is the same as the xy plane. So we can rephrase the problem to proving the center of mass must be on the xy plane. Because "V" has reflection symmetry and $\delta(x,y,z) = \delta(x,y,-z)$, I'm thinking that the opposite masses of P(x,y,z) and P*(x,y,-z) would cancel eachother out and lie in the plane that is equidistant to both points, and in this case it would be the xy plance which is z=0.
Yup, that's the intuitive idea. The more formal way of showing it is to prove directly from the integral definition and the change of variables formula for multi-variable integrals that $z_{CM} = 0$. Now, recall that by definition, if $\mu = \int_V \delta(x,y,z)\, dx \, dy \, dz $ denotes the total mass of the object then \begin{align} z_{CM} &:= \dfrac{1}{\mu} \int_V z \cdot \delta(x,y,z) \, dx \, dy \, dz \end{align} Let's call $f(x,y,z) = z \cdot \delta(x,y,z)$. Now, we consider the change of variables function $\phi(x,y,z) = (x,y,-z)$. Note that $|\det D \phi| = |(1)(1)(-1)| = 1$, and \begin{align} (f \circ \phi)(x,y,z) = f(x,y,-z) = (-z)\cdot\delta(x,y,-z) = -z \cdot\delta(x,y,z) = - f(x,y,z) \end{align} Hence, $f \circ \phi = -f$. Also, note that by assumption, $\phi[V] = V$ (i.e the volume is symmmetric about the $xy$ plane). Therefore, \begin{align} z_{CM} &= \dfrac{1}{\mu}\int_V f \\ &= \dfrac{1}{\mu}\int_{\phi[V]} f \tag{since $\phi[V] = V$}\\ &=\dfrac{1}{\mu}\int_{V} (f \circ \phi) \cdot \left| \det D \phi\right| \\ &= \dfrac{1}{\mu} \int_V -f \cdot 1 \\ &= - z_{CM} \end{align} It follows that $z_{CM} = 0$.