I was trying to help a dude with the inertia tensor and I couldn't explain quite well why the components $I_{xy} = I_{yx} = 0$ when we have symmetry around the $z$-axis.
We were talking about a planar body on the $xy$ plane, symmetric around $z$-axis, with the general expression of the tensor being
$I = \pmatrix{I_{xx} && I_{xy} && I_{xz} \\ I_{yx} && I_{yy} && I_{yz} \\ I_{zx} && I_{zy} && I_{zz}} = \pmatrix{\int_S \sigma(r^2-x^2) \: dS && -\int_S \sigma xy \: dS && -\int_S \sigma xz \: dS \\ -\int_S \sigma xy \: dS && \int_S \sigma(r^2-y^2) \: dS && -\int_S \sigma yz \: dS \\ -\int_S \sigma xz \: dS && -\int_S \sigma yz \: dS && \int_S \sigma(r^2-z^2) \: dS}$,
with $r^2 = x^2 + y^2 + z^2$ being the position of some point in the body. Being $z = 0$, then
$I = \pmatrix{\int_S \sigma(r^2-x^2) \: dS && -\int_S \sigma xy \: dS && 0 \\ -\int_S \sigma xy \: dS && \int_S \sigma(r^2-y^2) \: dS && 0 \\ 0 && 0 && \int_S \sigma r^2 \: dS}$.
So how can I see, mathematically, that the symmetry argument makes $I_{xy}$ and $I_{yx}$ vanish? Apparently this holds to any planar body with symmetry around some axis, and I wish to know how the integral of $xy$ is zero.
Thanks in advance!
Recall that the moment-of-inertia tensor of a three-dimensional domain $V$ can be expressed as
$$\mathbf I=\int_{V}\rho(\boldsymbol x)\big(|\boldsymbol x|^2\boldsymbol \delta -\boldsymbol x\otimes \boldsymbol x\big)\mathrm d^3\boldsymbol x$$
Where $\boldsymbol \delta$ denotes the identity tensor, $\delta_{ij}=\begin{cases}1 & i=j \\ 0 & i\neq j\end{cases}$.
Let's suppose that our density is constant ($\rho(\boldsymbol x)\equiv \rho=\text{constant}$ and equal to $1$, say), thus
$$\mathbf I=\int_{V}\big(|\boldsymbol x|^2\boldsymbol \delta -\boldsymbol x\otimes \boldsymbol x\big)\mathrm d^3\boldsymbol x\tag{1}$$
Let's assume also that our domain $V$ has rotational symmetry, which we can rigorously define as meaning that $V$ is invariant under rotations on the $x_3$ axis, i.e,
$$V=\operatorname{image}(V,\mathbf R)\equiv\left\{\mathbf R\boldsymbol x \mid \boldsymbol x\in V\right\}\equiv \mathbf R[V]$$ Where $$\mathbf R=\begin{bmatrix}\cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}\in\mathrm O(3)$$
Is a rotation of angle $\alpha$ around the $x_3$ axis. (The above should be true for all $\alpha$ !) This means that
$$\mathbf I=\int_V \big(|\boldsymbol x|^2\boldsymbol \delta -\boldsymbol x\otimes \boldsymbol x\big)\mathrm d^3\boldsymbol x=\int_{\mathbf R[V]}\big(|\boldsymbol x|^2\boldsymbol \delta -\boldsymbol x\otimes \boldsymbol x\big)\mathrm d^3\boldsymbol x$$ Which, by variable substitution, is equal to $$\mathbf I=\int_{V}\big(|\mathbf R\boldsymbol x|^2\boldsymbol \delta -(\mathbf R\boldsymbol x)\otimes(\mathbf R\boldsymbol x)\big)\mathrm d^3\boldsymbol x$$
Since $\mathbf R\in\mathrm O(3)$, we know that $|\mathbf R\boldsymbol x|^2=(\mathbf R\boldsymbol x)\cdot(\mathbf R\boldsymbol x)=\boldsymbol x\cdot \boldsymbol x=|\boldsymbol x|^2$ and hence
$$\mathbf I=\int_V\big(|\boldsymbol x|^2\boldsymbol \delta -(\mathbf R\boldsymbol x)\otimes(\mathbf R\boldsymbol x)\big)\mathrm d^3\boldsymbol x\tag{2}$$
For convenience, let's now define $$\mathbf J=\int_V\boldsymbol x\otimes \boldsymbol x~\mathrm d^3\boldsymbol x$$
$(1)$ implies that
$$I_{11}=J_{22}+J_{33} \\ I_{22}=J_{11}+J_{33} \\ I_{12}=I_{21}=-J_{12}=-J_{21}$$
On the other hand, $(2)$ implies that
$$I_{11}=\big(1-\cos(\alpha)^2\big)J_{11}+\big(1-\sin(\alpha)^2\big)J_{22}+J_{33}+\sin(2\alpha)J_{12} \\ I_{22}=\big(1-\sin(\alpha)^2\big)J_{11}+\big(1-\cos(\alpha)^2\big)J_{22}+J_{33}-\sin(2\alpha)J_{12} \\ I_{12}=\frac{1}{2}\sin(2\alpha)J_{11}-\frac{1}{2}\sin(2\alpha)J_{22}+\cos(2\alpha)J_{12}$$
Substituting in for $J_{11},J_{22},J_{12}$, we get
$$I_{11}=\big(1-\cos(\alpha)^2\big)(I_{22}-J_{33})+\big(1-\sin(\alpha)^2\big)(I_{11}-J_{33})+J_{33}+\sin(2\alpha)(-I_{12}) \\ I_{22}=\big(1-\sin(\alpha)^2\big)(I_{22}-J_{33})+\big(1-\cos(\alpha)^2\big)(I_{11}-J_{33})+J_{33}-\sin(2\alpha)(-I_{12}) \\ I_{12}=\frac{1}{2}\sin(2\alpha)(I_{22}-J_{33})-\frac{1}{2}\sin(2\alpha)(I_{11}-J_{33})+\cos(2\alpha)(-I_{12})$$
Simplifying, $$0=-\sin(\alpha)^2I_{11}+\big(1-\cos(\alpha)^2\big)I_{22}+\sin(2\alpha)I_{12} \\ 0=\big(1-\cos(\alpha)^2\big)I_{11}-\sin(\alpha)^2I_{22}-\sin(2\alpha)I_{12} \\ 0=-\frac{1}{2}\sin(2\alpha)I_{11}+\frac{1}{2}\sin(2\alpha)I_{22}-(1+\cos(2\alpha))I_{12}$$
Or, in matrix form, $$\begin{bmatrix}-\sin(\alpha)^2 & \sin(\alpha)^2 & \sin(2\alpha) \\ \sin(\alpha)^2 & -\sin(\alpha)^2 & -\sin(2\alpha) \\ -\frac{1}{2}\sin(2\alpha) & \frac{1}{2}\sin(2\alpha)& \big(1+\cos(2\alpha)\big)\end{bmatrix}\begin{bmatrix}I_{11} \\ I_{22} \\ I_{12}\end{bmatrix}=\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$$
Applying Gauss-Jordan elimination, this is equivalent to $$\begin{bmatrix}1 & -1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}I_{11} \\ I_{22} \\ I_{12}\end{bmatrix}=\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$$
In other words, the only possible solutions are when $I_{11}=I_{22}$ (which is intuitively obvious), and more importantly, that $I_{12}=0$.
QED.