How can we show that $I_{yy}=I_{zz}=I_{xx}$ and $I_{xy}=0$ using symmetry arguments?

Question

Consider two integrals of the form $$I_{xx}=\int x^2 f(r)d^3r,~~I^\prime=\int xy ~f(r)d^3r$$ where $f(r)$ is a function of $r=|\vec{r}|$ only and has no dependence on $\theta,\phi$ in pherical polar coordinates. I have two questions. Can we show that $I_{yy}=I_{zz}=I_{xx}$ and $I_{xy}=I_{yz}=I_{zx}=0$ using symmetry arguments?

Answer 1

Since $f = f(|{\bf r}|)$ it does not depend on the angles $(\theta, \phi)$, so at a given radius $f$ is a constant over the sphere defined by such radius. Now consider the term $x$, over the same sphere, for each $x$ there will be a $-x$, so for each $x f$ there will be a a $-x f$, the result after adding them up all together is zero. You can extend the argument to show that

$$ I_{ij} = \int x_i x_j f(r) {\rm d}^3{\bf r} = 0 ~~~\mbox{for}~~i\not= j $$

And from this is also easy to see why $I_{ii} \not = 0$

Answer 2

$f(r)$ has two relevant symmetries: rotational and reflection. From the rotational symmetry we can conclude that $\int x^n f(r) d^3r=\int y^n f(r) d^3r=\int z^n f(r) d^3r$. explicitly $f(\sqrt{x^2+y^2+z^2})=f(\sqrt{y^2+x^2+z^2})$ etc (in fact we didn't use the rotational symmetry but a permutation symmetry).

From the reflection symmetry we conclude that $\int xy f(r) d^3r=\int (-x)y f(r) d^3r=-\int xy f(r) d^3r$ so that $I_{xy}=-I_{xy}=0$, since $f(\sqrt{x^2+y^2+z^2})=f(\sqrt{(-x)^2+y^2+z^2}) $ etc.