Consider a function $f:\mathbb{R}^3\to \mathbb{R}$ which is odd in one variable. If $f(x,y,z)$ is odd in $x$, for example and without loss of generality, then $f(x,y,z)=-f(-x,y,z)$.
Consider the integral of $f$ over a spherical surface $S$ centered at the origin with arbitrary radius, i.e, $$ \iint_S f(x,y,z)\,dS. $$ How do I show, by symmetry, that the integral is $0$?
My attempt: Since $f$ is odd, and considering the change of variables $t=-x$, thus $dx=-dt$, we have that $$ I=\iint_S f(x,y,z)\,dS=-\iint_S f(-x,y,z)\,dS=-\iint_S \left(-f(t,y,z)\right)\,dS. $$ Then, using the fact that $f$ is odd, again, we get $$ -\iint_S \left(-f(t,y,z)\right)\,dS\equiv -\iint_S f(x,y,z)\,dS=-I. $$ Thus, the only solution to $I=-I$ is $I=0$. Is this correct? Is this change of variables what is meant by 'symmetry'?
We can consider the two hemispheres $S_1$ and $S_2$ separeted by $y-z$ plane then
$$I=\iint_S f(x,y,z)\,dS=\iint_{S_1} f(x,y,z)\,dS+\iint_{S_2} f(x,y,z)\,dS$$
and by $t=-x$
$$\iint_{S_2} f(x,y,z)\,dS=\iint_{S_1} f(-t,y,z)\,dS=-\iint_{S_1} f(t,y,z)\,dS$$