Suppose we have a minimization problem $F(u(x))=\int_A \sqrt{1+|\nabla u|^2}dA$, where $A$ is a ring such that $1<|x|<2$. After going to the polar coordinates we obtained: $$ F(\varphi(r,\theta))=\int_A \sqrt{1+\varphi_r^2+\frac{1}{r^2}\varphi_\theta^2}\ rdrd\theta $$ and now because of the symmetry the solution is independent of the angle $\theta$, but I do not know how to prove it rigorously to get: $$ F(\varphi(r))=2\pi\int_1^2\sqrt{1+\varphi_r^2}\ rdr $$
I suppose that the idea is to show that if we add some variation of the angle it will make things worse:
$$ F(\varphi(r))\leq F(\varphi(r)+d(\varphi(\theta))) $$ But I do not know if it is a way to do this and how to proceed.
Upd: To clarify. I understand that if $\varphi$ is independent of $\theta$, then we have $F(\varphi(r))=2\pi\int_1^2\sqrt{1+\varphi_r^2}\ rdr$, but I do not know how to prove that it is indeed independent, even though it is kind of obvious