Fourier transform of a radial function is radial

Let $f\in S(R^2)$ radial. Its Fourier transform is radial because the Lebeasgue measure is invariant under rotation?

Ultimately, the answer is affirmative, but that's not a proof. The rigorous proof is an easy exercise and you can find it in most books on harmonic analysis (for instance, on Stein and Weiss "Introduction to harmonic analysis on Euclidean spaces").

If you want to prove it yourself, here's a hint. Suppose that $f\in L^1(\mathbb R^d)$ is radially symmetric, that is, $$ f(Rx)=f(x), \qquad \forall R\in SO(d).$$ Then prove that $\hat{f}(R\xi)=\hat{f}(\xi)$ for all $R\in SO(d)$ by changing variable in the integral $$ \hat{f}(R\xi)=\int_{\mathbb R^d} f(x)e^{-2\pi i \langle x, R\xi\rangle}\, dx.$$