Let $U \subseteq \mathbb{R^2}$ open and $f: U \mapsto \mathbb{R^2}$, $f \in C^1$. $f(x,y) := (x,y)$. And $\gamma: [a,b] \mapsto \partial B $ parameterizes the boundary. $B\subseteq U$ has a smooth boundary and is compact.
I now want to express the surface area in terms of an integral over $[a,b]$.
I currently have, using the divergence theorem:
$$\mu(B)=\frac{1}{2}\int_B \operatorname{div}(f)\,dxdy=\frac{1}{2}\int_{\partial B} f\,dn=\frac{1}{2}\int_{\gamma([a,b])} f\,ds=\frac{1}{2}\int_{a}^{b} \langle f(\gamma(t),\dot{\gamma(t)}\rangle\,dt$$
Is this right and can I simplify the integral further?