Let the field, $\vec{F}$ be class $C¹$ in the open $Ω$. And $\sigma_1$ and $\sigma_2$ are portions of regular surfaces with boundaries $\Gamma_1$ and $\Gamma_2$ oriented positively in relation to the normal $\vec{n_1}$ and $\vec{n_2}$. And such that $\operatorname{img}({\sigma_1})$ and $\operatorname{img}({\sigma_2})$ are contained in $Ω$. Suppose further that $\Gamma_1$ is obtained from $\Gamma_2$ by a change of parameters that keeps the orientation. Prove that:
$$\int\int_{\sigma_1} \operatorname{rot}(\vec{F}) \cdot \vec{n_1}\mathrm{d}S = \int\int_{\sigma_2} \operatorname{rot}(\vec{F}) \cdot \vec{n_2}\mathrm{d}S $$
Observation:
Let $\gamma_1: [a, b] \rightarrow \Bbb{R}^n$and $\gamma_2: [c, d] \rightarrow \Bbb{R}^n$ both curves of class $C^1$; suppose further that there is a function $g: [c, d] → \Bbb{R}$, of class $C^1$ with $g '(u)>0$ ; $\operatorname{dom}(g '(u)) \in ] c, d [$ and $\operatorname{img} g = [a, b]$, such that, for $\forall u \in [c, d]$. it'll conserves the orientation if: $\boldsymbol{g '(u)>0}$
How about this:
Use stokes to turn the surface integral of $\sigma_1$ into a line integral
Change coordinates in which you write the integral, such that new boundary is same of the line integral as the one for $\sigma_2$
Now, since stokes is true for all surfaces attached to a given boundary. Use the $\sigma_2$ as a surface attached to the new boundary, hence the required is shown.