I'm having trouble proving that the integral of a $2$-form over a surface does not depend on the chosen frame.
Let $S$ be a surface in $\mathbb R^3$ and $F_a \colon U_a \to S$, $F_b \colon U_b \to S$ two parametrizations of the patch $S_a \cap S_b = F(U_a) \cap F(U_b)$. The $2$-form $\omega$ can be written as $\omega = f_a \mathrm du_{F_a} \wedge \mathrm dv_{F_b} = f_b \mathrm du_{F_b} \wedge \mathrm dv_{F_b}$.
We define the integral of $\omega$ with respect to the local parametrization $(U_a, F_a)$ as follows: $$\int_{(U_a, F_a)} \omega = \int_{U_a} f_a \circ F_a\,\mathrm du\mathrm dv$$ I want to prove that if $\operatorname{supp} \omega \subset S_a$ and $\operatorname{supp} \omega \subset S_b$, then the integrals of $\omega$ over $(U_a, F_a)$ and $(U_b, F_b)$ are equal.
Let $h_{ba} = F_b^{-1} \circ F_a$ and also $D_a = F_a^{-1}(\operatorname{supp} \omega)$, $D_b = F_b^{-1}(\operatorname{supp}\omega)$. We (should) have: $$\begin{align} \int_{(U_b, F_b)} \omega &= \int_{U_b} (f_b \circ F_b)\,\mathrm du\mathrm dv = \int_{D_b} (f_b \circ F_b)\,\mathrm du\mathrm dv = \tag{$\star$}\\ &= \int_{D_a} (f_b \circ F_b) \circ h_{ba}~|\;\!\!\det Jh_{ba}|\,\mathrm du\mathrm dv =\\ &= \int_{D_a} (f_b \circ F_a) \det Jh_{ba}\,\mathrm du\mathrm dv \color{red}{\stackrel{??}{=}} \int_{D_a} (f_a \circ F_a)\,\mathrm du\mathrm dv =\\ &= \int_{(U_a, F_a)} \omega \end{align}$$
In $(\star)$ I used the change of variables theorem (Jacobi theorem I think it's called?). I don't know how to bridge the red equals sign. I tinkered with the transformation law for $2$-forms, which is $$\tilde \omega|_P = \det(Jh_{ba})|_{(u, v)} \omega|_P$$ where $P = F(u, v)$ and $J$ represents the Jacobian matrix. Since in our case $\tilde\omega$ is $\omega$ with a different local frame we can write $$f_a(P) \mathrm du_{F_a} \wedge \mathrm dv_{F_a} = \omega|_P = \tilde\omega|_P = f_b(P) \mathrm du_{F_b} \wedge \mathrm dv_{F_b} = \det Jh_{ba}|_{(u, v)} f_a(P) \mathrm du_{F_a} \wedge \mathrm dv_{F_a}$$ Moreover $P = F_a(u, v)$, so over $S_a \cap S_b$ the following holds: $$f_a \mathrm du_{F_a} \wedge \mathrm dv_{F_a} = f_a (\det(Jh_{ba}) \circ F_a^{-1}) \mathrm du_{F_a} \wedge \mathrm dv_{F_a}$$
However, I don't see how this helps in the last steps of the proof above. Did I make a mistake somewhere?