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Let $P, Q: D \rightarrow \mathbb{R}$ be continuous functions on an open set $D \subset \mathbb{R}$, let a curve $\gamma: [a,b] \rightarrow \mathbb{R}^2$ be of class $C^1$ such that $\gamma([a,b]) \subset D$.

Let $\Phi=(\phi, \psi): D' \rightarrow D$ be a diffeomorphism of class $C^1$ from open $D' \subset \mathbb{R}^2$ onto $D$. Let $x=\phi(x',y')$, $y=\psi(x',y')$.

I search a formula for a line integral $I:=\int_{\gamma} P(x,y)dx+Q(x,y)dy$ in coordinates $x',y'$. If we calculate differentials $dx$ and $dy$ and formally put it in integral $I$ we obtain that

$\int_{\gamma} P(x,y)dx+Q(x,y)dy=\int_{\Phi^{-1} \circ \gamma} [P(\phi(x',y'), \psi(x',y')) \frac{\partial \phi}{\partial x'}+Q(\phi(x',y'), \psi(x',y')) \frac{\partial \psi}{\partial x'}] dx'$+$[P(\phi(x',y'), \psi(x',y')) \frac{\partial \phi}{\partial y'}+Q(\phi(x',y'), \psi(x',y')) \frac{\partial \psi}{\partial y'}]dy' $.

Does it formula indeed hold? How to show it? (maybe by changing the both line integrals on Riemann integrals?)

Thanks.

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1 Answer 1

Maybe you should use Green theorem to evaluate $I:=\int_{\gamma} P(x,y)dx+Q(x,y)dy$, and I seriously doubt the validity of your formula. I'm not sure if it is wrong though.

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