Let $\Omega$ be a bounded open in $\mathbb{R}^{2}$ such that $\partial \Omega$ is a $C^{1}$ curve, $F: \mathbb{R}^{2} \to \mathbb{R}^{2}$ a twice differentiable function. For $x = (x_{1},x_{2}) \in \mathbb{R}^{2}$, consider $$DF(x) = \left(\begin{array}{cc} \frac{\partial F_{1}}{\partial x_{1}} & \frac{\partial F_{1}}{\partial x_{2}} \\ \frac{\partial F_{2}}{\partial x_{1}} & \frac{\partial F_{2}}{\partial x_{2}} \end{array}\right).$$ Suppose that $F(x) = Mx$ for all $x \in \Omega$, where $M$ is a $2 \times 2$ matrice. Show that $$\int_{\Omega}\det(DF(x))dx = \det(M)\mathrm{area}(\Omega).\tag{1}$$

Prove that the same result in (1) is true if $F(x) = Mx$ for all $x \in \partial \Omega$. For this, use the Green's Theorem and show $$\int_{\Omega}\det(DF(x))dx = \frac{1}{2}\oint_{\partial \Omega}\left(\left[F_{1}\frac{\partial F_{2}}{dx_{1}} - F_{2}\frac{\partial F_{1}}{dx_{1}}\right]dx_{1} + \left[F_{1}\frac{\partial F_{2}}{dx_{2}} - F_{2}\frac{\partial F_{1}}{dx_{2}}\right]dx_{2}\right)$$ and use this identity.

I prove the first part and the identity. But I'm confuse about the case $x \in \partial \Omega$. Using the identity without looking at the details, I conclude that $\int_{\Omega}\det(DF(x))dx = 0$. I think that make sense, since $\mathrm{area}(\partial \Omega) = 0$. But this is not $(1)$. Also, $(1)$ make sense for $x \in \partial \Omega$?

  • $\begingroup$ $\int_{\Omega}\det(DF(x))dx $ doesn't appear anywhere. You should make clear for $\int_\Omega$ you are integrating $h(x)d^2 (x_1,x_2)$ while for $\int_{\partial \Omega}$ you are integrating $h_1(x)dx_1+h_2(x)dx_2$ for some functions $h,h_1,h_2$ and go back to the definition of those $\endgroup$ – reuns Mar 17 at 20:35
  • $\begingroup$ Are you aware that using the change of variable theorem is more natural? $\endgroup$ – Martín-Blas Pérez Pinilla Mar 17 at 20:40
  • $\begingroup$ @Martín-BlasPérezPinilla I don't think about it. Anyway, the question asks for the identity. $\endgroup$ – Lucas Corrêa Mar 17 at 20:43
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    $\begingroup$ In any case, the COV only works for the first part, obviously. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 17 at 20:47
  • $\begingroup$ @reuns yes. I just wrote the question exactly as it is. I proved the identity, but I don't know how to use. My question about the second part is: why is enough consider $x \in \partial \Omega$? The equality (1) is about $\Omega$. Is seems strange to me consider only $x \in \partial \Omega$. $\endgroup$ – Lucas Corrêa Mar 17 at 20:50

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