# Show that $\int_{\Omega}\det(DF(x))dx = \det(M)\mathrm{area}(\Omega)$.

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 $$x \in \partial \Omega$$?

• $\int_{\Omega}\det(DF(x))dx$ 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 – reuns Mar 17 at 20:35
• Are you aware that using the change of variable theorem is more natural? – Martín-Blas Pérez Pinilla Mar 17 at 20:40
• @Martín-BlasPérezPinilla I don't think about it. Anyway, the question asks for the identity. – Lucas Corrêa Mar 17 at 20:43
• In any case, the COV only works for the first part, obviously. – Martín-Blas Pérez Pinilla Mar 17 at 20:47
• @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 $x \in \partial \Omega$. – Lucas Corrêa Mar 17 at 20:50