# Proof - Substitution, Jacobian, etc.

1. Jun 22, 2011

### stanley.st

Hello!

I recently tried to prove following theorem: Let $$\phi:B\to\mathbb{R}^2$$ be a diffeomorphism (regular, injective mapping). Then

$$\int_{\phi(B)}f(\mathbf{x})\,\mathrm{d}x=\int_{B}f(\phi(\mathbf{t}))\left|{\mathrm{det}}\mathbf{J}_{\phi}\right|\mathrm{d}t$$

With following I can't proof this theorem. Look, I start with integral sums

$$(*)\quad\sum_{i=1}^{n}f(x_i,y_j)(x_{i+1}-x_i)(y_{j+1}-y_{j})$$

According to the transformation phi, we have

$$x_i=\phi_x(r(x_i,y_j),t(x_i,y_j))$$$$y_i=\phi_y(r(x_i,y_j),t(x_i,y_j))$$

We can imagine phi as a polar coordinate system transformation, so I use notation with variables r,t. Then we have using Taylor formula

$$\begin{array}{ll}x_{i+1}-x_{i}&=\phi_x(r(x_{i+1},y_j),t(x_{i+1},y_j))-\phi_x(r(x_{i},y_j),t(x_i,y_j))\\&=\frac{\partial \phi_x}{\partial r}(\xi,\eta)(r(x_{i+1},y_j)-r(x_{i},y_j))+\frac{\partial \phi_x}{\partial r}(\xi,\eta)(t(x_{i+1},y_{j})-t(x_{i},y_j))\\&=\frac{\partial \phi_x}{\partial r}\delta r_{i+1,j}+\frac{\partial \phi_x}{\partial r}\delta t_{i+1,j}\quad(\textrm{short form})\end{array}$$

In the same way I can derive

$$y_{j+1}-y_j=\frac{\partial \phi_y}{\partial r}\delta r_{i,j+1}+\frac{\partial \phi_y}{\partial r}\delta t_{i,j+1}$$

If I put this into (*) I get

$$\sum_{i,j=1}^{n}f(\phi(r_{ij},t_{ij}))\left(\frac{\partial \phi_x}{\partial r}\delta r_{i+1,j}+\frac{\partial \phi_x}{\partial r}\delta t_{i+1,j}\right)\left(\frac{\partial \phi_y}{\partial r}\delta r_{i,j+1}+\frac{\partial \phi_y}{\partial r}\delta t_{i,j+1}\right)$$

But this is different than I expected. I expected it in the form like

$$\sum_{i,j=1}^{n}f(\phi(r_{ij},t_{ij}))\left(\frac{\partial \phi_x}{\partial r}\frac{\partial \phi_y}{\partial t}-\frac{\partial \phi_y}{\partial t}\frac{\partial \phi_x}{\partial r}\right)\delta r\delta t$$

Do I something wrong? Thanks a lot..