# A Find the determinant of the metric on some graph

1. Feb 27, 2017

### Jonsson

Hello there,

Suppose $f$ smoothly maps a domain $U$ of $\mathbb{R}^2$ into $\mathbb{R}^3$ by the formula $f(x,y) = (x,y,F(x,y))$. We know that $M = f(U)$ is a smooth manifold if $U$ is open in $\mathbb{R}^2$. Now I want to find the determinant of the metric in order to compute the area of the manifold
$$I = \int 1 |g|^{1/2} d^2x$$
I guess that the metric on $\mathbb{R}^n$ is the Kronecker delta, so that
$$g_{ij} = \frac{d\xi^a}{dx^i} \frac{d\xi^b}{dx^j} \delta_{ab}$$
So if I can find $\xi^a$, my task is easy. How do I determine $\xi^a$. Any hints/help/solutions? Thanks

2. Feb 28, 2017

### davidge

Maybe I'm wrong, but I think your $\xi^\rho$ are given by the transformation law between the Cartesian and your new coordinate system. Maybe some mentor can correct me if I'm mistaken it.

3. Feb 28, 2017

### Staff: Mentor

The metric of what?

What do $\xi^a$ represent?

4. Mar 1, 2017

### Ben Niehoff

This formula gives the "pullback" of the Euclidean metric on $\mathbb{R}^3$ ($\delta_{ab}$ where $a,b \in \{1,2,3\}$) to the metric on your embedded surface ($g_{ij}$ where $i,j \in \{1,2\}$). So the $\xi^a$ are just the coordinates of $\mathbb{R}^3$, in this case

$$\xi^1 = x, \qquad \xi^2 = y, \qquad \xi^3 = F(x,y).$$
Note that your formula ought to have partial derivatives:

$$g_{ij} = \frac{\partial \xi^a}{\partial x^i} \frac{\partial \xi^b}{\partial x^j} \delta_{ab}.$$