Show a geometry can be embedded in 3D euclidean space Consider the 2D space with line element $ds^2=\frac{dr^2}{1-\frac{2\mu}{r}}+r^2d\phi^2$. How can I prove that this geometry can be embedded in a 3D Euclidean space and also find the equations for the corresponding 2-dimensional surface?
On a first attempt, I tried using the induced metric functions equation: $h_{ij}=g_{ab}\frac{\partial x^a}{\partial u^i}\frac{\partial x^b}{\partial u^j}$, so that parametrizing the surface in terms of $r$ and $\theta$, I could try to satisfy the relation between the euclidean metric tensor $g_{ab}=\mathbb{I}$ and the induced metric functions, which I found to be: $h_{11}=\frac{1}{1-\frac{2\mu}{r}}$ and $h_{22}=r^2$.
Anyway, I couldn't find any way forward, so any help will be much appreciated!
 A: Using @Arctic Char' suggestion: it make sence to use the rotational symmetry.
If we embed the required surface into 3-dymensional Euclidean space, we can require the rotational symmetry over axis Z (the coordinate system in 3D space - this is our choice). We can also use the angle $\phi$ for parametrization in 3D, so this angle will be the same for both surface and 3D space.
To define the surface it would be enough to draw its cross-section (a curve), for instance, in X-Z plane.
$ds^2=\frac{dr^2}{1-\frac{r_s}{r}}+r^2d\phi^2$, where $r_s$ is the "Schwarzschild radius".
In 3D space we introduce $z$ and $\rho$ coordinates: $ds^2=dz^2+d\rho^2+\rho^2d\phi^2$. On the other hand, if we stay in the X-Z plane, $d\phi=0$, so
$ds^2=\frac{dr^2}{1-\frac{r_s}{r}}\Rightarrow dz^2+d\rho^2=\frac{1}{1-\frac{r_s}{r}}dr^2$
$(\frac{dz}{dr})^2+(\frac{d\rho}{dr})^2=\frac{r}{r-r_s}$ - the parameterized equation for $z(r)$ and $\rho(r)$
Considering $z(r)$ as a function $z(\rho(r))$ we get
$\Bigr((\frac{dz}{d\rho})^2+1\Bigl)(\frac{d\rho}{dr})^2=\frac{r}{r-r_s}$, or $\Bigr((\frac{dz}{d\rho})^2+1\Bigl)=\frac{r}{r-r_s}(\frac{dr}{d\rho})^2$.
But we are allowed to transform the variable , switching from $r$ to $r(\rho)$. We can choose simply $r=\rho$.
$(\frac{dz}{dr})^2+1=\frac{r}{r-r_s}\Rightarrow \frac{dz}{dr}=\pm\sqrt{\frac{r_s}{r-r_s}}$
$z(r)=\pm2\sqrt{r_s(r-r_s)}+const$. A constant here defines the origine for the variable $z$ and can be chosen zero.
We got a parabola lying "on the side" with the beginning at the point $r=r_s$. Rotating this parabola around axis Z we get the required surface.
So, in 3D space we got the surface defined by the coordinates
Polar angle $\phi$
Radius r
$z$ coordinate depending on $r$: $z(r)=\pm2\sqrt{r_s(r-r_s)}$
At $r>>r_s$ $z(r)\approx2\sqrt{r_sr}<<r$, and $ds=\sqrt{1+(\frac{dz}{dr})^2} $ $dr\approx(1+\frac{r_s}{2r})$ $ dr\approx dr$, so at $r\to\infty$ the surface has two-dimension Euclidean metrics.
At $r\to{r}_s$ the surface is curved and $ds\to\infty$, but $S=\int_{r_1}^{r_s}ds$ is finite, since the singularity at $r=r_s$ is integrable.
If we suppose that the time in this system is absolute, a two-dimensional creature living on the surface and moving along the radius $r$  with a constant velocity $v$ will reach $r_s$ ($r_1>r_s$) in finite time
$$T=-\frac{1}{v}\int_{r_1}^{r_s}\sqrt{\frac{r}{r-r_s}}dr=\frac{1}{v}\Bigl(\sqrt{r_1(r_1-r_s)}+r_s\log\frac{\sqrt{r_1}+\sqrt{r_1-r_s}}{\sqrt{r_s}}\Bigr)$$
At $r_1>>r_s$ $$ T=\frac{r_1}{v}(1-\frac{r_s}{2r_1})+\frac{r_s}{v}\log(2\sqrt{\frac{r_1}{r_s}})\geqslant\frac{r_1}{v}$$
At $(r_1-r_s)<<r_s$ $$ T=2\frac{\sqrt{r_s}}{v}\sqrt{r_1-r_s}>>\frac{r_1-r_s}{v}$$
