covariant metric tensor in case of polar transformation Assume usual polar coordinates equations:
$$ x=r\cos(\theta) $$
$$ y=r\sin(\theta) $$
In what I think is called covariant we have a Jacobian of:
$$
J=\begin{pmatrix}
\frac{\partial{x}}{\partial{r}} & \frac{\partial{x}}{\partial{\theta}} \\
\frac{\partial{y}}{\partial{r}} & \frac{\partial{y}}{\partial{\theta}} \\
\end{pmatrix}=\begin{pmatrix}
\cos(\theta) & -r\sin(\theta) \\
\sin(\theta) & r\cos(\theta) \\
\end{pmatrix}
$$
and a metric tensor of:
$$
g_{ij}=J^T J=\begin{pmatrix}
1 & 0 \\
0 & r^2 \\
\end{pmatrix}
$$
My problems appears when I try the contravariant part:
$$
J=\begin{pmatrix}
\frac{\partial{r}}{\partial{x}} & \frac{\partial{r}}{\partial{y}} \\
\frac{\partial{\theta}}{\partial{x}} & \frac{\partial{\theta}}{\partial{y}} \\
\end{pmatrix}=\begin{pmatrix}
\cos(\theta) & \sin(\theta) \\
-\frac{\sin(\theta)}{r} & \frac{\cos(\theta)}{r} \\
\end{pmatrix}
$$
because the metric tensor:
$$
g^{ij}=J^T J=\begin{pmatrix}
\cos^2(\theta)+\frac{\sin^2(\theta)}{r} & ... \\
... & ... \\
\end{pmatrix}
$$
results different of the expected one:
$$\begin{pmatrix}
1 & 0 \\
0 & \frac{1}{r^2} \\
\end{pmatrix}
$$
why the expected result is not the one I obtain?
( I obtain the expected only if I change the product from $J^T J$ to $J J^T$ ). 
 A: Maybe a bit of a preamble will be useful here. If $\xi$ are some coordinates defining the local metric $\eta$ then, under the transformation $x^\alpha = x^{\alpha}(\xi)$ the metric becomes
$$
g_{\mu\nu} = \frac{\partial \xi^{\alpha}}{\partial x^\mu}\frac{\partial \xi^{\beta}}{\partial x^\nu} \eta_{\mu\nu} \tag{1}
$$
So for example, if you take $\xi^1 = x$ and $\xi^2 = y$ the cartesian coordinates, then the local matrix is the identity $\eta_{\mu\nu} = \delta_{\mu\nu}$ and Eq. (1) becomes
$$
g_{\mu\nu} = J^{\alpha}_{\;\mu}J^{\beta}_{\;\nu}\delta_{\alpha\beta} = (J^T J)_{\mu\nu} \tag{2}
$$
where the entries of the matrix $J$ are defined as 
$$
J^{\alpha}_{\;\beta} = \frac{\partial \xi^\alpha}{\partial x^\beta} \tag{3}
$$
As an example take $x^1 = r$ and $x^2 = \theta$ (your case), so that $\xi^1 = r\cos\theta = x^1 \cos x^2$ and $\xi^2 = x^1\sin x^2$, it is easy to calculate 
$$
\frac{\partial \xi^1}{\partial x^1} = \frac{\partial }{\partial x^1}(x^1 \cos x^2) = \cos x^2 ~~~(\cdots)
$$
so when you evaluate that in (1) you get
$$
g_{11} = 1, ~ g_{12} = g_{21} = 0 ~\mbox{and}~ g_{22} = (x^1)^2
$$
It is important to keep track of the order of things here. In this representation we are labeling the rows with the super-index $\alpha$  and the columns with the sub-index $\beta$.
Now, the contravariant version of Eq. (2) is
$$
g^{\mu\nu} = \eta^{\alpha \beta}\frac{\partial x^\mu}{\partial \xi^\alpha} \frac{\partial x^\nu}{\partial \xi^\beta} \tag{4}
$$
and again, under the same assumptions as before, this transforms to
$$
g^{\mu\nu} = \delta^{\alpha\beta}\mathcal{J}^\mu_{\;\alpha}\mathcal{J}^\nu_{\;\beta} = (\mathcal{J}\mathcal{J}^T)^{\mu\nu} \tag{5}
$$
where 
$$
\mathcal{J}^{\alpha}_{\;\beta} = \frac{\partial x^\alpha}{\partial \xi^\beta} \tag{6}
$$
Taking the same changes of coordinates you will get $x^1 = [(\xi^1)^2 + (\xi^2)^2]^{1/2}$ and $x^2 = \arctan(\xi^2/\xi^1)$ so that
$$
\frac{\partial x^1}{\partial \xi^1} = \frac{\xi^1}{[(\xi^1)^2 + (\xi^2)^2]^{1/2}} ~(\cdots)
$$
which results in 
$$
g^{11} = 1, ~ g^{12} = g^{21} = 0, ~\mbox{and}~ g^{22} = \frac{1}{(\xi^1)^2 + (\xi^2)^2} = \frac{1}{(x^1)^2}
$$
as expected
