http://mathinsight.org/parametrized_surface_area_examples
In reading through the example in the above link, it's straightforward to find the surface area for a cone as follows.
Find the surface area of the cone $S$
$\mathbf{\Phi}(r,\theta) = (r \cos \theta, r \sin\theta, r)$
for $0 \le \theta \le 2\pi$ and $0 \le r \le 1$
The area of cone $S$ is
$\begin{align*} A(S) &=\int_0^1\int_0^{2\pi} \left\| {{\partial \mathbf{\Phi}} \over { \partial r}} \times {{\partial \mathbf{\Phi}} \over { \partial \theta}}\right\| d\theta\, dr \end{align*} $
However, I am trying to gain a deeper understanding of how the Jacobian matrix works. I realize that for this problem there is no change of coordinate system, but is it possible to use the Jacobian to calculate the surface area?
In attempting this, I get the following matrix. To get it to match up with the cross product method, it looks like I would need to add the $<i,j,k>$ unit vector to the left side, transpose the matrix, and then take the determinant. I'm not sure if this is a valid approach.
$J(r,\theta)= \begin{bmatrix} & \cos(\theta) & -r \sin(\theta)\\ & \sin(\theta) & r \cos(\theta) \\ & 1 & 0 \\ \end{bmatrix}$