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.

$J(r,\theta)= \begin{bmatrix} & \cos(\theta) & -r \sin(\theta)\\ & \sin(\theta) & r \cos(\theta) \\ & 1 & 0 \\ \end{bmatrix}$

  • $\begingroup$ in essence you are calculating the norm of $\frac{\partial \Phi}{\partial r}\times \frac{\partial\Phi }{\partial\theta}$ because this is related with the area of the parallelogram spawned by $\frac{\partial \Phi}{\partial r}$ and $ \frac{\partial\Phi }{\partial\theta}$, which is tangent information $\endgroup$ – janmarqz Dec 1 '15 at 23:25
  • $\begingroup$ Note that the determinant of a matrix can be interpreted as the volume of a parallelepiped (in two dimensions area of a parallelogram) with sides defined by the columns of the matrix. The connection with cross products is that the length of the resulting vector is equal to the area of the parallelogram spanned by the multiplicands, i.e., to a determinant. $\endgroup$ – amd Dec 1 '15 at 23:36

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