How does one apply the Jacobian Matrix in practice?

Question

I learned how to use the Jacobian a long time ago (but only in various problems and not preexisting coordinate systems) and was merely demonstrating it for a friend asking about in the case of Parabolic Coordinates: https://en.m.wikipedia.org/wiki/Parabolic_coordinates

I used the functions:

$f_1(x,y) = xy$

$f_2(x,y) = \frac 12(x^2-y^2)$

I took the partial derivatives like described in the equations (each x one and each y one) and triple checked the matrix (so no arithmetic errors).

I didn't get the area coefficient "$x^2 + y^2$". I told my friend that I suspected either I made a subtle mistake in comprehension altogether or I simply used the wrong functions. I suspect that this would be if I converted a multivariate integral to Cartesian rather than the reverse.

Is my suspicion correct or is it some no longer known algebra error?

My answer was:

$(\frac 12 yx^2 - y^2) - (x^2 - \frac 12 xy^2)$

Answer 1

$$\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}$$

Let's compute the Jacobian determinant of your functions.

First, we need all the partial derivatives: \begin{align*} \pd{f_1}{x} & = y \\ \pd{f_1}{y} & = x \\ \pd{f_2}{x} & = x \\ \pd{f_2}{y} & = -y \end{align*} Now, we arrange them into the jacobian matrix: $$J = \begin{pmatrix}y & x \\ x & -y\end{pmatrix}$$ Now, we compute the 'area element', which is $|\det(J)|$: $$|\det J| = |(-y^2-x^2)| = x^2+y^2$$

I suspect you left off the absolute value. Remember the Jacobian determinant is the magnitude of the determinant of the jacobian.

Answer 2

The Jacobian is given by: $$\begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{bmatrix}$$ In this case, we get: $$\begin{bmatrix} y& x\\ x & -y \end{bmatrix}$$ Taking the determinant gives us $-y^2-x^2$; since we take the absolute value for the area coefficient, we indeed get $x^2+y^2$.