Inverting Jacobian trick: does it always work, or did professor get lucky? I am tutoring a multivariable calculus student, and his teacher was solving
$$\int_R f(x, y) dA$$
using the Change of Variable Theorem, where $f$ is an unimportant function and $R$ is the region bounded by
$$\begin{align*}
x^2-y^2 &= -1\\ x^2 - y^2 &= 1\\ x+y &= 1\\ x+y &= 3
\end{align*}$$
The he made the change of variable
$$(s, t) = F(x, y) = (x+y, x^2 - y^2)$$
For change of variables, we need $\det J_{F^{-1}}(s, t)$.
However, the professor found
$$\det J_F(x, y) = -2x - 2y = -2s$$
and proclaimed that $$\det J_{F^{-1}}(s, t) = -\frac {1}{2s}.$$
I think this shouldn't work, because the inverse function theorem says that in general,  $$J_{F^{-1}}(s, t) = [J_F(F^{-1}(s, t))]^{-1}$$
and to me it seems like the professor did
$$J_{f^{-1}}(x) = [J_f((x))]^{-1}.$$
However, I calculated
$$F^{-1}(s, t) = \left(\frac 12s + \frac {t}{2s}, \frac {12}s - \frac{t}{2s} \right)$$
and used this to calculate $\det J^{-1}(s, t)$, and I also got $-\frac {1}{2s}$.
Was the professor wrong but lucky, or am I misunderstanding something?
 A: There are two ways to do change of variable problems analytically. The "standard" way involves finding the inverse transformation of your mapping from the original variables to the transformed variables, and then directly computing the Jacobian determinant of this inverse mapping (which is the quantity that appears in the expression for the integral after the change of variable, similar to the 1D case).
The "alternate" way is to notice that the Jacobian of the inverse mapping is the inverse matrix of the Jacobian of the forward mapping, and so in particular their determinants are reciprocals of one another. (This is the multivariate version of the inverse function theorem. They may have seen the 1D counterpart of this when answering questions like "if $f(x)=x^3+x+1$, compute $\frac{d}{dy} f^{-1}(y)$ at $y=3$".)
Therefore you can write the new integrand in the form $f(x,y) \left ( \begin{vmatrix} u_x(x,y) & u_y(x,y) \\ v_x(x,y) & v_y(x,y) \end{vmatrix} \right )^{-1}$ and then hope that this simplifies in such a way that you can write the whole thing in terms of $u,v$, despite the fact that you never actually wrote down $x(u,v),y(u,v)$ in general.
In this method (unlike the "standard" method) the original integrand $f(x,y)$ is not an "unimportant function"; whether you can finish setting up the problem as a double integral in the $uv$ plane depends crucially on the precise way that it depends on $x$ and $y$.
(I presented everything in 2D language since multivariable calc books pretty much exclusively do these "general" change of variable problems in 2D, but the same concepts apply in higher dimensions.)

Addendum since you mentioned you're asking this as a tutor:
Here's a 1D problem done using the 1D analogues of these methods. This may help a student understand the distinction between them.
Consider $\int_0^2 x^3 dx$. Let's say we're using $u=x^2$ to solve this.
The standard method says that the image of the mapping is $[0,4]$, the inverse of the mapping is $x=\sqrt{u}$, the derivative of the inverse is $\frac{dx}{du}=\frac{1}{2} u^{-1/2}$, so the new integral is $\int_0^4 u^{3/2} \frac{1}{2} u^{-1/2} du = \int_0^4 \frac{1}{2} u du$.
The alternate method says the image of the mapping is $[0,4]$, the derivative of the mapping is $\frac{du}{dx}=2x$, the reciprocal of that is $\frac{1}{2x}$, and so the new integral looks like $\int_0^4 x^3 \frac{1}{2x} du$. To finish setting up we need to realize that $x^3 \frac{1}{2x}=\frac{x^2}{2}$ and then identify that as $\frac{u}{2}$. So you have $\int_0^4 \frac{u}{2} du$.
A: Look carefully at the equation that the professor wrote. We have $\det J_F(x,y) = -2(x+y) = -2s = -2F_1(x,y)$ (by which I mean the first component of $F$). Note that $$J_F(F^{-1}(s,t)) = -2F_1(F^{-1}(s,t)) = -2s,$$ since $F(F^{-1}(s,t)) = (s,t)$.
Then
$$\det J_{F^{-1}}(s,t) = \frac 1{J_F(F^{-1}(s,t))} = \frac 1{-2s},$$
as desired. This is a useful and powerful technique to master.
