Let's say there's a region $R$ with image $S$ with transformations $x=x (u,v)$ and $y=y (u,v) $.
I that the integral of $f(x,y) $ over $R $ is the integral of $f(x(u,v),y(u,v)) $ times the absolute value of the jacobian of x and y with respect to u and v $dudv $. (I'm writing this on a phone.) If I switch the bounds so it's $dvdu $, do I change the jacobian.(For example, changing it to x, y with respect to v and u?)
Edit (Fixed question based on answer and adding example)
For example: Let's say $R$ is the region defined by the curves:
$y=x$, $xy=2$, $xy=4$ and $y=\frac{1}{8}x^3$
If $y=v$ and $x=\frac{u}{v}$, then the new curves are:
$v=u^{1/2}$, $u=2$, $u=4$ and $v=\frac{1}{8^{1/4}}u^{3/4}$
Now, I have $\int\int_{R}^{}x^2 y^2 dA = \int_{2}^{4}\int_{\frac{1}{8^{1/4}}u^{3/4}}^{u^{1/2}}f(u,v)dvdu$.
The equation for change of variable I know about has the bounds in the opposite order, $du dv$ How do I do this in the $dvdu$ order?
Whether we use $(u,v)$ or $(v,u)$, we have to multiply by the same thing. A proof is as follows:
The factor by which you multiply is given by the absolute value of the determinant of the Jacobian,
$$\begin{vmatrix}\displaystyle\frac{\partial x}{\partial u}&\displaystyle\frac{\partial x}{\partial v}\\ \displaystyle\frac{\partial y}{\partial u}&\displaystyle\frac{\partial y}{\partial v}\end{vmatrix}.$$
If we switch $u$ and $v$, that switches the two columns of this matrix, which multiplies the determinant by $-1$. Since we're taking the absolute value, it doesn't change the value that we multiply. This proof generalizes to a transformation in $n$ variables in the same way.