I would like to change some variables in a integral and encountered to an issue. I create here 2 simple examples to describe my questions:
Exp 1. Suppose we want to change $(x,y)$ to $(u,v)$ such that:
$x= u + v$ and $y= -u -2v$.
Using the Chain Rule: $$dx dy = (du + dv) (-du -2dv) = -2du dv -dv du = -2du dv + du dv = -du dv.$$
On the other hand, using the Jacobian determinant formula, we need the absolute value of Jacobian determinant, which is $|-1| = 1$. So:
$$dx dy = 1. du dv = du dv \ne -du dv.$$
(Ref: http://www.math24.net/change-of-variables-in-double-integrals.html)
What was wrong here?
Exp 2. With one variable, we have $dx = x^'_u du$ – so no absolute value is needed. What is the main difference between the one variable case and the multi-variables case?
I really appreciate if anyone could give me good references for this. Thanks in advance!
Best Answer
Consider a simple case of univariate integral $\int_0^1 x \mathrm{d} x$ and apply to it a change of variables $x = 1-y$. It is easy to see that it flips integration range, and this is reflected in the Jacobian being negative.
$$ \int_0^1 x \mathrm{d} x = \int_1^0 (1-y) (- \mathrm{d} y) = -\int_0^1 (1-y) (-\mathrm{d} y) = \int_0^1 (1-y) \mathrm{d} y $$
The sign of the Jacobian indicates where the change of variable is, or is not, orientation preserving (i.e. whether it flips integration limits or not).
One usually write $\vert J \vert$ keeping in mind that the orientation is being preserved.
The same story holds in the multivariate settings as well. Consider a line, defined by $F(x,y) = x + y - 2$. The normal vector is $\vec{n} = \left\{ \partial_x F, \partial_y F \right\} = \left\{ 1, 1 \right\}$. When your change of variables is applied, the direction of the normal is flipped: $$ F(u,v) = (u+v) + (-u-2v) - 2 = -v - 2 \implies \vec{n}_2 = \{\partial_u F, \partial_v F\} = \{0, -1\} $$ Notice that $\langle \vec{n}, \vec{n}_2 \rangle = -1$.