substitution for a multiple integral

Question

I came across a concept which I have never dealt with before. I had an idea but it is flawed and I would like to know why it is flawed. To those who are familiar with that subject it will seem somewhat obvious but I would still appreciate an answer for dummies here.
Suppose: $$ \iint f(x, y) dxdy $$ and I want to transform the variables to $ x=g(u,v) $ and $ y=h(u,v)$. First, I apply the total differential, so that: $$ dx = {\partial g\over \partial u }du + {\partial g\over \partial v }dv; $$ $$ dy = {\partial h\over \partial u }du + {\partial h\over \partial v }dv. $$ Then, I multiply the two: $$ dxdy = ({\partial g\over \partial u }du + {\partial g\over \partial v }dv)*({\partial h\over \partial u }du + {\partial h\over \partial v }dv)\neq\vert Jacobian\vert. $$ Finally, I plug it into the above integral instead of the Jacobian. Why is my reasoning wrong? Thanks in advance.

Answer 1

The notation $dxdy$ in a double integral is just a formal notation. It doesn't mean you can literally make a substitution by multiplying together the expressions for $dx$ and $dy$ as you have done. Instead, you need to use the Jacobian.

(In fact, there is a correct way to think of $dxdy$ as $dx$ multiplied by $dy$, but you must use the exterior product of differential forms. This product, usually written $\wedge$, has the property that $du\wedge du=dv\wedge dv=0$ and $dv\wedge du=-du\wedge dv$. You then have $$\begin{align*} dx\wedge dy &= \left({\partial g\over \partial u }du + {\partial g\over \partial v }dv\right)\wedge\left({\partial h\over \partial u }du + {\partial h\over \partial v }dv\right) \\ &={\partial g\over \partial u }{\partial h\over \partial u }du\wedge du + {\partial g\over \partial u }{\partial h\over \partial v } du\wedge dv + {\partial g\over \partial v }{\partial h\over \partial u } dv\wedge du + {\partial g\over \partial v }{\partial h\over \partial v } dv\wedge dv \\ &= 0 + {\partial g\over \partial u }{\partial h\over \partial v } du\wedge dv - {\partial g\over \partial v }{\partial h\over \partial u } dv\wedge du \\ &= \left({\partial g\over \partial u }{\partial h\over \partial v } - {\partial g\over \partial v }{\partial h\over \partial u }\right) du\wedge dv \end{align*}$$ which is exactly the Jacobian times $du\wedge dv$. The use of the absolute value of the Jacobian in the usual formula is a result of not keeping track of an orientation on the region of integration when you make the change of variables.)