confusion about change of variable If you are integrating $f(x,y)$ over a region and you do a change of variable to $f(u,v)$. The jacobian gives $dx\,dy = du\,dv (\partial x/ \partial u\ \partial y/\partial v - \partial x/\partial v\ \partial y/\partial v)$
$$dx = \frac{\partial x}{\partial u} du + \frac{\partial x}{\partial v} dv.$$
$$dy = \frac{\partial y}{\partial u} du + \frac{\partial y}{\partial v} dv.$$
If I simply multiply out the $dx\,dy$ terms I would get $du^2$ terms, etc. Can someone advise as to what I'm doing wrong?
 A: You have to interpret the situation correctly. The formula is 
$\int \int _{D_{xy}}f(x,y)dxdy=\int \int _{D_{uv}}f(x(u,v),y(u,v))\det \vert \mathcal J\vert dudv$ where $\det \vert \mathcal J\vert $ is the $\textit {determinant}$ of the Jacobian matrix. 
Let's derive this formula: 
You want to calculate $\int \int_{D} f(x,y)dA_{xy}$ where $D$ is a planar region. This integral is the result of a limiting process on sums of the form $f(x^{*}_{i},y^{*}_{j})\Delta A_{ij}$ so when you make a change of variables, you get the equations $x=x(u,v)$ and $y=y(u,v)$ and you want to calculate sums of the form $f(u^{*}_{i},v^{*}_{j})\Delta A'_{ij}$. 
So consider a point $(u,v)$ in the $u-v$ plane. 
As $u$ moves to $u+\Delta u$, $x$ moves to $x(u+\Delta u,v)$ and $y$ moves to $y(u+\Delta u,v)$.
Similarly, As $v$ moves to $v+\Delta v$, $x$ moves to $x(u,v+\Delta v)$ and $y$ moves to $y(u,v+\Delta v)$.
Thus, the rectangular region determined by moving $u$ and $v$ as described above, in the $u-v$ plane corresponds to a planar region $D_{1}$ in the $x-y$ plane. So to do our integral in the variables $u, v$ we need to see how the areas are related.
So you argue as follows:
We know that $\left | \frac{x(u+\Delta u,v)-x(u,v)}{\Delta u}-\frac{\partial x}{\partial u} \right |\to 0$ as $\Delta u\to 0$. This means that 
$\tag 1x(u+\Delta u,v)-x(u,v)\approx \frac{\partial x}{\partial u}\Delta u$. 
Similarly, 
$\tag2y(u+\Delta u,v)-y(u,v)\approx \frac{\partial y}{\partial u}\Delta u$.
This also is true for the variable $v$:
$\left | \frac{x(u,v+\Delta v)-x(u,v)}{\Delta v}-\frac{\partial x}{\partial v} \right |\to 0$ as $\Delta v\to 0$ so that 
$\tag3x(u,v+\Delta v)-x(u,v)\approx \frac{\partial x}{\partial v}\Delta v$. 
Similarly, 
$\tag4y(u,v+\Delta v)-y(u,v)\approx \frac{\partial y}{\partial v}\Delta v$. 
Notice now that the LHS of $(1),(2),(3),(4)$ determine a closed region we might as well call $A_{ij}$ and whose area is approximately equal that of $D_{1}$. 
Finally, observe that the RHS of these expressions also determine a parallelogram whose area is precisely $\det \vert \mathcal J\vert \Delta u\Delta v $ so we have now 
$A_{ij}\approx D_{1}\approx \det \vert \mathcal J\vert  \Delta u\Delta v$ as desired as soon as we set  $A'_{ij}=\det \vert \mathcal J\vert  \Delta u\Delta v$
