Inquiring about change of parameters in double (OR Triple) integrals It's known that while replacing $x$ and $y$ coordinates in double integration, by another coordinate system $u(x,y)$ and $v(x,y)$, then
$$ \int_{a_2}^{b_2} \int_{a_1(y)}^{b_1(y)} f\left(x,y \right) \: dx \;dy=\int_{c_2}^{d_2} \int_{c_1(v)}^{d_1(v)} f\left(x(u,v),y(u,v) \right) \:\left|J\right| du \;dv=$$
Where $\left|J\right|$ is called: Jacobian determinant which is defined as:
$$\left|J \right|= \left |\begin{matrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ 
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}
\end{matrix}  \right | $$
In an attempt to prove that $dx\; dy=\left|J\right| du\;dv$, I firstly get the total differentation of $x$ and $y$ since they are functions of $u$ and $v$, so:
$$dx=\frac{\partial x}{\partial u} du+\frac{\partial x}{\partial v} dv$$
And
$$dy=\frac{\partial y}{\partial u} du+\frac{\partial y}{\partial v} dv$$
Multipling $dx$ and $dy$
$$\therefore dx \; dy=\left(\frac{\partial x}{\partial u} du+\frac{\partial x}{\partial v} dv \right)\left(\frac{\partial y}{\partial u} du+\frac{\partial y}{\partial v} dv \right) \\= \frac{\partial x}{\partial u} \frac{\partial u}{\partial u} du\;du+\frac{\partial x}{\partial v}\frac{\partial y}{\partial v}dv\;dv+du\;dv\left( \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} + \frac{\partial x}{\partial v} \frac{\partial y}{\partial u}\right) $$
Which is too far from the formula $\left|J\right| du\;dv$.
So, what's the wrong?
 A: You're manipulating the differentials via the wrong product. Anyways, the differential vectors are written, via the Jacobian matrix $J$, $\vec{dx} = \left( \begin{array}{ccc} \frac{\partial x}{\partial u}&\frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} \right)\left( \begin{array}{ccc} du \\ 0 \end{array} \right)$ and $\vec{dy}=\left( \begin{array}{ccc} \frac{\partial x}{\partial u}&\frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} \right)\left( \begin{array}{ccc} 0 \\ dv \end{array} \right)$. Thus, $$dA=|\vec{dx}\times \vec{dy}|
 \\ =\left( \begin{array}{ccc} \frac{\partial x}{\partial u} \\ \frac{\partial y}{\partial u} \end{array} \right)du\times \left( \begin{array}{ccc} \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial v} \end{array} \right)dv \\ =\left| \begin{array}{ccc} \frac{\partial x}{\partial u}&\frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} \right|dudv \\ = |J|dudv.$$
A: That's because you are treating $dx, dy$ as symmetric product, but they are actually antisymmetric: 
$$dx\, dy = -dy\, dx$$
(Think of them as infinitesimal signed area). In particular, 
$$du\,du = 0 = dv\,dv$$
and so 
$$\left(\frac{\partial x}{\partial u} du+\frac{\partial x}{\partial v} dv \right)\left(\frac{\partial y}{\partial u} du+\frac{\partial y}{\partial v} dv \right) = \left( \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u}\right) du \,dv$$
(The extra negative sign comes from $dv\,du = -du\, dv$)
