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Let $\Omega\subset \mathbb R^2$ closed and bounded. If we make the substitution $x=g(u,v)$ and $y=h(u,v)$, Why $$\iint_\Omega f(x,y)dxdy=\iint_{\Sigma}f(g(u,v),h(u,v))|J|dudv$$ where $|J|$ is the jacobian of $(u,v)\longmapsto (g(u,v),h(u,v))$ and $\Sigma=\{(u,v)\mid (x(u,v),y(u,v))\in \Omega \}$. It's in fact the Jacobian that I don't understand. Indeed,

$$dx=\frac{\partial g}{\partial u}du+\frac{\partial g}{\partial v}dv$$ $$dy=\frac{\partial h}{\partial u}du+\frac{\partial h}{\partial v}dv$$ and thus, using the convention that $dudu=dvdv=0$ (because it's very small), I get $$dxdy= \frac{\partial g}{\partial u}\frac{\partial h}{\partial v}dudv+\frac{\partial g}{\partial v}\frac{\partial h}{\partial u}dvdu=\det\begin{pmatrix}\frac{\partial g}{\partial u}&-\frac{\partial g}{\partial v}\\\frac{\partial h}{\partial u}&\frac{\partial h}{\partial v}\end{pmatrix}dudv,$$ which is not exactly the Jacobian (but very near to be). Is there any explanation ?

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    The exterior product of differential forms is not commutative. In fact one has $dudv=du\wedge dv=-dv\wedge du=-dvdu$. – 2017-01-22
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    @FrankLu: But I always say for example (in Fubini), $$\iint f(x,y)dxdy=\int\left(\int f(x,y)dx\right)dy=\int\left(\int f(x,y)dy\right)dx=\iint f(x,y)dydx...$$ – 2017-01-22
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    Informally(a little slopy here) speaking, when you switch the order of integration, the domain of integration is to be regarded as having the opposite orietation. When the oriatation is reversed, there comes up a factor $-1$, and two $-1$'s cancel out. – 2017-01-22
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    So you mean that for $(x,y)\in [a,b]\times [c,d]$, we have $$\int_{a}^b dx\int_{c}^ddy=\int_c^d dx\int_a^b dy=\int_{c}^d dy\int_a^b dx$$ ??? – 2017-01-22
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    When you switch $dx$ and $dy$, the integrals should also switch. – 2017-01-22
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    @FrankLu: Sorry, after my first equality, I wanted to put a $-$, i.e. $$\int_a^b dx\int_c^d dy=-\int_c^d dx\int_a^b dy=\int_c^d dy\int_a^b dx.$$ Now, is correct ? – 2017-01-22
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    No, what I mean is the integrals and $dx,dy$ should switch at the same time, because $x\in [a,b]$ and $y\in [c,d]$. Thus $\int_c^d dx$ may not be well defined. – 2017-01-22
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    @FrankLu: Ok, so I still don't understand why $dxdy=-dydx$... It doesn't look natural ! – 2017-01-22
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    Because the operation $dxdy$ here is not the regular multiplication, it's something called the exterior/wedge product, and this has different properties. I don't know whether I would be able to give you a convincing/comprehensive answer, if you are curious about this, you may google "differential forms". – 2017-01-22
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    BTW, when you pass to iterated integrals, you no longer multiply those differential forms. For example if you integrate $x$ first, then $\int f(x,y)dx$ will give you something only depend on $y$, and then you integrate this w.r.t $y$. – 2017-01-22

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