Simple proof of integration in polar coordinates? In every example I saw of integration in polar coordinates the 
Jacobian determinant is used, not that I have a problem with the Jacobian,
but I wondered if there's a simpler way to show this which will also give
me some more intuition about the Jacobian.
If I try to simply write the differentials:
\begin{align}
x & = r \cos \theta\\
y & = r \sin \theta\\
dx & = dr \cos \theta - r \sin \theta\ d\theta\\
dy & = dr \sin \theta + r \cos \theta\ d\theta\\
\end{align}
In a double integral you integrate $dxdy$, so if I try to plug in the results I'll get something which is not $r d\theta dr$
\begin{align}
dxdy & = \left(dr \cos \theta - r \sin \theta\ d\theta \right) \left( dr \sin \theta + r \cos \theta\ d\theta\right)\\
& = dr^2 \cos \theta \sin \theta  - r^2 d\theta^2 \cos \theta\ \sin\ \theta + r\ dr\ d\theta\ (\cos^2 \theta\ - \sin^2\theta )
\end{align}
I don't think I can go anywhere from here,
I'm not sure if it's just a calculation mistake or the entire logic is bad.
How do I get this right?
Thanx :)
 A: The area element needs to be computed carefully.
This is very informal, but perhaps you should think of the volume element as a pair, rather than just a product, as in:
$$\binom{dx}{dy} = \begin{bmatrix} \cos \theta & - r \sin \theta \\
\sin \theta &  r \cos \theta \end{bmatrix}\binom{dr}{d\theta}$$
Multiplying a set by a matrix $A$ corresponds to changing the volume by a factor $\det A$. In this case, $\det A = r$, so the volume element computation becomes, informally,  $dxdy = r dr d\theta$.
Addendum: The two volume elements are, informally, $[r,r+dr]\times[\theta,\theta+d\theta]$, and $[x,x+dx]\times[y,y+dy]$.
A: The short answer is that you need to consider the wedge product between the differentials, not a symmetric product as you have written.  The reason for this is addressed in Robjohn's excellent answer. 
The volume element, $dV$, is formally given by the wedge product of $dx_1, \dots, dx_n$.  This means that in $\mathbb{R}^2$ the volume element in Cartesian coordinates should technically be written as 
$$
dV = dx \wedge dy.
$$
Note that the wedge product is antisymmetric, which means that in particular $dx \wedge dy = - dy \wedge dx.$  Taking this into consideration, if you perfrom this wedge product between the $dx$ and $dy$ you calculated, we have the following 
$$\begin{align*}
dV &= dx \wedge dy \\
   &= (\cos\theta ~dr - r\sin\theta ~d\theta) \wedge (\sin\theta ~dr + r\cos\theta~ d\theta) \\
   &= \cos\theta \sin\theta ~dr\wedge dr + r \cos^2\theta ~dr \wedge d\theta - r\sin^2\theta~ d\theta \wedge dr -r \sin\theta\cos\theta~ d\theta \wedge d\theta\\
   &= r(\cos^2\theta + \sin^2\theta) dr \wedge d\theta\\
   &= r~ dr \wedge d\theta,
\end{align*} $$
where we have also used the fact that $dr \wedge dr = d\theta \wedge d\theta = 0.$  If you calculate the determinant of the Jacobian you'll find $$\det\left(\dfrac{\partial(x,y)}{\partial{(r,\theta)}}\right) = r,$$ which concurs with the above calculation.  
A: Of course, if you break $\mathbb{R}^2$ into a polar grid
$\hspace{3.5cm}$
the small slightly curved rectangles have area $r\,\mathrm{d}\theta\,\mathrm{d}r$.
However, it seems that you are interested in looking at
$$
\begin{align}
\mathrm{d}y\,\mathrm{d}x
&=(\sin(\theta)\,\mathrm{d}r+r\cos(\theta)\,\mathrm{d}\theta)(\cos(\theta)\,\mathrm{d}r-r\sin(\theta)\,\mathrm{d}\theta)\\
&=r\,\mathrm{d}\theta\,\mathrm{d}r
\end{align}
$$
and why the $\mathrm{d}r^2$ and $\mathrm{d}\theta^2$ terms disappear and the $\mathrm{d}r\,\mathrm{d}\theta$ and $\mathrm{d}\theta\,\mathrm{d}r$ have different signs.
Let's start with
$$
\begin{align}
\mathrm{d}x&=\cos(\theta)\,\mathrm{d}r-r\sin(\theta)\,\mathrm{d}\theta\\
\mathrm{d}y&=\sin(\theta)\,\mathrm{d}r+r\cos(\theta)\,\mathrm{d}\theta
\end{align}
$$
rewritten as
$$
\begin{bmatrix}\mathrm{d}x\\\mathrm{d}y\end{bmatrix}
=\begin{bmatrix}\cos(\theta)\\\sin(\theta)\end{bmatrix}\mathrm{d}r
+\begin{bmatrix}-r\sin(\theta)\\r\cos(\theta)\end{bmatrix}\mathrm{d}\theta
$$
Therefore, the displacements $\color{green}{\mathrm{d}r}$ and $\color{red}{\mathrm{d}\theta}$ get mapped to  $\color{green}{\begin{bmatrix}\cos(\theta)\\\sin(\theta)\end{bmatrix}\mathrm{d}r}$ and $\color{red}{\begin{bmatrix}-r\sin(\theta)\\r\cos(\theta)\end{bmatrix}\mathrm{d}\theta}$ in $\mathbb{R}^2$:
$\hspace{3cm}$
where the area in gray is given by $\color{green}{\begin{bmatrix}\cos(\theta)\\\sin(\theta)\end{bmatrix}\mathrm{d}r}\times\color{red}{\begin{bmatrix}-r\sin(\theta)\\r\cos(\theta)\end{bmatrix}\mathrm{d}\theta}=r\,\mathrm{d}r\,\mathrm{d}\theta$.
The fact that the cross product is involved is the reason that the $\mathrm{d}r^2$ and $\mathrm{d}\theta^2$ terms disappear and the $\mathrm{d}r\,\mathrm{d}\theta$ and $\mathrm{d}\theta\,\mathrm{d}r$ have different signs. This, and its $n$-dimensional analogs, are why we use wedge products and differential forms when changing variables.
A: We know that
$$x=r\cos\theta,\quad y= r\sin\theta$$ 
Differentiate b.s respectively, we get
$$dx= -r\sin\theta\, d\theta+ \cos\theta\, dr,\quad dy= r\cos\theta\, d\theta + \sin\theta\, dr$$
Then
  $$\begin{align}dx\cdot dy &= (-r\sin\theta\, d\theta+ \cos\theta\, dr).(r\cos\theta\, d\theta + \sin\theta\, dr)\\ &= - r\cdot r \sin\theta\cdot\cos\theta\, d\theta\cdot d\theta + r\cos\theta\cdot\cos\theta\, dr\cdot d\theta - r \sin\theta\sin\theta\, dr\,d\theta +\sin\theta\cdot\cos\theta\,dr\,dr\end{align}$$
