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 &amp; = r \cos \theta\\
y &amp; = r \sin \theta\\
dx &amp; = dr \cos \theta - r \sin \theta\ d\theta\\
dy &amp; = 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 &amp; = \left(dr \cos \theta - r \sin \theta\ d\theta \right) \left( dr \sin \theta + r \cos \theta\ d\theta\right)\\
&amp; = 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 :)
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
$$
\begin{align}
\mathrm{d}y\,\mathrm{d}x
&amp;=(\sin(\theta)\,\mathrm{d}r+r\cos(\theta)\,\mathrm{d}\theta)(\cos(\theta)\,\mathrm{d}r-r\sin(\theta)\,\mathrm{d}\theta)\\
&amp;=r\,\mathrm{d}\theta\,\mathrm{d}r
\end{align}
$$
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
$$
\begin{align}
\mathrm{d}x&amp;=\cos(\theta)\,\mathrm{d}r-r\sin(\theta)\,\mathrm{d}\theta\\
\mathrm{d}y&amp;=\sin(\theta)\,\mathrm{d}r+r\cos(\theta)\,\mathrm{d}\theta
\end{align}
$$
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.
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
  $$\begin{align}dx\cdot dy &amp;= (-r\sin\theta\, d\theta+ \cos\theta\, dr).(r\cos\theta\, d\theta + \sin\theta\, dr)\\ &amp;= - r\cdot r \sin\theta\cdot\cos\theta\, d\theta\cdot d\theta + r\cos\theta\cdot\cos\theta\, dr\cdot d\theta - r \sin\theta\sin\theta\, dr\,d\theta +\sin\theta\cdot\cos\theta\,dr\,dr\end{align}$$