TITLE: Why is $dy dx = r dr d \theta$
QUESTION [10 upvotes]: Possible Duplicate:
Explain $\iint \mathrm dx\mathrm dy = \iint r \mathrm d\alpha\mathrm dr$ 

I'm reading the proof of Gaussian integration. When we change to polar coordinates, why do we get an "extra" r in there? 

\begin{align}
\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)}\ dx dy
&= \int_0^{2\pi} \int_0^{\infty} e^{-r^2}r\ dr\ d\theta\\
\end{align}

I've looked at a few different proofs:

http://www.math.uconn.edu/~kconrad/blurbs/analysis/probint.pdf
"The differential dx dy represents an element of area in cartesian
coordinates, with the domain of integration extending over the
entire xy-plane. An alternative representation of the last integral
can be expressed in plane polar coordinates r, $\theta$"
http://www.umich.edu/~chem461/Gaussian%20Integrals.pdf

but none explain this step fully enough for me to really see why this happened.

REPLY [8 votes]: When you make the change of variables $x=r\cos \theta,y=r\sin \theta$, the integral becomes
$$ \int_D f(x,y)dxdy=\int_{D'} f(r\cos \theta,r\sin\theta) J(r,\theta) drd\theta \ \ (F)$$
where $D'$ is the changed domain, where $r,\theta$ belong, and $\displaystyle J(r,\theta)=\left| \begin{matrix} \frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta}\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta}\end{matrix}\right| $ is the jacobian matrix.
The formula $(F)$ stays true even when you make different change of variables for $x,y$. This is the theory behind $dxdy=rdrd\theta$. 
For a proof of $(F)$ you need to use Jordan measurable sets (I think ) and the definition of the double integral. Of course, this works in higher dimensions, with more intricate computations.
You can take a look in the wikipedia article: http://en.wikipedia.org/wiki/Multiple_integral

REPLY [2 votes]: Changing coordinates in multiple integrals requires adding the Jacobian as a factor. The Jacobian in this case is $r (\cos^2 \theta + \sin^2 \theta) = r$.