# Why is $dy dx = r dr d \theta$ [duplicate]

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:

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

## marked as duplicate by Aryabhata, t.b., Ross Millikan, Asaf Karagila♦, Américo TavaresJul 6 '11 at 19:57

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

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