Integral of $x^2$ over $x^2+y^2 ≤ a^2$ 
This is in my lecture notes, I understand that $x=rcos\theta$ so therefore its the integral of $x^2=(rcos\theta)^2$ but why is there an r in $rdrd\theta$ in there?
Any help would be appreciated.
 A: Here's the underlying geometric intuition (not a rigorous argument).
In Cartesian coordinates the size of an infinitesimal rectangle with sides $dx$ and $dy$ is just the product $dxdy$, independent of where it is in the plane.
In polar coordinates, changing the angle changes the area more when you're farther from the origin. The area of a small not quite rectangular patch determined by $d\theta$ and $dr$ is $  dr \times r d\theta = r  dr d\theta$.
There's a picture here: http://citadel.sjfc.edu/faculty/kgreen/vector/block3/jacob/node4.html
A: When we convert co-ordinates $(x,y)$ to $(r,\theta)$, 
$dx\ dy = |J|\ dr \ d\theta$
where $J$ is the Jacobian of transformation.
$x = r \cos\theta \ , \ y = r  \sin\theta$ 
Taking partial derivatives,
$x_r = \cos\theta \ , y_r = \sin\theta$
$x_{\theta} =  - r \sin\theta \ , \ y_{\theta} = r \cos\theta $
$J = \begin{vmatrix} 
\cos\theta &  \sin\theta \\
- r \sin\theta  &  r \cos\theta \end{vmatrix}$
$J = r(\cos\theta)(\cos\theta) + r(\sin\theta)(\sin\theta) = r(\cos^2\theta + \sin^2\theta) = r$
$$|J| = r$$
Thus, $$dx \ dy = r \ dr \ d\theta$$
A: That is the formula for changing variables from cartesian coordinates to polar cordinates in double integrals $\mathrm d x\,\mathrm d y$ is replaced with $r\,\mathrm d r\,\mathrm d\theta$.
Intuitively, this come from the fact that the area of a small circular sector corresponding to a small increment  $\mathrm d r$ of the radius and a small increment of the angle $\mathrm d\theta$  is approximately $\;\mathrm dr\cdot r\,\mathrm d\theta$
