Double / Surface Integral over a circular area To compute is the following integral:
$$\int_{A} dx \ dy \ (x + y)^2$$
over the area $A$, which is the circular area between: $r_0 ^2 < x^2 + y^2 < r_1^2$
My approach:
1) Parametrize: 
$$x( \theta) = r \cos(\theta)  \\ y(\theta) = r \sin(\theta) $$
2) Set up the double Integral: 
$$\int_0^{2\pi} d\theta \int_{r_o}^{r_1} dr \ (r\cos(\theta) + r\sin(\theta))^2$$
Now I encounter a Problem, the Integral at point 2) is wrong, apparently I should multiply the whole thing by the radius. and the Integral should look like this: 
$$\int_0^{2\pi} d\theta \int_{r_o}^{r_1} dr \ r(r\cos(\theta) + r\sin(\theta))^2$$
Where does that $r$ come from? 
 A: As was said in the comments, $r$ is the Jacobian of the change of variables from Cartesian to Polar. You can find more about this in this Wikipedia page. 
An intuitive reason is that while $dxdy$ represents the area spanned by an infinitesimal change in each variable $x$ and $y$, the area spanned by an infinitesimal change of $r$ and $\theta$ is an infinitesimal section of an annulus.
The area of that section of an annulus is "close" (up to higher order infinitesimals) to the area of a rectangle of sides $dr$ and $rd\theta$. This $rd\theta$ is the distance yielded by a change of angle by $d\theta$ at a distance $r$ from the origin.
A: The Jacobian is like a transfer matrix when switching between coordinate systems and you have to take its determinant
This case has $r$ constant but in general, it doesn't have to be
$\displaystyle x(r,\theta)=r\cos\theta\\y(r,\theta)=r\sin\theta$
$$\iint_Df(x,y)dxdy=\iint_Df\left(x(r,\theta),y(r,\theta)\right)Jdrd\theta$$
Where J is the Jacobian
So
$$J=\frac{\partial(x,y)}{\partial(r,\theta)}=\begin{vmatrix}\frac{\partial x}{\partial r}&\frac{\partial x}{\partial \theta}\\\frac{\partial y}{\partial r}&\frac{\partial y}{\partial \theta}\end{vmatrix}=\frac{\partial x}{\partial r}\frac{\partial y}{\partial \theta}-\frac{\partial x}{\partial\theta}\frac{\partial y}{\partial r}=\cos\theta\left(r\cos\theta\right)-\left(-r\sin\theta\sin\theta\right)$$
Simplifying gives that
$$J = r\left(\cos^2\theta+\sin^2\theta\right)=r$$
