Double Integral, Change of Variables to Polar Coordinates Quick question on Polar Coordinates.
When evaluating the double integral and changing variables, I'm not sure if the limits are correct.
The question is as follows:

Evaluate $$\int\!\!\!\int_D xy\sqrt{x^2 + y^2}\,dxdy $$where 
  $D = \{(x,y) \mid 1 \leq x^2 + y^2 \leq 4,\  x \geq 0,\ y \geq 0\}$

So my question is when I change to polar coordinates, is the limit for the integral with respect to r from 1 to 2 or 1 to 4?
Instinctively, I would say it's 1 to 4 but the answer given out by the lecturer (which does not have all the steps) has the limts at 1 to 2.
Is it maybe because $x^2 + y^2 = a^2$?
Note: I have the new integral, in terms of r and $\theta$ as:
$\int$$\int$$r^4$cos$\theta$sin$\theta$drd$\theta$

Look at the actual variables in your integral. You have $\sqrt{x^2+y^2}=\sqrt{r^2}=|r|$, $x=r\cos\theta$, and $y=r\sin\theta$, and you have $dxdy=r dr d\theta$, so your integrand must be $|r|r^3 \sin\theta\cos\theta dr d\theta$. Now look at the region over which you’re integrating:
$$\begin{align*}&1\le x^2+y^2\le 4\;,\tag{1}\\
&x\ge 0\;,\tag{2}\\
&y\ge 0\;.\tag{e}
\end{align*}$$
What limits on $r$ and $\theta$ describe this region? $(1)$ says that $1\le r^2\le 4$, so $1\le |r|\le 2$. $(2)$ and $(3)$ say that the region is limited to the first quadrant. If you take $0\le\theta\le\pi/2$, you stay in the first quadrant provided that you also keep $r\ge 0$. Thus, $(1)-(3)$ translate to
$$\begin{align*}
&1\le r\le 2\;,\\\\
&0\le\theta\le\frac{\pi}2\;.
\end{align*}$$
Your integral should therefore be $$\int_0^{\pi/2}\int_1^2 r^4\sin\theta \cos\theta dr d\theta\;.$$
