$\iint_Ex\ dx\ dy$ over $E=\lbrace(x,y)\mid 0\le x, 0\le y\le 1, 1\le x^2+y^2\le 4\rbrace$ (Answer check and curious about alternative methods) $$\iint_Ex\ dx\ dy$$
$$E=\lbrace(x,y)\mid 0\le x, 0\le y\le 1, 1\le x^2+y^2\le 4\rbrace$$
Shape of region
Entirely in first quadrant of xy plane, between two circles, of r=1 and r=2 respectively (about the origin), and below the line y=1.
I split this up into two regions:
1) A nice segment, $\lbrace(r,\theta)\mid r\in[1,2],\theta\in[0,\frac{\pi}{6}]\rbrace$
2) A part I just called "Other part", which is: $\lbrace(r,\theta)\mid\theta\in[\frac{\pi}{6},\frac{\pi}{2}],1\le r\le\frac{1}{\sin(\theta)}\rbrace$
I made a small note about using two closed regions (rather than one open and one closed at $\theta=\frac{\pi}{6}$) being fine because of continuity.
Method
So now $\iint_Ex\ dA=\iint_Ex\ dx\ dy=\iint_Er\cos(\theta)r\ d\theta\ dr$ = $\iint_Er^2\cos(\theta)d\theta\ dr$
Which I just did over the two regions, it was nice and straight forward, I'll provide more details if anyone thinks I'm trying to scavenge answers.
To get $$\iint_Ex \ dx\ dy=\frac{3}{2}$$
Twist

Hints: It may be helpful to use polar-coordinates and to do the angular integration first, noting that the polar description of the line y=1 is $r\sin(\theta)=1$

When I did it I had two integrals (one for each of the regions, the segment and the other part) the segment with it's nice easy r from 1 to 2 made it clear I wanted to do the $\int^\frac{\pi}{6}_0\cos(\theta)d\theta$ first, so I did, the other integral having bounds from 1 to $\frac{1}{\sin(\theta)}$ basically required I do radial integration first.
I worked out the $\frac{\pi}{6}$ and whatnot just by looking at the triangle, created by the x-axis and the line dividing my two regions. it must have a point r=2 and y=1, that is $\sin(\theta)=\frac{1}{2}$ thus the $\frac{\pi}{6}$.
I also noticed that after $\theta=\frac{\pi}{6}$ that $r\sin(\theta)=1$, this is in the hints though.
Anyway I didn't adhere to the hints, I found this way, which (if my answer was right, which I believe it to be) seems nicer.
Can anyone see a faster way, or what the hint was expecting, that one integral required a radial evaluation first. (unless the change of order would have been easier than I thought, I decided not to bother looking to change for the sake of a hint when I had a clear path) 

Let's use Stoke's Theorem:
\begin{align}
\iint_{E}x\,{\rm d}x\,{\rm d}y
&=
\iint_{E}\,{\partial\left(x^{2}/2\right) \over \partial x}\,{\rm d}x\,{\rm dy}
=
\iint_{E}\nabla\times\left({1 \over 2}\,x^{2}\,\hat{y}\right)\,\cdot\hat{z}
\,{\rm d}x\,{\rm dy}
=
{1 \over 2}\oint x^{2}\,\hat{y}\cdot{\rm d}\vec{r}
\\[3mm]&=
{1 \over 2}\int_{\rm GA \bigcup RA} x^{2}\,{\rm d}y
\end{align}
${\rm GA}$ and ${\rm RA}$ stand for ${\rm G}$reen ${\rm A}$rc and
${\rm R}$ed ${\rm A}$rc, respectively. Also, for an arc of radius $a$ which spans angles in $\left(0, \beta\right)$:
\begin{align}
\int_{0}^{\beta}{\rm x}^{2}\left(\theta\right)\,
{{\rm d}{\rm y}\left(\theta\right) \over {\rm d}\theta}\,{\rm d}\theta
&=
\int_{0}^{\beta}a^{2}\cos^{2}\left(\theta\right)
\left\lbrack a\cos\left(\theta\right)\right\rbrack\,{\rm d}\theta
=
a^{3}\int_{0}^{\beta}\cos^{3}\left(\theta\right)\,{\rm d}\theta
\\[3mm]&=
{1 \over 12}\,a^{3}\left\lbrack%
9\sin\left(\beta\right) + \sin\left(3\beta\right)
\right\rbrack
\\[5mm]&
\end{align}
\begin{align}
{1 \over 2}\int_{\rm GA} x^{2}\,{\rm d}y
&=
{1 \over 2}\,2^{3}\int_{0}^{\pi/6}\cos^{3}\left(\theta\right)\,{\rm d}\theta
=
4\,{1 \over 12}\,\left\lbrack%
9\sin\left(\pi \over 6\right) + \sin\left(3\,{\pi \over 6}\right)
\right\rbrack
=
{11 \over 6}
\\[3mm]
{1 \over 2}\int_{\rm RA} x^{2}\,{\rm d}y
&=
{1 \over 2}\,1^{3}\int^{0}_{\pi/2}\cos^{3}\left(\theta\right)\,{\rm d}\theta
=
{1 \over 2}\,{-1 \over 12}\,\left\lbrack%
9\sin\left(\pi \over 2\right) + \sin\left(3\,{\pi \over 2}\right)
\right\rbrack
=
-\,{1 \over 3}
\\[1cm]&
\end{align}
$$
\begin{array}{|c|}\hline\\
\quad\iint_{E}x\,{\rm d}x\,{\rm d}y
=
{11 \over 6} + \left(-\,{1 \over 3}\right)
=
{\large{3 \over 2}}\quad
\\
\\
\hline
\end{array}
$$
