<s>[INST] How can I calculate the area of a circle centered at (2,2) with radius 2 using double integrals 
This is what the graph loosk like.
Obviously I know $\pi r^2$ but if I specifically wanted to find the area using double integrals in polar coordinates, how would I go about it?
My guess is that $\theta$ goes from $0$ to $\pi/2$
and for $r$ I need to do:
$(x-2)^2+(y-2)^2=4$
$x^2-4y+4+y^2-4y+4=4$
$(x^2+y^2)-4(x+y)=-4$
$r^2-4(r\cos\theta+r\sin\theta)=-4$
$r^2=4(r\cos\theta+r\sin\theta)-4$
Not exactly sure what to do from here.
 [/INST] After getting $r^2=4(r\cos\theta+r\sin\theta)-4$ you can solve for $r$ to get $r_{1,2}(\theta) = 2\left(\sin\theta+\cos\theta\pm\sqrt{2\sin\theta\cos\theta}\right)$
The two solutions are the two intersections of the ray from the origin with the circle:

When using polar coordinates, the area can be obtained as $\frac{1}{2}\int_a^b r^2(\theta)\;d\theta$
So in this case you'll need to calculate
\begin{align}
&\frac{1}{2}\int_0^{\pi/2}r_1^2(\theta)\;d\theta - \frac{1}{2}\int_0^{\pi/2}r_2^2(\theta)\;d\theta \\
& = \frac{1}{2}\int_0^{\pi/2}(r_1^2(\theta)-r_2^2(\theta))\;d\theta\\
& =  8 \int_0^{\pi/2}(\sin\theta + \cos\theta)\sqrt{2\sin\theta\cos\theta}\;d\theta
\end{align}
which, after some calculations, gives $4\pi$.
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