Changing the order of integration on a rectangular and polar region change the order of integration for the following integral from dydx to dxdy, and from dydx to polar coordinates.
$$ \int \int f(x,y) dydx$$ 
where 
$$ 0≤y≤(-x^2)+2 $$
$$ 0≤x≤1$$
From dydx to dxdy 
$$ \int \int f(x,y) dxdy + \int \int f(x,y) dxdy$$ 
First integral
$$ 0≤x≤1 $$
$$ 0≤y≤1 $$
Second integral
$$ 0≤x≤\sqrt(2-y) $$
$$ 0≤y≤1 $$
I'm not sure about the $\sqrt(2-y)$ for the bounds of x for the second integral. 
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I'm having more trouble converting this into polar coordinates though. I think I can leave the first integral as it is in terms of dxdy, because the region is a rectangle. Is there any way to switch this rectangular region into polar coordinates?
For the second integral
$$ 0≤r≤1 $$
$$ 0≤\theta≤\pi/2 $$
$$ \int \int f(x,y) dxdy + \int \int f(r,\theta) rdrd\theta$$ 
 A: 
Where the overlapping region of red and blue is the desired region.
Cut the region using the green line ($x-y=0$) instead.
The upper region becomes $\displaystyle\int_{\pi/4}^{\pi/2}\int_0^{(\sqrt{1+7\cos^2\theta}-\sin\theta)/(2\cos^2\theta)}f(r,\theta)\ r\ \mathrm dr\ \mathrm d\theta$.
The lower region becomes $\displaystyle\int_{0}^{\pi/4}\int_0^{\sec\theta}f(r,\theta)\ r\ \mathrm dr\ \mathrm d\theta$.
Note: $(\sqrt{1+7\cos^2\theta}-\sin\theta)/(2\cos^2\theta)$ is the positive solution of $(r\cos\theta)^2+(r\sin\theta)=2$
A: your work fipping the order of integration is correct.
coverting to polar -- it is going to get messy.
$x = r \cos \theta\\
y = r \sin \theta$
Inside the rectangular region.
$\theta \in [0,\frac{\pi}{4})\\
x =1\\ 
r \cos\theta = 1\\
r = \sec\theta$
Inside the parabola
$\theta \in [\frac{\pi}{4},\frac{\pi}{2}]\\
y = -x^2 + 2\\
r \sin \theta = - r^2 cos^2 \theta + 2\\
r^2 \cos^2 \theta  + r \sin \theta - 2 = 0\\
r = \frac {-\sin\theta + \sqrt{sin^2 \theta + 8 cos^2 \theta}}{2 cos^2 \theta}$
I wouldn't want to integrate that, but it would be a limit.
