calculus - 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

$$0≤r≤1$$
$$0≤\theta≤\pi/2$$

$$\int \int f(x,y) dxdy + \int \int f(r,\theta) rdrd\theta$$

Answer

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.