I'm calculating the area of the region $A$ defined by the following constraints:
$$ A = \{(x, y) \mid x^2 + y^2 \geq 2, \; x^2 + y^2 \leq 2x, \; y \geq 0\} $$
To calculate the area of the region $A$, we can set up the double integral in Cartesian coordinates as:
$$ A = \iint_A \ dx \ dy $$
The constraints $x^2 + y^2 \geq 2$ and $x^2 + y^2 \leq 2x$ define region $A$ in the $xy$-plane. Additionally, the constraint $y \geq 0$ ensures that the region is above the $x$-axis.
Transforming to cylindrical coordinates, where $x = r \cos\theta$ and $y = r \sin\theta$, the constraints are expressed as $r^2 \geq 2$ and $r^2 \leq 2r \cos\theta$. The angle $\theta$ varies between $0$ and $\pi$ to address the upper half of the $xy$-plane.
The integral to calculate the area is:
$$ A = \int_{0}^{\pi} \int_{\sqrt{2}}^{2\cos\theta} \ r \ dr \ d\theta $$
I appreciate any additional guidance or methods that can facilitate solving this integral and calculating the area of region $A$. Thank you in advance for your assistance!
The way to solve this type of problem is by understanding the region. $$x^2+y^2\ge 2$$ This means that is oytside of the circle of radius $\sqrt 2$ centered on origin $$x^2+y^2\le 2x$$We can rewrite this: $$x^2-2x+y^2\le 0\\x^2-2x+1+y^2\le 1\\(x-1)^2+y^2\le 1$$ This is the inside of the circle of radius $1$, centered on $(1,0)$. It looks like this
Since you are outside of the red circle, your $\theta$ varies only between $0$ and $\pi/4$. Otherwise $r=2\cos\theta<\sqrt 2$. Can you take it from here?