Find the area

(The area in question is the lower-right section of the circle shown above.)

Manipulating the equation of the circle, solving for $y$ and taking the negative square root gives the function for the lower half of the circle as $f(x) = 2 - \sqrt{4 - x^2}$ . The line $y = 1$ intersects the circle at $x = \sqrt{3}$ . This gives us the information we need to set up the integral to find the desired area:

$A = \int_0^{\sqrt{3}} \left(1 - \left(2 - \sqrt{4 - x^2}\right)\right) \: dx = \int_0^{\sqrt{3}} \left(\sqrt{4 - x^2} - 1\right) \: dx$ $= \left.\left(\frac{x}{2}\sqrt{4 - x^2} + 2\,\sin^{-1}\left(\frac{x}{2}\right) - x\right)\right|_0^{\sqrt{3}} = \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \approx \boxed{1.228}$