Area of a circle portion Say I have the following circle and I want to find the area bound between the axis and p, is there an easy way to do it? I tried using an integral but for some reason it doesn't have a nice clean form (or it times out of I use online integrators)

 A: As David says, it's easy to see that the area is $\frac{1}{2}(P\sqrt{1-P^2} + \arcsin P)$ by dividing your region into two parts.
Since you had trouble doing it with integration, here it is: the area is 
$$\int_0^P \sqrt{1-x^2}dx.$$
Now put $x=\sin \psi$, $dx = \cos \psi\:  d\psi$. Then the integral is
$$\int_0^{\arcsin P} \cos^2 \psi\: d\psi = \frac{1}{2}\int_0^{\arcsin P} (1+\cos{2\psi}) \: d\psi$$
$$= \frac{1}{2}(\arcsin P + \frac{1}{2}\sin(2\arcsin P)) = \frac{1}{2}(\arcsin P + P\sqrt{1-P^2}).$$
A: You can do this by elementary geometry. We'll divide up the first quadrant into various pieces, all of which can be readily computed. I'll assume your circle has radius 1.
The trick is to consider the radial wedge of the circle between the axis and the upper-right corner of your figure. The width of your figure will be $\cos \theta$ for some $0 \le \theta \le \pi/2$, and so this wedge will have area $\theta/2$. The total area of the circle in the first quadrant will be $\pi / 4$. The triangle formed by the radius of angle $\theta$ and the right edge of your figure has area $1/2 \cos \theta \sin \theta = 1/4 \sin 2\theta$. Thus, the portion of the circle in the first quadrant lying outside of your figure, is $\theta/2 - 1/4 \sin 2\theta$; the area of your figure is therefore $$1/4 \left( \pi + \sin 2 \theta - 2\theta \right).$$
