A circle of radius $1$ is tangent to $y=x^2$ at two points. Find the area bounded by the circle and the parabola. (Alternate solutions?) 
This is from Rambo's Math subject GRE book.
One solution to this problem is to note that the equation of the circle is $x^2+(y-k)^2=1$. By taking the derivative of this and solving for $y'$, and then setting this equal to the derivative of the parabola, $y'=2x$, we can solve for $x$ and $k$ and get all the information we need to set up the necessary integeral:
$2\int_0^{\frac{\sqrt{3}}{2}}\frac{5}{4}-\sqrt{1-x^2}-x^2dx=\frac{3\sqrt{3}}{4}-\frac{\pi}{3}$
Where the expression involving a square root can is evaluated with trig substitution and power reduction identities.
The author of the text mentions that this problem also be solved by using some trigonometry and the fact that:
$2\int_0^{\frac{\sqrt{3}}{2}}x^2dx=\frac{\sqrt{3}}{2}$
However I can't figure out what he means by this. Can anyone show me how to solve this problem using less calculus and more trigonometry? Also, if anyone has any other alternative methods, I'd greatly appreciate it. Thanks!
 A: 
Equation of the circle is $~x^2 + (y-k)^2 = 1$
Equation of the parabola is $~y = x^2$
If they are tangent to each other,
$y + (y-k)^2 = 1 \implies y^2 - (2k-1) y + (k^2-1) = 0~$ should have a double root.
So we must have, $(2k-1)^2 - 4 (k^2-1) = 0 \implies k = \dfrac 54$
At point of tangency, $ \displaystyle y = \frac{2k-1}{2} = \frac 34, x = \pm \sqrt y = \pm \frac{\sqrt{3}}{2}$
$ \displaystyle \cos \alpha = \frac{\sqrt3}{2} \implies \alpha = \frac {\pi}{3}$
We can just focus on the desired area to the right of y-axis and then using symmetry, we can multiply the result by $2$.
Half of the desired area $~ = [OABC] - [BCE] - [OAB]$ (note we use the curved path $BE$ and $OB$)
$ \displaystyle [OABC] = \frac 12 \left(\frac 54 + \frac 34\right) \cdot \frac {\sqrt3}{2} = \frac{\sqrt3}{2}$
$ \displaystyle [BCE] = \frac {\pi}{6}$
$ \displaystyle [OAB] = \int_0^{\sqrt3/2} x^2 ~ dx = \frac{\sqrt3}{8} $
So, the desired area $ ~ = \displaystyle 2 \cdot \left(\frac{\sqrt3}{2} - \frac{\sqrt3}{8} - \frac{\pi}{6}\right) = \frac{3 \sqrt3}{4} - \frac{\pi}{3}$
A: The area swept out by a segment of length $r$ as it rotates about one end point through the angle $\theta$ is $\frac{1}{2}r^2\theta$. Now, since you already know how to find the points of intersection, it's pretty easy to use "a little trigonometry" to show that the marked angle is $\pi/3$. Thus, the area of the shaded sector shown below is $\pi/3$.

A: This solution is not by me, but available in Charles Rambo book,



