Finding area of section of a circle So my calculus teacher made a test in which she made a mistake. 
Their was a piecewise function she graphed and we were supposed to do a definite integral on. 
Part of this function was a semicircle with a radius of $2$ and a center at the origin. 
Now if I had to calculate the integral from $0$ to $2$ that would be easy! It's just $\pi$. Or if I had to do $0$ to $-2$ that would be easy as well it is just $-\pi$. 
However she misspelled the question...  AND as a result we had to calculate the integral from $1$ to $2$... Which really confused me. 
While she meant to write "$0$ to $2$" I am still curious..... How would one calculate the area of a section of the semicircle?
 A: Well, the function of a semicircle above the $x$-axis is simply this:
$$y=\sqrt{ r^2 - x^2 }$$
where $r$ is the radius of the circle (in your case, it is $2$).
You would then find the definite integral from $1$ to $2$ of that function.
A: Lets first look at this using classical geometry and a little trig.
The area area in question is the area of a section of the circle (the pie slice including the origin) less the area of a triangle.
The section of a cirlce: $A = \frac {1}{2} r^2 \theta$.
The triangle: $A = \frac{1}{2} bh$
$b = r\cos\theta, h=r\sin \theta$
$A = \frac{1}{2} r^2 (\theta - \sin \theta cos \theta)$
and in this case $r=2$ and $\theta = \pi/3$
Using calculus:
$x^2 + y^2 = 4$
$\int_1^2 (4-x^2)^{1/2} dx$
$x = 2 \cos \theta$
$dx = -2 \sin \theta \, d\theta$
limits of integration:
$2 \cos \theta = 1; \theta = \pi/3$
$2 \cos \theta = 2; \theta = 0$
$\int_{\pi/3}^{0} (4-4\cos^2\theta)^{1/2} (-2\sin\theta \, d\theta)$
$\int_{0}^{\pi/3} 4 sin^2 \theta \, d\theta$
Half angle indentity:
$2 sin^2\theta = 1 - \cos\,2\theta$
$\int_{0}^{\pi/3} 2(1- \cos\,2\theta) \, d\theta$
$2(\theta - \frac{1}{2} \sin\,2\theta) |_0^{\pi/3}$
$2(\theta -  \sin\theta\cos\theta) |_0^{\pi/3}$
$2(\pi/3 -  \frac{\sqrt 3}{4})$
