Find the area of the smallest region bounded by the graphs of $y=|x|$ and $x^2+y^2=4$.
$\textbf{(A) }\frac{\pi}{4}\qquad \textbf{(B) }\frac{3\pi}{4}\qquad \textbf{(C) }\pi\qquad \textbf{(D) }\frac{3\pi}{2}\qquad \textbf{(E) }2\pi$
Solution by e_power_pi_times_i
The graph of $x^2+y^2 = 4$ is a circle with radius $2$ centered at the origin. The graph of $y=|x|$ is the combined graphs of $y=x$ and $y=-x$ with a nonnegative y. Because the arguments of $y=x$ and $y=-x$ are $135^\circ$ and $45^\circ$ respectively, the angle between the graphs of $y=x$ and $y=-x$ is $90^\circ$. Thus, the smallest region bounded by the graphs is $\frac{1}{4}\cdot2^2\cdot\pi = \boxed{\textbf{(C) } \pi}$.