Problem (Geometry):
Compute the area of the region defined by $x^{2}+y^{2} \leq|x|+|y|$.

Solution:
Call the region $R$, and let $R_{q}$ be the portion of $R$ in the $q^{\text {th }}$ quadrant. Noting that the point $(x, y)$ is in $R$ if and only if $( \pm x, \pm y)$ is in $R$, it follows that $\left[R_{1}\right]=\left[R_{2}\right]=\left[R_{3}\right]=\left[R_{4}\right]$, and so $[R]=4\left[R_{1}\right]$. So it suffices to determine $\left[R_{1}\right]$.

In the first quadrant, the boundary equation is just $x^{2}+y^{2}=x+y \Rightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{2}$. This equation describes a circle of radius $\frac{\sqrt{2}}{2}$ centered at $\left(\frac{1}{2}, \frac{1}{2}\right)$. The portion of the circle's interior which is inside the first quadrant can be decomposed into a right isosceles triangle with side length 1 and half a circle of radius $\frac{\sqrt{2}}{2}$. Thus $\left[R_{1}\right]=\frac{1}{2}+\frac{\pi}{4}$, hence $[R]=\mathbf{2}+\boldsymbol{\pi}$.

Final answer: $2+\pi$