[Geometry]
Problem: Let $\mathcal{R}$ denote the circular region bounded by $x^{2}+y^{2}=36$. The lines $x=4$ and $y=3$ partition $\mathcal{R}$ into four regions $\mathcal{R}_{1}, \mathcal{R}_{2}, \mathcal{R}_{3}$, and $\mathcal{R}_{4}$. $\left[\mathcal{R}_{i}\right]$ denotes the area of region $\mathcal{R}_{i}$. If $\left[\mathcal{R}_{1}\right]>\left[\mathcal{R}_{2}\right]>\left[\mathcal{R}_{3}\right]>\left[\mathcal{R}_{4}\right]$, compute $\left[\mathcal{R}_{1}\right]-\left[\mathcal{R}_{2}\right]-\left[\mathcal{R}_{3}\right]+\left[\mathcal{R}_{4}\right]$.

Solution: Draw the lines $x=-4$ and $y=-3$, creating regions $\mathcal{R}_{21}, \mathcal{R}_{22}, \mathcal{R}_{11}, \mathcal{R}_{12}, \mathcal{R}_{13}, \mathcal{R}_{14}$ as shown below.

<img_3593>



Then $\left[\mathcal{R}_{21}\right]=\left[\mathcal{R}_{4}\right]=\left[\mathcal{R}_{13}\right],\left[\mathcal{R}_{22}\right]=\left[\mathcal{R}_{14}\right]$, and $\left[\mathcal{R}_{3}\right]=\left[\mathcal{R}_{12}\right]+\left[\mathcal{R}_{13}\right]$. Therefore

$$
\begin{aligned}
{\left[\mathcal{R}_{1}\right]-\left[\mathcal{R}_{2}\right]-\left[\mathcal{R}_{3}\right]+\left[\mathcal{R}_{4}\right] } & =\left(\left[\mathcal{R}_{1}\right]-\left[\mathcal{R}_{2}\right]\right)-\left(\left[\mathcal{R}_{3}\right]-\left[\mathcal{R}_{4}\right]\right) \\
& =\left(\left[\mathcal{R}_{1}\right]-\left[\mathcal{R}_{13}\right]-\left[\mathcal{R}_{14}\right]\right)-\left(\left[\mathcal{R}_{12}\right]+\left[\mathcal{R}_{13}\right]-\left[\mathcal{R}_{21}\right]\right) \\
& =\left(\left[\mathcal{R}_{11}\right]+\left[\mathcal{R}_{12}\right]\right)-\left[\mathcal{R}_{12}\right] \\
& =\left[\mathcal{R}_{11}\right] .
\end{aligned}
$$

This last region is simply a rectangle of height 6 and width 8 , so its area is 48 .

Final answer: 48