Let $f(x) = x^2 + 6x + 1$, and let $R$ denote the [set](https://artofproblemsolving.com/wiki/index.php/Set) of [points](https://artofproblemsolving.com/wiki/index.php/Point) $(x,y)$ in the [coordinate plane](https://artofproblemsolving.com/wiki/index.php/Coordinate_plane) such that 
\[f(x) + f(y) \le 0 \qquad \text{and} \qquad f(x)-f(y) \le 0\]
The area of $R$ is closest to 
$\textbf{(A) } 21 \qquad\textbf{(B)}\ 22 \qquad\textbf{(C)}\ 23 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 25$
The first condition gives us that
\[x^2 + 6x + 1 + y^2 + 6y + 1 \le 0 \Longrightarrow (x+3)^2 + (y+3)^2 \le 16\]
which is a [circle](https://artofproblemsolving.com/wiki/index.php/Circle) centered at $(-3,-3)$ with [radius](https://artofproblemsolving.com/wiki/index.php/Radius) $4$. The second condition gives us that 
\[x^2 + 6x + 1 - y^2 - 6y - 1 \le 0 \Longrightarrow (x^2 - y^2) + 6(x-y) \le 0 \Longrightarrow (x-y)(x+y+6) \le 0\]
Thus either
\[x - y \ge 0,\quad x+y+6 \le 0\]
or
\[x - y \le 0,\quad x+y+6 \ge 0\]

[2002 12B AMC-25.png](https://artofproblemsolving.com/wiki/index.php/File:2002_12B_AMC-25.png)
Each of those lines passes through $(-3,-3)$ and has slope $\pm 1$, as shown above. Therefore, the area of $R$ is half of the area of the circle, which is $\frac{1}{2} (\pi \cdot 4^2) = 8\pi \approx \boxed{\textbf{(E) }25}$.