Let $f(x) = x^2 + 6x + 1$, and let $R$ denote the set of points $(x, y)$ in the coordinate plane such that $$
f(x) + f(y) \le 0\quad {\text{and}}\quad f(x) - f(y) \le 0.
$$What integer is the area of $R$ closest to?
Note that $$
f(x) + f(y) = x^2 + 6x + y^2 + 6y + 2 = (x+3)^2 + (y+3)^2 -
16
$$and $$
f(x) - f(y) = x^2-y^2 + 6(x-y) = (x-y)(x+y+6).
$$The given conditions can be written as $$
(x+3)^2 + (y+3)^2 \le 16 \quad {\text{and}}\quad (x-y)(x+y+6) \le
0.
$$The first inequality describes the region on and inside the circle of radius 4 with center $(-3,-3)$. The second inequality can be rewritten as $$
(x-y \ge 0 \text{ and } x+y+6 \le 0) \quad \text{or} \quad (x-y
\le 0 \text{ and } x+y+6 \ge 0).
$$Each of these inequalities describes a half-plane bounded by a line that passes through $(-3,-3)$ and has slope 1 or $-1$. Thus, the set $R$ is the shaded region in the following diagram, and its area is half the area of the circle, which is $8\pi \approx 25.13.$, so the closest integer to $R$ is $\boxed{25}$. [asy]
fill((-3,1.2)..(0,0)--(-3,-3)--(-6,0)..cycle,gray(0.7));
fill((-3,-7.2)..(0,-6)--(-3,-3)--(-6,-6)..cycle,gray(0.7));
draw(Circle((-3,-3),4.2),linewidth(0.7));
for (int i=-7; i<3; ++i) {
draw((i,-0.2)--(i,0.2));
draw((-0.2,i)--(0.2,i));
}
draw((-7,1)--(1,-7),linewidth(1));
draw((1,1)--(-7,-7),linewidth(1));
draw((-8,0)--(3,0),Arrow);
draw((0,-8)--(0,3),Arrow);
label("$x$",(3,0),S);
label("$y$",(0,3),E);
label("-6",(-6,0),N);
label("-6",(0,-6),E);
label("$(-3,-3)$",(-3,-3),W);
[/asy]