6. In the Cartesian coordinate system $x-o-y$, the figure defined by the inequalities
$$
y^{100}+\frac{1}{y^{100}} \leqslant x^{100}+\frac{1}{x^{100}}, x^{2}+y^{2} \leqslant 100
$$

has an area equal to ( ).
(A) $50 \pi$
(B) $40 \pi$
(C) $30 \pi$
(D) $20 \pi$
6. (A)

The region we are examining is symmetric with respect to the coordinate axes, so we only need to consider the part where $x>0, y>0$.
Let $z=x^{100}, t=y^{100}$, then we have
$$
\begin{array}{l}
z+\frac{1}{z} \geqslant t+\frac{1}{t} \Leftrightarrow\left(z^{2}+1\right) t \geqslant\left(t^{2}+1\right) z \\
\Leftrightarrow z^{2} t-t^{2} z+t-z \geqslant 0 \Leftrightarrow(z-t)(z t-1) \geqslant 0 .
\end{array}
$$

Therefore, the solution set of the inequality has the form of the shaded area in the figure. Since the shaded part within the circle $x^{2}+y^{2}=100$ is equal in area to the unshaded part, the area of the shaded part is $\frac{1}{2} \times 10^{2} \pi=50 \pi$.

Explanation: From $(z-t)(z t -1) \geqslant 0$ in the first quadrant, we have $z \geqslant t$ and $z t \geqslant 1$, i.e., $x \geqslant y$ and $x \geqslant \frac{1}{y}$. Therefore, in the first quadrant, the region can be drawn by $x \geqslant y, x \geqslant \frac{1}{y}$, and $x^{2}+y^{2}=10^{2}$. By symmetry, the graph in the other quadrants can be completed.