1. Let in the $x O y$ plane, the area enclosed by $0<y \leqslant x^{2}, 0 \leqslant x \leqslant 1$ be $\frac{1}{3}$. Then the intersection of the sets
$$
\begin{array}{l}
M=\{(x, y)|x \leqslant| y \mid\}, \\
N=\left\{(x, y) \mid x \geqslant y^{2}\right\}
\end{array}
$$

representing the area of $M \cap N$ is ( ).
(A) $\frac{2}{3}$
(B) $\frac{1}{3}$
(C) 1
(D) $\frac{1}{6}$
-1. .
$M \cap N$ is symmetric with respect to the $x$-axis on the $x O y$ plane, therefore, the area of the figure of $M \cap N$ can be calculated by multiplying the area of the figure in the first quadrant by 2. According to the problem, the area of the figure of $M \cap N$ in the first quadrant is $\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$.