6. The inner wall of a glass is formed by rotating the parabola
$$
y=x^{2}(-2 \leqslant x \leqslant 2)
$$

around the $y$-axis. Then the maximum radius of a sphere that can touch the bottom of the cup is $\qquad$
6. $\frac{1}{2}$.

Let the radius of the sphere be $r$. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
x^{2}+(y-r)^{2}=r^{2}, \\
y=x^{2}
\end{array}\right. \\
\Rightarrow x^{2}\left(1-2 r+x^{2}\right)=0 .
\end{array}
$$

According to the problem, the equation
$$
x^{2}+1-2 r=0
$$

has no non-zero real solutions.
Thus $x^{2}=2 r-1 \leqslant 0 \Rightarrow r \leqslant \frac{1}{2}$.