13. 3. 13 ${ }^{\star \star}$ The axial cross-section of a large wine glass is the graph of the function $y=x^{4}$. A small ball with radius $r$ is placed into the glass. When the radius $r$ is maximized, the ball can touch the lowest point of the bottom of the glass (i.e., what is the maximum value of $r$ such that a circle with radius $r$ located in the region $y \geqslant x^{4}$ can pass through the origin)?
The original problem is equivalent to the circle $(y-r)^{2}+x^{2}=r^{2}$ passing through the origin being completely within the region $y \geqslant x^{4}$, i.e., when $0 \leqslant x \leqslant r$, we have $r-\sqrt{r^{2}-x^{2}} \geqslant x^{4}$.

Let $x=r \sin \theta$. (1) becomes that when $0 \leqslant \theta \leqslant \frac{\pi}{2}$, we have $1-\cos \theta \geqslant r^{3} \sin ^{4} \theta$.
(2) When $\theta=0$, (2) is obviously true: when $0<\theta \leqslant \frac{\pi}{2}$, (2) transforms into $r^{3} \leqslant$ $\frac{1}{\sin ^{2} \theta(1+\cos \theta)}$.
(3) Therefore, we have
$$
\sin ^{2} \theta(1+\cos \theta)=8 \sin ^{2} \frac{\theta}{2} \cos ^{4} \frac{\theta}{2} \leqslant 4\left(\frac{2 \sin ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}}{3}\right)^{3}=\frac{32}{27} .
$$

Substituting into (3) gives $r \leqslant \sqrt[3]{\frac{27}{32}}=\frac{3}{1} \sqrt[3]{2}$.