4. If a cylinder is intersected by a plane, and the eccentricity of the resulting elliptical section is $\frac{2 \sqrt{2}}{3}$, then the acute dihedral angle between this section and the base of the cylinder is ( )
A: $\quad \arcsin \frac{1}{3}$
B: $\arccos \frac{1}{3}$
C: $\arcsin \frac{2}{3}$
D: $\quad \arccos \frac{2}{3}$
Answer B.
Analysis From the problem, we know that the short axis length $2b$ of the elliptical section is the same as the diameter of the cylinder's base, then the section and the acute dihedral angle $\theta$ satisfy: $\cos \theta=\frac{2 b}{2 a}=\frac{b}{a}$, where $2a$ is the length of the major axis of the ellipse. From the given information,
$$
e=\frac{2 \sqrt{2}}{3} \Rightarrow \frac{c}{a}=\frac{2 \sqrt{2}}{3} \Rightarrow \frac{b}{a}=\sqrt{\frac{a^{2}-c^{2}}{a^{2}}}=\sqrt{1-\left(\frac{b}{a}\right)^{2}}=\sqrt{1-\frac{8}{9}}=\frac{1}{3}
$$

Thus, $\cos \theta=\frac{1}{3}$, so $\theta=\arccos \frac{1}{3}$.