4. Given that the line $l$ passes through the focus of the parabola $C: x^{2}=4 y$ and is perpendicular to the $y$-axis. Then the area of the figure enclosed by the line $l$ and the parabola $C$ is ( ).
(A) $\frac{4}{3}$
(B) 2
(C) $\frac{8}{3}$
(D) $\frac{16 \sqrt{2}}{3}$
4. C.

From the problem, we know the coordinates of the focus are $(0,1)$. Therefore, by symmetry, the area of the required figure is $S=2 \int_{0}^{1} \sqrt{4 y} \, \mathrm{d} y=4 \times\left.\frac{2}{3} y^{\frac{3}{2}}\right|_{0} ^{1}=\frac{8}{3}$.