For the region bounded by the parabola $x^{2} = 8y$ and the line $y = 2$, which of the following statements are true?

I. The line meets the parabola at $(-4, 2)$ and $(4, 2)$.

II. Its area is $2\int_0^{4} \left(2 - \frac{x^{2}}{8}\right)dx$.

III. Its area is $\frac{32}{3}$ square units.

IV. Its area is $\int_0^{2} \sqrt{8y}\,dy$.

Explanation

1. 01

   I: $x^{2} = 16$, so $x = \pm 4$. II and III: the line lies above the parabola and the region is symmetric about the $y$-axis, so the area is $2\int_0^{4} \left(2 - \frac{x^{2}}{8}\right)dx = 2\left(8 - \frac{8}{3}\right) = \frac{32}{3}$.

2. 02

   IV: a horizontal strip runs from $-\sqrt{8y}$ to $\sqrt{8y}$, so its length is $2\sqrt{8y}$; $\int_0^{2} \sqrt{8y}\,dy = \frac{16}{3}$ is only half the area. False.