The area of the region enclosed between the curves $4 x^2=y$ and $y=4$ is:

$\frac{16}{3}$ sq. units

The correct answer is Option (4) → $\frac{16}{3}$ sq. units

$4 x^2=y$, $y=4$

at $y=4$, $4 x^2=4$

$x=±1$

area required = $4×2-\int\limits_{-1}^14x^2dx$

$=4×2-2\int\limits_0^14x^2dx$

$=8-8\left[\frac{x^3}{3}\right]_0^1$

$=\frac{16}{3}$ sq. units