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

First, let's find the points of intersection:

$4x^2 = 4$

$x^2 = 1$

$x = \pm 1$

So the curves intersect at $(-1,4)$ and $(1,4)$.

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Rearranging the parabola equation: $4x^2=y$ becomes $y=4x^2$

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To find the area, we integrate the difference between the upper curve ($y=4$) and lower curve ($y=4x^2$) from $x=-1$ to $x=1$:

$\text{Area} = \int_{-1}^{1} [4 - 4x^2] \, dx$

$= \int_{-1}^{1} 4 \, dx - \int_{-1}^{1} 4x^2 \, dx$

$= 4 \int_{-1}^{1} \, dx - 4 \int_{-1}^{1} x^2 \, dx$

$= 4[x]_{-1}^{1} - 4[\dfrac{x^3}{3}]_{-1}^{1}$

$= 4(1-(-1)) - 4(\dfrac{1^3}{3}-\dfrac{(-1)^3}{3})$

$= 4 \cdot 2 - 4(\dfrac{1}{3}-\dfrac{-1}{3})$

$= 8 - 4 \cdot \dfrac{2}{3}$

$= 8 - \dfrac{8}{3}$

$= \dfrac{24 - 8}{3}$

$= \dfrac{16}{3}$

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Therefore, the area of the region enclosed between the curves is $\dfrac{16}{3}$ square units.