The area of region bounded by the curve $y=x^2$ and the line $y=4$ is :

$\frac{32}{3}$ sq.unit

Curve: $y = x^2$, Line: $y = 4$

Intersection points: $x^2 = 4 \Rightarrow x = -2, 2$

Area = $\int_{-2}^{2} (4 - x^2) \, dx$

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

= $[4x]_{-2}^{2} - \left[\frac{x^3}{3}\right]_{-2}^{2}$

= $(4*2 - 4*(-2)) - \left(\frac{8}{3} - \frac{-8}{3}\right)$

= $(8 + 8) - \left(\frac{16}{3}\right)$

= $16 - \frac{16}{3} = \frac{48 - 16}{3} = \frac{32}{3}$

Answer: $\frac{32}{3}$