Area of the region bounded by $y = x^2$ and the line $y = 16$ is

$\frac{256}{3}$ square units

The correct answer is Option (1) → $\frac{256}{3}$ square units

Region bounded by: $y = x^{2}$ and $y = 16$

Intersection points: $x^{2} = 16 \Rightarrow x = \pm 4$

Area = $\displaystyle \int_{-4}^{4} (16 - x^{2})\,dx$

$= \left[16x - \frac{x^{3}}{3}\right]_{-4}^{4}$

$= \left(64 - \frac{64}{3}\right) - \left(-64 + \frac{64}{3}\right)$

$= 128 - \frac{128}{3}$

$= \frac{384 - 128}{3} = \frac{256}{3}$

The area of the region is $\frac{256}{3}$ square units.