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

The region is bounded by the parabola $y = x^2$ and the horizontal line $y = 16$.

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The curves intersect where:

$x^2 = 16$

$x = \pm 4$

The intersection points are at $x = -4$ and $x = 4$.

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The area between the curves is:

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

The upper curve is $y = 16$ and the lower curve is $y = x^2$.

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The region is symmetric about the y-axis:

$\text{Area} = 2 \int_{0}^{4} (16 - x^2) \, dx$

$= 2 \left[ 16x - \frac{x^3}{3} \right]_{0}^{4}$

$= 2 \left[ \left(16(4) - \frac{4^3}{3}\right) - \left(16(0) - \frac{0^3}{3}\right) \right]$

$= 2 \left[ 64 - \frac{64}{3} - 0 \right]$

$= 2 \left[ \frac{192}{3} - \frac{64}{3} \right]$

$= 2 \left[ \frac{128}{3} \right]$

$= \frac{256}{3}$

Therefore, the area of the region is $\dfrac{256}{3}$ square units.