The curve is a parabola $y = x^2$ opening upwards with vertex at $(0,0)$. The line $y = 16$ is a horizontal line intersecting the parabola at points where $x^2 = 16$, i.e., $x = -4$ and $x = 4$. The region is symmetric about the y-axis.
To find the area using integration with respect to $y$, we express $x$ in terms of $y$: $x = \pm\sqrt{y}$. The width of the region at any height $y$ is the distance between the two branches of the parabola, which is $\sqrt{y} - (-\sqrt{y}) = 2\sqrt{y}$. The limits for $y$ are from $0$ to $16$. Thus, the area is: $$Area = \int_{0}^{16} 2\sqrt{y} dy$$
Comparing our derived expression $2\int_{0}^{16}\sqrt{y}dy$ with the given options, it matches option (D).
Final Answer: D
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