Q.Find the area of the region bounded by the curve $x^2 = 16y$ and the line $y = 4$.

The region sits between the parabola $x^2 = 16y$ and the line $y = 4$. Since the boundary line is horizontal, we integrate $x$ with respect to $y$ and double it (the region is symmetric about the $y$-axis), giving an area of $\dfrac{128}{3}$ sq. units.

The parabola $x^2 = 16y$ opens upward with vertex at the origin, so $x = 4\sqrt{y}$ on its right half. The line $y = 4$ closes the region at the top; it meets the parabola where $x^2 = 16(4) = 64$, i.e. $x = \pm 8$.

Because the region is symmetric about the $y$-axis, the total area is twice the area to the right of it. Taking a horizontal strip and integrating with respect to $y$:

$A = 2\int_{0}^{4} x\, dy = 2\int_{0}^{4} 4\sqrt{y}\, dy$

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