Q.A student observes an open-air Honeybee nest on the branch of a tree, whose plane figure is parabolic shape given by $x^{2}=4y$. Then the area (in sq units) of the region bounded by parabola $x^{2}=4y$ and the line $y=4$ is
(A) $\frac{32}{3}$
(B) $\frac{64}{3}$
(C) $\frac{128}{3}$
(D) $\frac{256}{3}$

Concept: Area Under Parabola – The region is bounded by the parabola $x^2 = 4y$ and the horizontal line $y = 4$.

Step 1: The parabola opens upward with vertex at the origin. The line $y = 4$ cuts the parabola where $x^2 = 16$, so $x = \pm 4$.

Step 2: The required area is symmetric about the $y$-axis. Using vertical strips, area from $x = -4$ to $x = 4$ between the line and the parabola:

Step 3: By symmetry, compute twice the area from $0$ to $4$: …

The key insight here is that the parabola $x^2 = 4y$ opens upward, with its vertex at the origin. The line $y = 4$ is a horizontal line cutting across it. The region bounded between them is symmetric about the y-axis, so we can find the area in the right half and double it.

When a region is bounded by a curve and a horizontal line, it's often easier to integrate with respect to $y$ rather than $x$. Why? Because the boundaries become simple: the left and right boundaries are given by the parabola, and the top and bottom boundaries are horizontal lines. Integrating along $y$ means we slice the region into thin horizontal strips, each of which has a simple rectangular shape.

Let's work through it step by step.

1. Find the intersection points.

   The parabola is $x^2 = 4y$ and the line is $y = 4$. Substituting $y = 4$ into the parabola gives $x^2 = 16$, so $x = \pm 4$. The region runs from $x = -4$ to $x = 4$ horizontally, and from $y = 0$ (the vertex) to $y = 4$ vertically.

2. Set up the integral with respect to $y$.

   For a fixed $y$, the parabola gives $x = \pm 2\sqrt{y}$. The horizontal width of the region at that $y$ is the distance between the right and left branches:

The area is the sum (integral) of these widths over $y$ from $0$ to $4$:

3. Evaluate the integral.

Applying the limits:

$\left[ \frac{8}{3} y^{3/2} \right]_{0}^{4} = \frac{8}{3} (4^{3/2} - 0) = \frac{8}{3} \cdot 8 = \frac{64}{3}.$ …

Method: Area between a curve and a horizontal line

Use this whenever a region is enclosed by a parabola (or similar curve) and a horizontal line $y=k$. Choosing the right slicing direction turns a messy integral into a clean one.

Steps

Step 1: Find the intersection points.

Set the line's value into the curve. For $x^2=4y$ and $y=k$, that gives $x=\pm 2\sqrt{k}$ — these fix the horizontal extent of the region.

Step 2: Decide the slicing direction.

For a region capped by a horizontal line, integrating with respect to $y$ is usually cleanest: horizontal strips have width = (right branch) $-$ (left branch).

Step 3: Set up the definite integral of the strip width. …

Common Mistakes

Mistake 1: Integrating the parabola alone and forgetting to subtract.

Why it's wrong: $\int \tfrac{x^2}{4}\,dx$ gives the area under the parabola, not the region between it and $y=4$. Correct approach: the bounded area is (rectangle $8\times4=32$) $-$ (area under parabola $=\tfrac{32}{3}$) $=\tfrac{64}{3}$, or integrate the strip width directly.

Mistake 2: Wrong strip width from the parabola. …