Animated Solution for Mathematics - Definite Integration: The area between the parabolas $x^2=\frac{y}{4}$ and $x^2=9y$ and the straight line $y=2$ is:

We are tasked with finding the area trapped between two parabolas, $x^2=\frac{y}{4}$ and $x^2=9y$, and the horizontal line $y=2$. Both parabolas open upwards and are anchored at the origin.

The line $y=2$ acts as a ceiling, capping the region we need to measure. Visualizing this, we see a narrow region bounded by these two curves within the interval $y\in [0,2]$.

Before we dive into the integration, we must appreciate the elegance of the setup. Both parabolas are perfectly symmetric about the $y$-axis, and the line $y=2$ is also symmetric.

This symmetry is a gift. It means the area on the left side of the $y$-axis is a perfect mirror image of the area on the right.

In the world of JEE, time is your most precious resource. We can calculate the area in the first quadrant—where $x\ge 0$—and simply multiply by 2. Thus, our total area $A$ becomes:

To find the area, it is most efficient to integrate with respect to $y$. Consider an infinitesimal horizontal strip of thickness $dy$ at a height $y$.

The right end of this strip touches the outer curve $x^2=9y$, and the left end touches the inner curve $x^2=\frac{y}{4}$. Solving for $x$ in terms of $y$, we get:

The length of our strip is the difference $\Delta x=3\sqrt{y}-\frac{1}{2}\sqrt{y}=\frac{5}{2}\sqrt{y}$. The area of this tiny strip is $dA=\frac{5}{2}\sqrt{y}dy$.

We sum these strips from the origin ($y=0$) to the ceiling ($y=2$). Our integral becomes:

The constant $2$ cancels with the denominator, leaving us with:

Applying the power rule $\int y^{n}dy=\frac{y^{n+1}}{n+1}$, we evaluate the integral:

Since $2^{3/2}=2\sqrt{2}$, our final result is: