Animated Solution for Mathematics - Definite Integration: The area of the region enclosed by $y\le 4x^2,x^2\le 9y$ and $y\le 4$, is equal to:

A Geometric Journey

Imagine you are standing on the Cartesian plane, looking at a landscape defined by three simple constraints. We have two parabolas, one wide and one narrow, and a horizontal line acting as a ceiling.

It might look like a simple area problem, but it is actually a beautiful dance of curves. Let's break down this journey step by step.

Visualizing the Arena

First, we must see the shapes. We have $x^2\le 9y$, which is a wide, upward-opening parabola. The inequality tells us we are looking at the region 'inside' this curve.

Then, we have $y\le 4x^2$, a much narrower, steeper parabola. The inequality here tells us we are looking at the region 'outside' or below this narrow curve.

Finally, the line $y=4$ acts as a ceiling, capping our region. When you combine these, you see a symmetric, vase-like shape centered on the $y$-axis, bounded by the two parabolas and capped by the line $y=4$.

The Power of Symmetry

In JEE Advanced, time is your most precious resource. When you see a shape that is perfectly symmetric about the $y$-axis, you should immediately think: "Can I calculate half the area and double it?"

The answer here is a resounding yes. By focusing only on the first quadrant, where $x\ge 0$, we simplify our algebra and avoid the confusion of negative coordinates.

We will calculate the area of the right half and multiply by $2$ at the very end.

The Calculus Strategy

Now, how do we slice this region? We could use vertical strips ($dx$), but that would require us to break the integral into pieces because the top boundary changes.

Instead, let's use horizontal strips of thickness $dy$. This is the "pro" move. By integrating with respect to $y$, our boundaries remain consistent from $y=0$ to $y=4$.

For any horizontal strip at height $y$, the right end touches the wide parabola $x^2=9y$. Solving for $x$ in the first quadrant, we get $x_{right}=3\sqrt{y}$.

The left end touches the narrow parabola $y=4x^2$. Solving for $x$, we get $x^2=\frac{y}{4}$, so $x_{left}=\frac{\sqrt{y}}{2}$.

The length of our strip is simply $x_{right}-x_{left}=3\sqrt{y}-\frac{\sqrt{y}}{2}$.

The Execution

Now, we assemble the integral. The total area $A$ is given by:

Before we dive into the integration, let's simplify the integrand. Factoring out $\sqrt{y}$, we get $(3-\frac{1}{2})\sqrt{y}=\frac{5}{2}\sqrt{y}$.

Our integral becomes:

This is a standard power rule integral. The antiderivative of $y^{1/2}$ is $\frac{y^{3/2}}{3/2}=\frac{2}{3}{y}^{3/2}$.

Applying this to our integral:

Plugging in the limits, $4^{3/2}=(\sqrt{4}{)}^3=2^3=8$. The lower limit is $0$.

Thus, the final calculation is:

And there you have it! The area is $\frac{80}{3}$ square units.

Notice how the symmetry and the choice of horizontal strips turned a potentially messy problem into a clean, elegant calculation. This is the essence of physics and math in the JEE: it's not about brute force; it's about finding the most elegant path to the truth.

Keep practicing this mindset, and you will master these problems with ease.