Animated Solution for Mathematics - Definite Integration: The area of the region, inside the circle $(x-2\sqrt{3}{)}^2+y^2=12$ and outside the parabola $y^2=2\sqrt{3}x$ is:

A Geometric Odyssey

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an area problem; we are choreographing a dance between two fundamental shapes: the circle and the parabola.

Imagine standing on the Cartesian plane, watching these two curves emerge from the origin. It is a beautiful sight, and our goal is to find the hidden space between them.

The Anchor - Analyzing the Circle

First, let us look at our circle:

This is not just an equation; it is a perfect, symmetric entity. By comparing it to the standard form $(x-h{)}^2+(y-k{)}^2=R^2$, we immediately see the center is at $(2\sqrt{3},0)$ and the radius is $R=\sqrt{12}=2\sqrt{3}$.

Notice something profound? The $x$-coordinate of the center is exactly equal to the radius. This means the circle kisses the $y$-axis perfectly at the origin $(0,0)$. It is anchored there, waiting for the parabola to join it.

The Path - The Parabola

Now, consider the parabola:

This is a classic right-opening parabola with its vertex at the origin. It represents a path of constant acceleration, sweeping out from the origin and growing wider as it moves to the right.

As we draw this on our mental axes, we see the circle and the parabola starting together at the origin. But where do they meet again? This is the crucial moment of intersection.

The Collision - Finding Intersection Points

To find where they meet, we substitute the parabola's equation into the circle's equation. We replace $y^2$ with $2\sqrt{3}x$ in the circle's equation:

Expanding this, we get $x^2-4\sqrt{3}x+12+2\sqrt{3}x=12$. The $12$s cancel out, leaving us with $x^2-2\sqrt{3}x=0$.

Factoring this, we find $x(x-2\sqrt{3})=0$. The intersection points are at $x=0$ and $x=2\sqrt{3}$. These points define the boundaries of our world.

The Strategy - Carving the Region

We need the area inside the circle and outside the parabola. If you visualize the graph, you will see that for $x>2\sqrt{3}$, the circle is entirely swallowed by the parabola.

Thus, our region of interest is confined to the left side, between $x=0$ and $x=2\sqrt{3}$. Our strategy is elegant: we take the area of the left semi-circle and subtract the area under the parabola within that same interval.

It is like carving a piece of art out of a block of marble.

The Integration - The Heavy Lifting

The area of the left semi-circle is straightforward:

Now, for the parabola. The area under the curve is:

Do not let the square root intimidate you. We rewrite it as $2\sqrt{2\sqrt{3}}{\int }_0^{2\sqrt{3}}{x}^{1/2}dx$. Integrating $x^{1/2}$ gives $\frac{2}{3}{x}^{3/2}$.

Evaluating this from $0$ to $2\sqrt{3}$ yields:

The complexity vanishes, leaving us with a clean integer.

We have our two pieces: the semi-circle area of $6\pi$ and the parabolic area of $16$. Subtracting the latter from the former, we arrive at our final answer:

This problem is a testament to the beauty of coordinate geometry—where complex curves and integrals resolve into a simple, elegant expression. You have mastered the geometry, the algebra, and the calculus. Keep this confidence, for it is the key to conquering the JEE.