Animated Solution for Mathematics - Definite Integration: The area of the region, inside the ellipse $x^2+4y^2=4$ and outside the region bounded by the curves $y=|x|-1$ and $y=1-|x|$, is :

Welcome, future IITian! Today, we are going to master a beautiful problem that bridges the gap between coordinate geometry and pure, elegant intuition.

Imagine you are standing on the coordinate plane, looking at the equation $x^2+4y^2=4$. To truly understand its dimensions, we must standardize it.

By dividing the entire equation by $4$, we obtain:

Comparing this to the standard form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we immediately see that $a^2=4$ and $b^2=1$. This implies our semi-major axis $a=2$ and our semi-minor axis $b=1$.

The area of an ellipse is given by the classic result:

Substituting our values, we find the total area of this ellipse is $2\pi$. This is our canvas.

Absolute Values

Now, let us turn our attention to the boundaries we need to exclude. We are given two curves: $y=|x|-1$ and $y=1-|x|$.

These are transformations of the absolute value function. The first, $y=|x|-1$, is a V-shape shifted downwards by $1$ unit, with its vertex at $(0,-1)$.

The second, $y=1-|x|$, is an inverted V-shape shifted upwards by $1$ unit, with its vertex at $(0,1)$. If you sketch these, you will see they intersect at $(1,0)$ and $(-1,0)$.

The Geometry of Symmetry

When you look at the region enclosed by these two V-shapes, you see a perfect, symmetric diamond. This is a rhombus with vertices at $(1,0)$, $(0,1)$, $(-1,0)$, and $(0,-1)$.

The horizontal diagonal stretches from $-1$ to $1$, giving it a length of $2$. The vertical diagonal also stretches from $-1$ to $1$, giving it a length of $2$.

The area of a rhombus is given by the formula:

Substituting our diagonal lengths, we get:

We have successfully calculated the area of the region to be excluded.

Subtraction

The problem asks for the area inside the ellipse but outside the rhombus. This is the final, satisfying step.

We take the total area of our ellipse, $2\pi$, and subtract the area of the rhombus, $2$. The result is $2\pi -2$.

Factoring out the $2$, we arrive at our final answer:

It is a beautiful, clean result that rewards our geometric visualization. Keep practicing, and remember: every complex problem is just a collection of simple, elegant shapes waiting to be understood.