Consider a square with vertices at (1, 1), (-1, 1), (-1, -1) and (1, -1). Let S be the region consis

Q: Wait, why do we reject the negative root when solving $x^2 + 2x - 1 = 0$?

Great catch! The quadratic formula gives $x = -1 \pm \sqrt{2}$. We are working in the first octant where $x$ must be positive (since the region is inside the square and we are looking at the area near the origin). Since $-1 - \sqrt{2}$ is negative, it lies outside our region of interest. Thus, we only accept $x = \sqrt{2} - 1$.

Q: Why can't we just use simple geometry instead of calculus?

You absolutely can! The region is bounded by a parabola and a line. While you could use the formula for the area of a parabolic segment (which is $\frac{2}{3} \times \text{base} \times \text{height}$), using calculus is a robust, universal method that ensures you don't miss any subtle geometric constraints. It's a powerful tool to have in your JEE toolkit.

Q: Why do we multiply the final integral by 8?

The square has 8-fold symmetry (4 reflectional axes: the x-axis, y-axis, and the two diagonals $y=x$ and $y=-x$). By calculating the area of just one octant—the tiny slice where $0 \le y \le x$—we have captured exactly one-eighth of the total area. Multiplying by 8 is the most efficient way to cover the entire region $S$.

Question

Consider a square with vertices at $(1, 1),(-1, 1),(-1,-1)$ and $(1,-1)$. Let $S$ be the region consisting of all points inside the square which are nearer to the origin than to any edge. Sketch the region $S$ and find its area.

Sigma Percentile · Accepted Answer

### The Geometry of Proximity: Unveiling Region $S$

Imagine you are standing at the very center of a perfect square, the origin $(0,0)$. The edges of this square are your boundaries, located at $x = \pm 1$ and $y = \pm 1$.

You are tasked with finding a special region $S$—a sanctuary where every point is closer to you, the origin, than to any of the four walls. This isn't just a math problem; it's a journey into the heart of symmetry and the elegance of coordinate geometry.

### The Locus of Equidistance: The Parabola Emerges

Let's pick an arbitrary point $P(x,y)$ inside the square. We want to find the boundary where the distance to the origin is exactly equal to the distance to the nearest edge.

Let's focus on the right edge, defined by the line $x=1$. The distance from $P$ to the origin is $d_1 = \sqrt{x^2+y^2}$, and the distance to the right edge is $d_2 = 1-x$. Setting these equal gives us the boundary condition:

$$\sqrt{x^2+y^2} = 1-x$$

Squaring both sides, we get $x^2+y^2 = (1-x)^2$. Expanding the right side, we find $x^2+y^2 = 1 - 2x + x^2$.

The $x^2$ terms vanish, leaving us with the beautiful, simple equation: $y^2 = 1 - 2x$, or $x = \frac{1-y^2}{2}$. This is a parabola! It opens to the left, with its vertex at $(0.5, 0)$.

### Harnessing the Power of Symmetry

Now, consider the entire square. Because of its perfect symmetry, each of the four edges creates its own parabolic boundary. The region $S$ is the intersection of the interiors of these four parabolas.

Instead of trying to integrate over the entire complex shape, we can use the square's 8-fold symmetry. By drawing the axes ($x=0, y=0$) and the diagonals ($y=x, y=-x$), we divide the square into eight identical octants.

We only need to calculate the area of one such octant and multiply it by 8. Let's focus on the first octant, where $0 \le y \le x$.

### The Integration: Calculating the Area

In this first octant, our boundary is the parabola $x = \frac{1-y^2}{2}$. To find where this region ends, we look for the intersection with the diagonal line $y=x$.

Substituting $y=x$ into our parabolic equation, we get $x = \frac{1-x^2}{2}$, which rearranges to $x^2 + 2x - 1 = 0$. Using the quadratic formula, we find the intersection point at $x = \sqrt{2} - 1$.

Now, we set up our integral for the area of this octant ($A_1$):

$$A_1 = \int_{0}^{\sqrt{2}-1} \left( \frac{1-y^2}{2} - y ight) \, dy$$

This integral represents the area between the right-hand boundary (the parabola) and the left-hand boundary (the diagonal line $x=y$). Evaluating this, we get:

$$\left[ \frac{y}{2} - \frac{y^3}{6} - \frac{y^2}{2} ight]_0^{\sqrt{2}-1}$$

Substituting the upper limit $a = \sqrt{2}-1$, and performing the algebraic expansion, we find the area of this single octant to be $\frac{4\sqrt{2}-5}{6}$.

### The Final Triumph

With the area of one octant secured, the final step is simply to multiply by 8 to account for the entire region $S$:

$$A = 8 	imes \left( \frac{4\sqrt{2}-5}{6} ight) = \frac{4(4\sqrt{2}-5)}{3} = \frac{16\sqrt{2}-20}{3}$$

And there it is! The area of region $S$ is **$\frac{16\sqrt{2}-20}{3}$** square units. Through the lens of symmetry and the precision of calculus, we have tamed the geometry of the square.

Animated Solution for Mathematics - Definite Integration: Consider a square with vertices at $(1, 1), (- 1, 1), (- 1, - 1)$ and $(1, - 1)$ . Let $S$ be the region consisting of all points inside the square which are nearer to the origin than to any edge. Sketch the region $S$ and find its area.

Visualized Solution

Visualizing the Square

Distance to the Right Edge

Boundary Equation

Simplifying to a Parabola

Symmetry of the Region

Intersection of Boundaries

Solving for the Intersection

Area of One Octant

Evaluating the Integral

Total Area of Region $S$

The Sigma Insight: Area Bounded by Curves

The Geometry of Proximity

The Locus of Equidistance

Harnessing the Power of Symmetry

The Integration

The Final Triumph

Similar Questions

Find the area of the region bounded by the curves $y = x^{2}, y = 2 - x^{2}$ and $y = 2$ , which lies to the right of the line $x = 1$ .

Sketch the region bounded by the curves $y = 5 - x^{2}$ and $y = ∣ x - 1∣$ and find its area.

The area (in sq. units) of the region ${(x, y) : 0 \leq y \leq 2∣ x ∣ + 1, 0 \leq y \leq x^{2} + 1, ∣ x ∣ \leq 3}$ is

The area of the region $A = {(x, y) : 0 \leq y \leq x^{2} + 1 and - 1 \leq x \leq 1}$ in sq. units, is :

The area of the region $x, y : x^{2} \leq y \leq ∣ x^{2} - 4∣, y \geq 1$ is

The area of the region $S = {(x, y) : y^{2} \leq 8 x, y \geq 2 x, x \geq 1}$ is

Sketch the region bounded by the curves $y = x^{2}$ and $y = \frac{2}{1 + x ^{2}}$ . Find the area.

The area of the region described by $A = {(x, y) : x^{2} + y^{2} \leq 1 and y^{2} \leq 1 - x}$ is:

Find the area bounded by the curves, $x^{2} + y^{2} = 25, 4 y = 4 - x^{2}$ and $x = 0$ above the $x$ -axis.

The area of the region ${(x, y) : ∣ x - 1∣ \leq y \leq 5 - x^{2}}$ is equal to: