Let the area enclosed between the curves |y|=1-x² and x²+y²=1 be α. If 9α=βπ+γ;β, γ are integers, th

Q: Wait, why is the area $\alpha = \text{Area of Circle} - \text{Area of Parabolas}$? Shouldn't it be the other way around?

Great question! The problem asks for the area 'enclosed between' the curves. If you sketch the circle and the parabolas, you will see the parabolas cut through the circle. The region trapped between them is the part of the circle that is not covered by the parabolas. Thus, subtracting the parabola area from the circle area is the correct geometric approach.

Q: Why did we get $\gamma = -24$? Can an area coefficient be negative?

Remember, $\gamma$ is just an integer constant in the equation $9\alpha = \beta\pi + \gamma$. It is not the area itself. When we expand $9(\pi - 8/3)$, we get $9\pi - 24$. Comparing this to $\beta\pi + \gamma$, we identify $\beta = 9$ and $\gamma = -24$. The negative sign is perfectly valid for an integer coefficient!

Q: Is there a way to solve this without calculus?

While you could potentially use geometric formulas for circular segments, calculus is the most robust tool here. The integral $\int (1-x^2) dx$ is straightforward, and using symmetry makes it very fast. In JEE Advanced, always choose the method that minimizes the chance of calculation errors!

Question

Let the area enclosed between the curves $|y|=1-x^{2}$ and $x^{2}+y^{2}=1$ be $\alpha$. If $9\alpha=\beta\pi+\gamma;\beta, \gamma$ are integers, then the value of $|\beta-\gamma|$ equals.

Options:

(A) $27$

(B) $33$

(C) $15$

(D) $18$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today we are diving into a problem that perfectly marries the elegance of coordinate geometry with the power of calculus.

We are considering two distinct mathematical entities: the unit circle $x^2 + y^2 = 1$ and the modulus-defined curve $|y| = 1 - x^2$. The modulus sign forces us to split our perspective.

When $y \ge 0$, we have the downward-opening parabola $y = 1 - x^2$. When $y < 0$, we have the upward-opening parabola $y = x^2 - 1$.

By visualizing this, you see the parabolas acting like a pair of jaws biting into the circle. The area $\alpha$ we seek is the region trapped inside the circle but outside these parabolic jaws.

### The Power of Symmetry

To solve this, we utilize the symmetry of the shapes. The entire figure is symmetric about both the $x$-axis and the $y$-axis.

Instead of calculating the whole area, we focus on the first quadrant. The area of the circle in the first quadrant is:

$$\frac{\pi}{4}$$

The area under the parabola $y = 1 - x^2$ in the first quadrant is given by the integral:

$$\int_{0}^{1} (1 - x^2) dx$$

Calculating this integral is straightforward:

$$\left[ x - \frac{x^3}{3} ight]_0^1 = 1 - \frac{1}{3} = \frac{2}{3}$$

Since we have four such symmetric regions, the total area of the parabolas is:

$$4 	imes \frac{2}{3} = \frac{8}{3}$$

### The Final Comparison

The total area of the circle is $\pi$. Subtracting the parabolic area from the circle gives us the area $\alpha$:

$$\alpha = \pi - \frac{8}{3}$$

The problem asks us to relate this to the form $9\alpha = \beta\pi + \gamma$. Multiplying our result by $9$, we get:

$$9\left(\pi - \frac{8}{3}ight) = 9\pi - 24$$

By simple comparison, we identify $\beta = 9$ and $\gamma = -24$.

The final step is to find $|\beta - \gamma|$, which is:

$$|9 - (-24)| = |9 + 24| = \mathbf{33}$$

See how the complexity melts away when you break it down? Keep this clarity in your mind, and no problem will ever be too difficult for you.

Animated Solution for Mathematics - Definite Integration: Let the area enclosed between the curves $∣ y ∣ = 1 - x^{2}$ and $x^{2} + y^{2} = 1$ be $α$ . If $9 α = β π + γ; β, γ$ are integers, then the value of $∣ β - γ ∣$ equals.

Select Answer:

Visualized Solution

Analyze the Curves

Break Down the Modulus

The Unit Circle

Finding Intersection Points

Solving for $x$

Plotting Intersection Points

Identify the Enclosed Area $α$

Area of the Circle

Setup Integral for Parabolas

Evaluate the Integral

Calculate Parabola Area

Calculate $α$

Use the Given Relation

Compare Coefficients

Final Calculation

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Power of Symmetry

The Final Comparison

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