If the area of the region S = {(x, y) : 2y - y² ≤ x² ≤ 2y, x ≥ y} is equal to (n+2)/(n+1) - π/(n-1),

Q: Why do we split the integral at $x=1$? Can't we just integrate from $0$ to $2$?

Great question! Integration is all about the 'height' of the region. If you look at the graph, the 'ceiling' of your region changes at $x=1$. From $0$ to $1$, the ceiling is the circle, but from $1$ to $2$, the ceiling is the line $y=x$. If you don't split it, you'll be calculating the wrong area! Always check if your boundary functions change along the x-axis.

Q: How do I know if the region is inside or outside the circle?

Look at the inequality $x^2 + (y-1)^2 \ge 1$. Test a point like $(0, 2)$. Plugging it in gives $0^2 + (2-1)^2 = 1$, which is $\ge 1$. Since $(0, 2)$ is outside the circle, the inequality represents the region outside the circle. Always use a test point if you are unsure!

Q: Why is the lower boundary $1 - \sqrt{1-x^2}$ and not $1 + \sqrt{1-x^2}$?

Excellent observation. The circle equation is $x^2 + (y-1)^2 = 1$, which means $(y-1)^2 = 1 - x^2$, or $y = 1 \pm \sqrt{1-x^2}$. Since our region is below the center of the circle (which is at $y=1$), we must choose the negative root, $y = 1 - \sqrt{1-x^2}$.

Question

If the area of the region $S = \{(x, y) : 2y - y^2 \le x^2 \le 2y, x \ge y\}$ is equal to $\frac{n+2}{n+1} - \frac{\pi}{n-1}$, then the natural number $n$ is equal to _______

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The set $S$ is defined by the conditions $2y - y^2 \le x^2 \le 2y$ and $x \ge y$. To understand this region, we must translate these algebraic inequalities into geometric boundaries.

The first condition, $2y - y^2 \le x^2$, can be rearranged as:

$$x^2 + y^2 - 2y \ge 0$$

By completing the square, we obtain:

$$x^2 + (y-1)^2 \ge 1$$

This represents the region **outside** a circle centered at $(0, 1)$ with a radius of $1$.

The second condition, $x^2 \le 2y$, simplifies to:

$$y \ge \frac{x^2}{2}$$

This is the region **above** an upward-opening parabola. Finally, the condition $x \ge y$ restricts our region to the area **below** the line $y=x$.

### The Intersection Points

To determine the limits of integration, we find the intersection points of these boundaries. Solving $y=x$ and $y=\frac{x^2}{2}$ yields $x=0$ and $x=2$.

Solving $y=x$ and the circle $x^2 + (y-1)^2 = 1$ yields $x=0$ and $x=1$. These values, $x=0, 1, 2$, serve as the critical boundaries for our integration.

### The Integration

We split the total area into two distinct parts, $A_1$ and $A_2$. For $x \in [0, 1]$, the region is bounded between the lower arc of the circle, $y = 1 - \sqrt{1-x^2}$, and the parabola $y = \frac{x^2}{2}$.

The area $A_1$ is given by:

$$A_1 = \int_{0}^{1} \left( 1 - \sqrt{1-x^2} - \frac{x^2}{2} \right) dx$$

Evaluating this integral, we find:

$$A_1 = \left[ x - \frac{1}{2} \left( x\sqrt{1-x^2} + \sin^{-1}x \right) - \frac{x^3}{6} \right]_{0}^{1} = \frac{5}{6} - \frac{\pi}{4}$$

For $x \in [1, 2]$, the region is bounded by the line $y=x$ and the parabola $y = \frac{x^2}{2}$. The area $A_2$ is:

$$A_2 = \int_{1}^{2} \left( x - \frac{x^2}{2} \right) dx = \left[ \frac{x^2}{2} - \frac{x^3}{6} \right]_{1}^{2} = \left( 2 - \frac{8}{6} \right) - \left( \frac{1}{2} - \frac{1}{6} \right) = \frac{1}{3}$$

### Final Calculation

The total area $A$ is the sum of these two components:

$$A = A_1 + A_2 = \left( \frac{5}{6} - \frac{\pi}{4} \right) + \frac{1}{3} = \frac{7}{6} - \frac{\pi}{4}$$

We compare this result with the expression $\frac{n+2}{n+1} - \frac{\pi}{n-1}$. By matching the terms, we identify:

$$\frac{\pi}{n-1} = \frac{\pi}{4} \Rightarrow n-1 = 4$$

Thus, the final value is **$n=5$**.

Animated Solution for Mathematics - Definite Integration: If the area of the region $S = {(x, y) : 2 y - y^{2} \leq x^{2} \leq 2 y, x \geq y}$ is equal to $\frac{n + 2}{n + 1} - \frac{π}{n - 1}$ , then the natural number $n$ is equal to _______

Enter Numerical Value:

Visualized Solution

Analyze the Region $S$

Identify the Circle Boundary

Identify the Parabola and Line

Find Intersection Points

Analyze Relative Positions

Set Up Integrals

Evaluate First Integral ( $A_{1}$ )

Evaluate Second Integral ( $A_{2}$ )

Calculate Total Area

Compare and Solve for $n$

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Intersection Points

The Integration

Final Calculation

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Enter Numerical Value:

Visualized Solution

Analyze the Region .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Identify the Circle Boundary

Identify the Parabola and Line

Find Intersection Points

Analyze Relative Positions

Set Up Integrals

Evaluate First Integral ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​)

Evaluate Second Integral ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2​)

Calculate Total Area

Compare and Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Intersection Points

The Integration

Final Calculation

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