JEE MainMathematics

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Area Bounded by a Parabola, Circle, and Line

Q: Why is the circular region area evaluated as π/4 - 1/2?

The circle is x^2 + (y-1)^2 = 1 centered at (0,1). The line y=x passes through (0,0) and the point (1,1), dividing the right half of the circle. The area under y=x within the circle is a quadrant sector of radius 1 minus the right triangle with vertices (0,1), (0,0), (1,1), giving 1/4π(1)^2 - 1/2(1)(1) = π/4 - 1/2.

If the area of the region S={(x,y): 2y-y^2≤x^2≤2y, x≥y} is equal to (n+2)/(n+1)-π/(n-1), then the natural number n is equal to ________

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Correct answer

The natural number n is equal to 5.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Region $S = {(x, y) : 2 y - y^{2} \leq x^{2} \leq 2 y, x \geq y}$ and $Area (S) = \frac{n + 2}{n + 1} - \frac{π}{n - 1}$ for natural number $n$ .

02Goal

Compute the area of the region $S$ and match it to $\frac{n + 2}{n + 1} - \frac{π}{n - 1}$ to find the value of $n$ .

03Approach

Analyze the three boundary conditions in the first quadrant: 1. $x^{2} + y^{2} - 2 y \geq 0 ⟹ x^{2} + (y - 1)^{2} \geq 1$ (outside/on the circle centered at $(0, 1)$ with radius $1$ ). 2. $x^{2} \leq 2 y ⟹ y \geq \frac{x^{2}}{2}$ (above/on the parabola $x^{2} = 2 y$ ). 3. $x \geq y$ (below/on the line $y = x$ ). Calculate the area by decomposing the region below $y = x$ and bounded between the parabola and the circle.

04Execute

The line $y = x$ intersects the parabola $x^{2} = 2 y$ at $(0, 0)$ and $(2, 2)$ . The line $y = x$ intersects the circle $x^{2} + y^{2} - 2 y = 0$ at $(0, 0)$ and $(1, 1)$ . The total area under $y = x$ from $x = 0$ to $x = 2$ is the triangle with vertices $(0, 0), (2, 0), (2, 2)$ , having area $\frac{1}{2} \times 2 \times 2 = 2$ . The area under the parabola from $x = 0$ to $x = 2$ is $\int_{0}^{2} \frac{x^{2}}{2} d x = {[\frac{x^{3}}{6}]}_{0}^{2} = \frac{8}{6} = \frac{4}{3}$ . The circular sector region in the first quadrant below $y = x$ bounded by $x^{2} + y^{2} - 2 y \leq 0$ has area equal to: $(Sector of circle) - (Triangle) = \frac{π (1)^{2}}{4} - \frac{1}{2} (1) (1) = \frac{π}{4} - \frac{1}{2}$ . Thus, $Area (S) = Area (region between y = x and y = \frac{x^{2}}{2}) - Area (circle region below y = x)$ , $Area (S) = (2 - \frac{4}{3}) - (\frac{π}{4} - \frac{1}{2}) = \frac{2}{3} + \frac{1}{2} - \frac{π}{4} = \frac{7}{6} - \frac{π}{4}$ .

05Execute

Comparing $Area (S) = \frac{7}{6} - \frac{π}{4}$ with $\frac{n + 2}{n + 1} - \frac{π}{n - 1}$ : $\frac{n + 2}{n + 1} = \frac{7}{6} ⟹ 6 n + 12 = 7 n + 7 ⟹ n = 5$ . Also, $\frac{π}{n - 1} = \frac{π}{4} ⟹ n - 1 = 4 ⟹ n = 5$ .

✓Verify

Substitute $n = 5$ into the given formula: $\frac{5 + 2}{5 + 1} - \frac{π}{5 - 1} = \frac{7}{6} - \frac{π}{4}$ , which matches our computed area identically.

Hints that build this answer step by step

How can the area of region S be broken down using simpler geometric and calculus-based components?
Subtract the circular segment below the line y = x from the total area enclosed between the line y = x and the parabola y = x²/2.
What is the total area bounded between the line y = x and the parabola y = x²/2 from x = 0 to their intersection at x = 2?
2/3
What is the area of the circular segment of x² + (y - 1)² ≤ 1 lying below the line y = x?
π/4 - 1/2
Subtracting the segment area from the parabola-line area yields 7/6 - π/4. Equating this to (n + 2)/(n + 1) - π/(n - 1), what is n?
5

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 September 2026

Quick checks

Students also ask

Why is the circular region area evaluated as

\frac{π}{4} - \frac{1}{2}

The circle is $x^{2} + (y - 1)^{2} = 1$ centered at $(0, 1)$ . The line $y = x$ passes through $(0, 0)$ and the point $(1, 1)$ , dividing the right half of the circle. The area under $y = x$ within the circle is a quadrant sector of radius $1$ minus the right triangle with vertices $(0, 1), (0, 0), (1, 1)$ , giving $\frac{1}{4} π (1)^{2} - \frac{1}{2} (1) (1) = \frac{π}{4} - \frac{1}{2}$ .

Answer

The natural number n is equal to 5.

Why each option works or fails

Step-by-step solution

Given: Region $S = {(x, y) : 2 y - y^{2} \leq x^{2} \leq 2 y, x \geq y}$ and $Area (S) = \frac{n + 2}{n + 1} - \frac{π}{n - 1}$ for natural number $n$ .
Goal: Compute the area of the region $S$ and match it to $\frac{n + 2}{n + 1} - \frac{π}{n - 1}$ to find the value of $n$ .
Approach: Analyze the three boundary conditions in the first quadrant: 1. $x^{2} + y^{2} - 2 y \geq 0 ⟹ x^{2} + (y - 1)^{2} \geq 1$ (outside/on the circle centered at $(0, 1)$ with radius $1$ ). 2. $x^{2} \leq 2 y ⟹ y \geq \frac{x^{2}}{2}$ (above/on the parabola $x^{2} = 2 y$ ). 3. $x \geq y$ (below/on the line $y = x$ ). Calculate the area by decomposing the region below $y = x$ and bounded between the parabola and the circle.
Execute: The line $y = x$ intersects the parabola $x^{2} = 2 y$ at $(0, 0)$ and $(2, 2)$ . The line $y = x$ intersects the circle $x^{2} + y^{2} - 2 y = 0$ at $(0, 0)$ and $(1, 1)$ . The total area under $y = x$ from $x = 0$ to $x = 2$ is the triangle with vertices $(0, 0), (2, 0), (2, 2)$ , having area $\frac{1}{2} \times 2 \times 2 = 2$ . The area under the parabola from $x = 0$ to $x = 2$ is $\int_{0}^{2} \frac{x^{2}}{2} d x = {[\frac{x^{3}}{6}]}_{0}^{2} = \frac{8}{6} = \frac{4}{3}$ . The circular sector region in the first quadrant below $y = x$ bounded by $x^{2} + y^{2} - 2 y \leq 0$ has area equal to: $(Sector of circle) - (Triangle) = \frac{π (1)^{2}}{4} - \frac{1}{2} (1) (1) = \frac{π}{4} - \frac{1}{2}$ . Thus, $Area (S) = Area (region between y = x and y = \frac{x^{2}}{2}) - Area (circle region below y = x)$ , $Area (S) = (2 - \frac{4}{3}) - (\frac{π}{4} - \frac{1}{2}) = \frac{2}{3} + \frac{1}{2} - \frac{π}{4} = \frac{7}{6} - \frac{π}{4}$ .
Execute: Comparing $Area (S) = \frac{7}{6} - \frac{π}{4}$ with $\frac{n + 2}{n + 1} - \frac{π}{n - 1}$ : $\frac{n + 2}{n + 1} = \frac{7}{6} ⟹ 6 n + 12 = 7 n + 7 ⟹ n = 5$ . Also, $\frac{π}{n - 1} = \frac{π}{4} ⟹ n - 1 = 4 ⟹ n = 5$ .
Verify: Substitute $n = 5$ into the given formula: $\frac{n + 2}{n + 1} - \frac{π}{n - 1}$ , which matches our computed area identically.

Area Bounded by a Parabola, Circle, and Line

What feels right?

The natural number n is equal to 5.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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