Integral Calculus: JEE Main Mathematics Question with Solution

Q: Why is the circular region area evaluated as ?

The circle is centered at . The line passes through and the point , dividing the right half of the circle. The area under within the circle is a quadrant sector of radius minus the right triangle with vertices , giving .

If the area of the region

\mathrm{S}=\left\{(\mathrm{x},\mathrm{y}): 2\mathrm{y}-\mathrm{y}^{2}\le\mathrm{x}^{2}\le2\mathrm{y}, \mathrm{x}\ge\mathrm{y}\right\}

is equal to

\frac{\mathrm{n}+2}{\mathrm{n}+1}-\frac{\pi}{\mathrm{n}-1}

, then the natural number

\mathrm{n}

is equal to ________

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Step-by-step solutionView

Correct answer

The natural number n is equal to 5.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Region $S = \{(x,y): 2y - y^2 \le x^2 \le 2y, \; x \ge y\}$ and $\text{Area}(S) = \frac{n+2}{n+1} - \frac{\pi}{n-1}$ for natural number $n$ .

02goal

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

03approach

Analyze the three boundary conditions in the first quadrant: 1. $x^2 + y^2 - 2y \ge 0 \implies x^2 + (y-1)^2 \ge 1$ (outside/on the circle centered at $(0,1)$ with radius $1$ ). 2. $x^2 \le 2y \implies y \ge \frac{x^2}{2}$ (above/on the parabola $x^2 = 2y$ ). 3. $x \ge y$ (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 $x^2 = 2y$ at $(0,0)$ and $(2,2)$ . The line $y = x$ intersects the circle $x^2 + y^2 - 2y = 0$ 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 $\int_0^2 \frac{x^2}{2}\,dx = \left[\frac{x^3}{6}\right]_0^2 = \frac{8}{6} = \frac{4}{3}$ . The circular sector region in the first quadrant below $y=x$ bounded by $x^2 + y^2 - 2y \le 0$ has area equal to: $(\text{Sector of circle}) - (\text{Triangle}) = \frac{\pi(1)^2}{4} - \frac{1}{2}(1)(1) = \frac{\pi}{4} - \frac{1}{2}$ . Thus, $\text{Area}(S) = \text{Area}(\text{region between } y=x \text{ and } y=\frac{x^2}{2}) - \text{Area}(\text{circle region below } y=x)$ , $\text{Area}(S) = \left(2 - \frac{4}{3}\right) - \left(\frac{\pi}{4} - \frac{1}{2}\right) = \frac{2}{3} + \frac{1}{2} - \frac{\pi}{4} = \frac{7}{6} - \frac{\pi}{4}$ .

05execute

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

✓verify

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

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

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Why is the circular region area evaluated as

\frac{\pi}{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 $\frac{1}{4}\pi(1)^2 - \frac{1}{2}(1)(1) = \frac{\pi}{4} - \frac{1}{2}$ .

Answer

The natural number n is equal to 5.

Why each option works or fails

Step-by-step solution

given: Region $S = \{(x,y): 2y - y^2 \le x^2 \le 2y, \; x \ge y\}$ and $\text{Area}(S) = \frac{n+2}{n+1} - \frac{\pi}{n-1}$ for natural number $n$ .
goal: Compute the area of the region $S$ and match it to $\frac{n+2}{n+1} - \frac{\pi}{n-1}$ to find the value of $n$ .
approach: Analyze the three boundary conditions in the first quadrant: 1. $x^2 + y^2 - 2y \ge 0 \implies x^2 + (y-1)^2 \ge 1$ (outside/on the circle centered at $(0,1)$ with radius $1$ ). 2. $x^2 \le 2y \implies y \ge \frac{x^2}{2}$ (above/on the parabola $x^2 = 2y$ ). 3. $x \ge y$ (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 $x^2 = 2y$ at $(0,0)$ and $(2,2)$ . The line $y = x$ intersects the circle $x^2 + y^2 - 2y = 0$ 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 $\int_0^2 \frac{x^2}{2}\,dx = \left[\frac{x^3}{6}\right]_0^2 = \frac{8}{6} = \frac{4}{3}$ . The circular sector region in the first quadrant below $y=x$ bounded by $x^2 + y^2 - 2y \le 0$ has area equal to: $(\text{Sector of circle}) - (\text{Triangle}) = \frac{\pi(1)^2}{4} - \frac{1}{2}(1)(1) = \frac{\pi}{4} - \frac{1}{2}$ . Thus, $\text{Area}(S) = \text{Area}(\text{region between } y=x \text{ and } y=\frac{x^2}{2}) - \text{Area}(\text{circle region below } y=x)$ , $\text{Area}(S) = \left(2 - \frac{4}{3}\right) - \left(\frac{\pi}{4} - \frac{1}{2}\right) = \frac{2}{3} + \frac{1}{2} - \frac{\pi}{4} = \frac{7}{6} - \frac{\pi}{4}$ .
execute: Comparing $\text{Area}(S) = \frac{7}{6} - \frac{\pi}{4}$ with $\frac{n+2}{n+1} - \frac{\pi}{n-1}$ : $\frac{n+2}{n+1} = \frac{7}{6} \implies 6n + 12 = 7n + 7 \implies n = 5$ . Also, $\frac{\pi}{n-1} = \frac{\pi}{4} \implies n - 1 = 4 \implies n = 5$ .
verify: Substitute $n = 5$ into the given formula: $\frac{5+2}{5+1} - \frac{\pi}{5-1} = \frac{7}{6} - \frac{\pi}{4}$ , which matches our computed area identically.