JEE MainMathematics

Answer verified against the official NTA key

+4 marks−1 if incorrectSingle correctPrevious-year question

Area Between a Circle and a Parabola

Area of the region {(x,y): x^2 + (y-2)^2 ≤ 4, x^2 ≥ 2y} is

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

A2π - 16/3Bπ - 8/3Cπ + 8/3D2π + 16/3

Step-by-step solutionView

Correct answer

The area of the given region is $2 π - \frac{16}{3}$ .

Option analysis

Why each option works or fails

A · 2π - 16/3

This option correctly finds the intersection points at $y = 2$ and subtracts the region inside the parabola below $y = 2$ from the lower semicircle. Correctly identify the region as the lower semicircle of radius $2$ minus the area inside the parabola $x^{2} \leq 2 y$ bounded by $y = 2$ .

B · π - 8/3

The student computed the area for only one half of the symmetric region ( $x \geq 0$ ) instead of the full region. Remember that both curves are symmetric about the $y$ -axis, so double the area in the first quadrant or integrate from $x = - 2$ to $x = 2$ .

C · π + 8/3

The student computed half the region and incorrectly added the parabolic area instead of subtracting it from the circular sector. The condition $x^{2} \geq 2 y$ means points lie outside or below the parabola, requiring subtraction from the semicircle rather than addition.

D · 2π + 16/3

The student added the area under the parabola to the semicircle instead of subtracting it. Check the inequality: $x^{2} \geq 2 y$ defines the region outside the parabola, so subtract the area between the parabola and the diameter from the semicircle.

Reviewed route

Solution

StepWorking

01Given

The region is defined by $x^{2} + (y - 2)^{2} \leq 4$ (interior and boundary of a circle centered at $(0, 2)$ with radius $2$ ) and $x^{2} \geq 2 y$ (region outside/below the parabola $y = x^{2} / 2$ ).

02Goal

Find the total area of the bounded region satisfying both inequalities.

03Approach

Find the intersection points of $x^{2} + (y - 2)^{2} = 4$ and $x^{2} = 2 y$ . Due to symmetry about the y-axis, calculate the area in the first quadrant and multiply by 2.

04Execute

Substitute $x^{2} = 2 y$ into the circle equation: $2 y + y^{2} - 4 y + 4 = 4 ⟹ y^{2} - 2 y = 0 ⟹ y = 0$ or $y = 2$ . For $y = 2$ , $x = \pm 2$ . Thus, the intersection points are $(0, 0)$ , $(2, 2)$ , and $(- 2, 2)$ .

05Execute

In the first quadrant, the region lies between $y = 0$ and $y = 2$ . For a given $y$ , the horizontal width goes from the parabola $x = \sqrt{2 y}$ to the circle $x = \sqrt{4 - (y - 2)^{2}}$ . Alternatively, integrating with respect to $y$ : $Area = 2 \int_{0}^{2} (\sqrt{4 - (y - 2)^{2}} - \sqrt{2 y}) d y$ . $\int_{0}^{2} \sqrt{4 - (y - 2)^{2}} d y$ is the area of a quarter circle of radius $2$ , which is $\frac{1}{4} π (2^{2}) = π$ . $\int_{0}^{2} \sqrt{2 y} d y = \sqrt{2} {[\frac{2}{3} y^{3 / 2}]}_{0}^{2} = \sqrt{2} \cdot \frac{2}{3} \cdot 2 \sqrt{2} = \frac{8}{3}$ . Therefore, $Total Area = 2 (π - \frac{8}{3}) = 2 π - \frac{16}{3}$ .

✓Verify

The total area of the circle is $π (2^{2}) = 4 π \approx 12.57$ . The parabola removes the upper interior portion. The computed area is $2 π - \frac{16}{3} \approx 6.28 - 5.33 = 0.95 > 0$ , which is positive and strictly less than the semicircle area $2 π \approx 6.28$ .

Hints that build this answer step by step

What are the intersection points of the circle $x^{2} + (y - 2)^{2} = 4$ and the parabola $x^{2} = 2 y$ ?
$(- 2, 2)$ and $(2, 2)$
How can the area of the region $x^{2} + (y - 2)^{2} \leq 4$ with $x^{2} \geq 2 y$ be decomposed using geometry and integration?
Area of the lower semicircle of radius $2$ minus the area bounded by $y = \frac{x^{2}}{2}$ and $y = 2$
What is the value of the integral for the area under $y = 2$ and above the parabola $y = \frac{x^{2}}{2}$ between $x = - 2$ and $x = 2$ ?
$\frac{16}{3}$
Subtracting this area from the lower semicircle of radius $r = 2$ , what is the final area?
$2 π - \frac{16}{3}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why is integrating with respect to y simpler than integrating with respect to x?

Integrating with respect to y is simpler. It takes the circle's right arc minus the parabola's right branch. This avoids breaking the integral at $x = 2$ , where the upper boundary curve changes.

Answer

The area of the given region is $2 π - \frac{16}{3}$ .

Why each option works or fails

A · correct: 2π - 16/3 - This option correctly finds the intersection points at $y = 2$ and subtracts the region inside the parabola below $y = 2$ from the lower semicircle. Correctly identify the region as the lower semicircle of radius $2$ minus the area inside the parabola $x^{2} \leq 2 y$ bounded by $y = 2$ .
B: π - 8/3 - The student computed the area for only one half of the symmetric region ( $x \geq 0$ ) instead of the full region. Remember that both curves are symmetric about the $y$ -axis, so double the area in the first quadrant or integrate from $x = - 2$ to $x = 2$ .
C: π + 8/3 - The student computed half the region and incorrectly added the parabolic area instead of subtracting it from the circular sector. The condition $x^{2} \geq 2 y$ means points lie outside or below the parabola, requiring subtraction from the semicircle rather than addition.
D: 2π + 16/3 - The student added the area under the parabola to the semicircle instead of subtracting it. Check the inequality: $x^{2} \geq 2 y$ defines the region outside the parabola, so subtract the area between the parabola and the diameter from the semicircle.

Step-by-step solution

Given: The region is defined by $x^{2} + (y - 2)^{2} \leq 4$ (interior and boundary of a circle centered at $(0, 2)$ with radius $2$ ) and $x^{2} \geq 2 y$ (region outside/below the parabola $y = x^{2} / 2$ ).
Goal: Find the total area of the bounded region satisfying both inequalities.
Approach: Find the intersection points of $x^{2} + (y - 2)^{2} = 4$ and $x^{2} = 2 y$ . Due to symmetry about the y-axis, calculate the area in the first quadrant and multiply by 2.
Execute: Substitute $x^{2} = 2 y$ into the circle equation: $2 y + y^{2} - 4 y + 4 = 4 ⟹ y^{2} - 2 y = 0 ⟹ y = 0$ or $y = 2$ . For $y = 2$ , $x = \pm 2$ . Thus, the intersection points are $(0, 0)$ , $(2, 2)$ , and $(- 2, 2)$ .
Execute: In the first quadrant, the region lies between $y = 0$ and $y = 2$ . For a given $y$ , the horizontal width goes from the parabola $x = \sqrt{2 y}$ to the circle $x = \sqrt{4 - (y - 2)^{2}}$ . Alternatively, integrating with respect to $y$ : $Area = 2 \int_{0}^{2} (\sqrt{4 - (y - 2)^{2}} - \sqrt{2 y}) d y$ . $\int_{0}^{2} \sqrt{4 - (y - 2)^{2}} d y$ is the area of a quarter circle of radius $2$ , which is $\frac{1}{4} π (2^{2}) = π$ . $\int_{0}^{2} \sqrt{2 y} d y = \sqrt{2} {[\frac{2}{3} y^{3 / 2}]}_{0}^{2} = \sqrt{2} \cdot \frac{2}{3} \cdot 2 \sqrt{2} = \frac{8}{3}$ . Therefore, $Total Area = 2 (π - \frac{8}{3}) = 2 π - \frac{16}{3}$ .
Verify: The total area of the circle is $π (2^{2}) = 4 π \approx 12.57$ . The parabola removes the upper interior portion. The computed area is $2 π - \frac{16}{3} \approx 6.28 - 5.33 = 0.95 > 0$ , which is positive and strictly less than the semicircle area $2 π \approx 6.28$ .

Area Between a Circle and a Parabola

What feels right?

The area of the given region is $2 π - \frac{16}{3}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

We think you should solve this next.

If $f : R \to R$ be a continuous function satisfying…

The area of the region…

'A' and 'B' in the below reactions are…

Students also ask

Area Between a Circle and a Parabola

The area of the given region is 2π−1632π−316​.

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

If f:R→Rf:R→R be a continuous function satisfying…

The area of the region…

'A' and 'B' in the below reactions are…

Students also ask

The area of the given region is $2 π - \frac{16}{3}$ .

If $f : R \to R$ be a continuous function satisfying…