Question 442 marksSection E

(a) Using integration find the area of region covered by the roundabout. OR (b) Using integration, find the area of region covered by circular pond.

Application of Integrals
Area of a Circle Using Definite Integration

Official answer

Using integration, the area enclosed by a circle of radius r works out to πr2\pi r^2, giving 64π sq. units for the roundabout (option a) and 4π sq. units for the pond (option b).


(a) Area of the roundabout (C1:x2+y2=64,r=8C_1: x^2+y^2=64, r=8):

By symmetry about both axes, total area = 4 × (area in the first quadrant):

Area=4∫0864−x2 dx=4×[x264−x2+32sin⁡−1(x8)]08=4×16π\text{Area} = 4\int_0^8 \sqrt{64-x^2}\,dx = 4\times\left[\frac{x}{2}\sqrt{64-x^2}+32\sin^{-1}\left(\frac{x}{8}\right)\right]_0^8 = 4\times16\pi = 64π sq. units (≈ 201.06 sq. units)


(b) Area of the circular pond (C2:x2+y2=4,r=2C_2: x^2+y^2=4, r=2):

Area=4∫024−x2 dx=4×[x24−x2+2sin⁡−1(x2)]02=4×π\text{Area} = 4\int_0^2 \sqrt{4-x^2}\,dx = 4\times\left[\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\left(\frac{x}{2}\right)\right]_0^2 = 4\times\pi = 4π sq. units (≈ 12.57 sq. units)

Answer keywords:area by integrationdefinite integralcircle areaπr²symmetry about axesstandard integral formularoundaboutcircular pond

Marking scheme

  1. 1.1 mark: correct integral set-up, Area = 4∫0rr2−x2 dx4\int_0^r \sqrt{r^2-x^2}\,dx using symmetry, with the correct standard antiderivative x2r2−x2+r22sin⁡−1(xr)\frac{x}{2}\sqrt{r^2-x^2} + \frac{r^2}{2} \sin^{-1}\left(\frac{x}{r}\right).
  2. 2.1 mark: correct evaluation and final area — 64π sq. units for the roundabout (option a) or 4π sq. units for the pond (option b); direct use of πr2\pi r^2 alone (without the integral) does not earn full method marks since the question specifically says 'using integration'.

Hint

Write the circle's area as 4∫0rr2−x2 dx4\int_0^r \sqrt{r^2-x^2}\,dx and use the standard integral formula ∫a2−x2 dx=x2a2−x2+a22sin⁡−1(xa)\int \sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right).

Quick oral answer

Using Area = 4∫0rr2−x2 dx4\int_0^r \sqrt{r^2-x^2}\,dx and the standard antiderivative x2r2−x2+r22sin⁡−1(xr)\frac{x}{2}\sqrt{r^2-x^2}+\frac{r^2}{2}\sin^{-1}\left(\frac{x}{r}\right), the roundabout (r=8r=8) has area 64π sq. units and the pond (r=2r=2) has area 4π sq. units.

Common mistakes

  1. 1.Using the direct geometry formula πr2\pi r^2 without actually performing the integration, even though the question explicitly says 'using integration' — this loses method marks.
  2. 2.Forgetting the factor of 4 for quadrant symmetry, or integrating from −r to r and forgetting to multiply by only 2.
  3. 3.Errors in evaluating sin⁡−1(1)=π2\sin^{-1}(1) = \frac{\pi}{2}, or misapplying the standard integral formula for a2−x2\sqrt{a^2-x^2}.

Analysis and explanation

This sub-question is the culminating computational step of the case study, applying definite integration to find the area enclosed by a circle — one of the most important 'Application of Integrals' problems in the CBSE syllabus.


Concept: Using the function y=r2−x2y=\sqrt{r^2-x^2} derived in Q43 and the circle's symmetry about both axes, total area = 4∫0rr2−x2 dx4\int_0^r \sqrt{r^2-x^2}\,dx. Evaluating this using the standard formula ∫a2−x2 dx=x2a2−x2+a22sin⁡−1(xa)+C\int \sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C reproduces the familiar geometric result πr2\pi r^2, confirming the calculus method against known geometry.


Exam trap: Because the question insists on 'using integration', simply quoting Area=πr2\pi r^2 without deriving it through the integral loses method marks — CBSE markers specifically check for the integral set-up and the sin⁻¹ evaluation. Forgetting the factor of 4 (for one quadrant) or misevaluating sin⁻¹(1)=π/2 are the most common slips.


Real-world application: This is exactly how the paved circulating area of a real roundabout would be computed by a civil engineer — by subtracting the pond's area from the roundabout's total area (64π−4π=60π64\pi - 4\pi = 60\pi sq. units) to determine how much road surface material is needed.

Interesting facts

The result Area=πr2\text{Area} = \pi r^2 obtained here via integration matches the formula known since antiquity — Archimedes derived circle-area relationships around 250 BCE using the method of exhaustion, a precursor to integral calculus.

This exact problem type — deriving πr2\pi r^2 using ∫r2−x2 dx\int \sqrt{r^2-x^2}\,dx — appears as a worked example in the NCERT Class 12 textbook's Application of Integrals chapter, making it one of the most frequently repeated integration problems in CBSE board exams.

Real roundabouts are sized using precisely this kind of area calculation — traffic engineers compute the paved circulating area as the difference between the outer and central-island circle areas, i.e. 64π−4π=60π64\pi - 4\pi = 60\pi sq. units here.

Question details

Q44 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section E
Marks
2
Chapter
Application of Integrals
Topic
Area of a Circle Using Definite Integration
Bloom's level
apply
Syllabus topic
Unit 7: Integrals — Application of Integrals: Area Enclosed by a Circle x2+y2=r2x^2+y^2=r^2 via Definite Integrals
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why use integration when Area=πr2\text{Area}=\pi r^2 is already known?

The question specifically instructs 'using integration' to test the application of definite integrals; the geometric formula πr2\pi r^2 is only used here to verify that the integration reproduces the correct known result.

What is the area between the roundabout and the pond?

It is the difference of the two circle areas: 64π−4π=60π64\pi - 4\pi = 60\pi sq. units, representing the actual drivable road area of the roundabout.