CBSE Board Examination 2026 · Section E · 4 marks

Q38: Case study: a roundabout and its pond, areas by integration

Board 2026Ch 8 · Application of Integrals Ch 7 · Integrals

The question

Roundabouts are often made on busy roads to ease the traffic and avoid red lights.

One such round-about is made such that equation representing its boundary is given by $C_1$ ; $x^2 + y^2 = 64$ ⁠.

There is a circular pond with a fountain in the middle of the roundabout whose equation is given by $C_2 : x^2 + y^2 = 4$ ⁠.

Based on the given information, answer the following questions :

(i) Represent the given equations $C_1$ and $C_2$ with the help of a diagram. (1 mark)

(ii) Express y as a function of $x$ ⁠, ( $y = f(x)$ ⁠), for both $C_1$ an $C_2$ ⁠. (1 mark)

(iii) (a) Using integration find the area of region covered by the roundabout. (2 marks)

(iii) (b) Using integration, find the area of region covered by circular pond. (2 marks)

Exam answer

(i) $C_1$ is the circle with centre $O(0, 0)$ and radius 8; $C_2$ is the circle with centre $O(0, 0)$ and radius 2.

Why?

The circle with centre at the origin and radius $r$ is

x^2 + y^2 = r^2

So $r^2 = 64$ gives $r = 8$ ⁠, and $r^2 = 4$ gives $r = 2$ ⁠. Both circles have the same centre, so the pond sits in the middle of the roundabout.

The circles

C_1

(radius 8) and

C_2

(radius 2), both with centre O.

(ii)

$C_1$ ⁠: $y = \sqrt{64 - x^2}$ or $y = -\sqrt{64 - x^2}$ ⁠, $-8 \le x \le 8$

Why?

From $x^2 + y^2 = 64$ ⁠, $y^2 = 64 - x^2$ ⁠, so $y$ is one of the two square roots. The $+$ root is the upper half of the circle and the $-$ root the lower half. A whole circle is not the graph of one function.

The root is real only when $64 - x^2 \ge 0$ ⁠, that is, $-8 \le x \le 8$ ⁠.

$C_2$ ⁠: $y = \sqrt{4 - x^2}$ or $y = -\sqrt{4 - x^2}$ ⁠, $-2 \le x \le 2$

Why?

The same steps for $C_2$ ⁠:

\begin{aligned} x^2 + y^2 &= 4 \\ y^2 &= 4 - x^2 \\ y &= \pm\sqrt{4 - x^2} \end{aligned}

The root is real only when $4 - x^2 \ge 0$ ⁠, that is, $x^2 \le 4$ ⁠, so $-2 \le x \le 2$ ⁠.

(iii) (a) The region inside $C_1$ is symmetric about both axes, so

\text{Area} = 4\int_0^8\sqrt{64 - x^2}\,dx

Why?

The circle is symmetric about both axes, so its four quarters have equal areas. In the first quadrant the quarter lies under the upper half, $y = \sqrt{64 - x^2}$ ⁠, from $x = 0$ to $x = 8$ ⁠.

So the whole area is 4 times the area of that quarter.

Using $\displaystyle\int\sqrt{a^2 - x^2}\,dx$ $= \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$ ⁠,

\begin{aligned} \text{Area} &= 4\bigg[\frac{x}{2}\sqrt{64 - x^2} \\ &\qquad + 32\sin^{-1}\frac{x}{8}\bigg]_0^8 \\ &= 2\bigg[x\sqrt{64 - x^2} \\ &\qquad + 64\sin^{-1}\frac{x}{8}\bigg]_0^8 \\ &= 2\left(0 + 64\cdot\frac{\pi}{2}\right) - 2(0 + 0) \end{aligned}

Why?

Here $a = 8$ ⁠, so $\frac{a^2}{2} = \frac{64}{2} = 32$ ⁠.

Taking 2 out of the bracket: $4\cdot\frac{x}{2} = 2x$ and $4 \cdot 32 = 2 \cdot 64$ ⁠.

At $x = 8$ ⁠: $\sqrt{64 - 64} = 0$ and $\sin^{-1}1 = \frac{\pi}{2}$ ⁠. At $x = 0$ ⁠: both terms are 0, since $\sin^{-1}0 = 0$ ⁠.

\begin{aligned} x = 8&: 8\sqrt{64 - 8^2} + 64\sin^{-1}\tfrac88 \\ &= 8(0) + 64\cdot\tfrac{\pi}{2} \\ &= 0 + 32\pi = 32\pi \\[4pt] x = 0&: 0\sqrt{64 - 0} + 64\sin^{-1}0 \\ &= 0 + 64(0) = 0 \end{aligned}

So the area is

\begin{aligned} 2(32\pi - 0) &= 2 \times 32\pi = 64\pi \end{aligned}

$\therefore$ Area of the region covered by the roundabout $= 64\pi$ sq. units.

Why?

Check: the area of a circle is $\pi r^2 = \pi\cdot 8^2 = 64\pi$ ⁠.

(iii) (b) The region inside $C_2$ is symmetric about both axes, so

\text{Area} = 4\int_0^2\sqrt{4 - x^2}\,dx

Why?

As in (a): 4 times the first-quadrant quarter, which lies under $y = \sqrt{4 - x^2}$ from $x = 0$ to $x = 2$ ⁠.

Using the same standard integral with $a = 2$ ⁠,

\begin{aligned} \text{Area} &= 4\Big[\frac{x}{2}\sqrt{4 - x^2} \\ &\qquad + 2\sin^{-1}\frac{x}{2}\Big]_0^2 \end{aligned}

\begin{aligned} \text{Area} &= 2\Big[x\sqrt{4 - x^2} + 4\sin^{-1}\frac{x}{2}\Big]_0^2 \\ &= 2\left(0 + 4\cdot\frac{\pi}{2}\right) - 2(0 + 0) \end{aligned}

Why?

With $a = 2$ ⁠, $\frac{a^2}{2} = 2$ ⁠. Taking 2 out of the bracket: $4\cdot\frac{x}{2} = 2x$ and $4 \cdot 2 = 2 \cdot 4$ ⁠.

At $x = 2$ ⁠: $\sqrt{4 - 4} = 0$ and $\sin^{-1}1 = \frac{\pi}{2}$ ⁠. At $x = 0$ both terms are 0.

\begin{aligned} x = 2&: 2\sqrt{4 - 2^2} + 4\sin^{-1}\tfrac22 \\ &= 2(0) + 4\cdot\tfrac{\pi}{2} \\ &= 0 + 2\pi = 2\pi \\[4pt] x = 0&: 0\sqrt{4 - 0} + 4\sin^{-1}0 \\ &= 0 + 4(0) = 0 \\[4pt] \text{Area} &= 2(2\pi - 0) = 4\pi \end{aligned}

Check: $\pi\cdot 2^2 = 4\pi$ ⁠.

$\therefore$ Area of the region covered by the pond $= 4\pi$ sq. units.

Where this comes from

Chapter 8, Application of Integrals: Area of a circle by integration.

Question type: Circles

How to spot it: click to see

Type 3. Circles: the square-root card, one quarter at a time

Spot it: $x^2 + y^2 = a^2$ ⁠, a circle with its centre at the origin and radius $a$ ⁠.

Here: (a) $C_1$ has radius 8: one quarter is $\int_0^8\sqrt{64 - x^2}\,dx$ ⁠, and times four gives $64\pi$ ⁠. (b) $C_2$ has radius 2: the same with $\sqrt{4 - x^2}$ ⁠, giving $4\pi$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 7, Integrals.

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