CBSE Sample Paper 2026-27 · Section D · 5 marks

Q34: Area between the circle $y = \sqrt{4 - x^2}$ and $x + y = 2$ ⁠; or between $y^2 = 2x$ and $x - 2y = 0$

Sample 2026-27Ch 8 · Application of Integrals Ch 7 · Integrals

The question

(a) Using Integration, find the area of the region

$\left\{(x, y) : 0 \le y \le \sqrt{4 - x^2},\ x \ge 2 - y\right\}$ ⁠.

(b) Using Integration, find the area of the region bounded by the curve $y^2 = 2x$ and the line $x - 2y = 0$ ⁠.

For Visually Impaired:

(a) Using Integration, find the area of the region enclosed within $25x^2 + 16y^2 = 400$

(b) Using Integration find the area of the region in the first quadrant bounded by the curve $y^2 = 8x$ ⁠, and the lines $x = 2$ ⁠, $x = 5$ and x-axis.

Exam answer

(a)

$y = \sqrt{4 - x^2}$ is the upper semicircle with centre $(0, 0)$ and radius 2. $x \ge 2 - y$ is $x + y \ge 2$ ⁠, the side of the line $x + y = 2$ away from O. The line meets the circle at $(0, 2)$ and $(2, 0)$ ⁠.

Why?

Squaring $y = \sqrt{4 - x^2}$ gives $x^2 + y^2 = 4$ ⁠, a circle of radius 2. The root is never negative, so only the upper half $y \ge 0$ is used.

At O, $0 + 0 \ge 2$ is false, so $x + y \ge 2$ is the side of the line away from O.

Putting $y = 2 - x$ in $x^2 + y^2 = 4$ ⁠: $2x^2 - 4x = 0$ ⁠, so $x = 0$ or $x = 2$ ⁠.

Why?

$x^2 + (2 - x)^2 = 4$ gives $2x^2 - 4x + 4 = 4$ ⁠, that is $2x(x - 2) = 0$ ⁠. The points are $(0, 2)$ and $(2, 0)$ ⁠.

\begin{aligned} (2 - x)^2 &= 4 - 4x + x^2 \\[4pt] x^2 + 4 - 4x + x^2 &= 4 \\ 2x^2 - 4x + 4 &= 4 \\ 2x^2 - 4x &= 0 \\ 2x(x - 2) &= 0 \end{aligned}

So $x = 0$ or $x = 2$ ⁠. Then $y = 2 - x$ gives $y = 2 - 0 = 2$ and $y = 2 - 2 = 0$ ⁠.

The region (shaded) between the line

x + y = 2

and the circle, from

(0, 2)

(2, 0)

⁠.

\begin{aligned} \text{Area} &= \int_0^2\bigg(\sqrt{4 - x^2} \\ &\qquad - (2 - x)\bigg)dx \end{aligned}

Why?

The area between two curves is $\int(\text{upper} - \text{lower})\,dx$ ⁠. From $x = 0$ to $x = 2$ ⁠, the arc is above the line. For example, at $x = 1$ ⁠: $\sqrt3 \approx 1.73$ is above $2 - 1 = 1$ ⁠.

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

Why?

Standard integral, here with $a = 2$ ⁠, so $\frac{a^2}{2} = 2$ ⁠:

\begin{aligned} &\int\sqrt{a^2 - x^2}\,dx \\ &= \frac{x}{2}\sqrt{a^2 - x^2} \\ &\quad + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C \end{aligned}

Also $\int(2 - x)\,dx = 2x - \frac{x^2}{2}$ ⁠, and it is subtracted, so its signs flip.

-\left(2x - \frac{x^2}{2}\right) = -2x + \frac{x^2}{2}

At $x = 2$ ⁠: the root is 0 and $\sin^{-1}1 = \frac{\pi}{2}$ ⁠. At $x = 0$ ⁠: every term is 0.

\begin{aligned} \frac{2}{2}\sqrt{4 - 2^2} &= 1 \times \sqrt{0} = 0 \\ 2\sin^{-1}\frac{2}{2} &= 2\sin^{-1}1 = 2 \cdot \frac{\pi}{2} \\ -2(2) &= -4 \\ \frac{(2)^2}{2} &= \frac{4}{2} = 2 \end{aligned}

= \pi - 2

Why?

Inside the bracket, $2 \cdot \frac{\pi}{2} = \pi$ ⁠, and $-4 + 2 = -2$ ⁠. So the bracket is $0 + \pi - 2$ ⁠.

Check with geometry: the quarter circle has area $\frac14\pi(2)^2 = \pi$ ⁠, and the triangle under the line has area $\frac12 \times 2 \times 2 = 2$ ⁠.

\begin{aligned} \frac14\pi(2)^2 &= \frac14 \times 4\pi = \pi \\ \text{quarter} - \text{triangle} &= \pi - 2 \end{aligned}

$\therefore$ Required area $= (\pi - 2)$ sq. units

(b)

$y^2 = 2x$ is a parabola and $x - 2y = 0$ is the line $y = \frac{x}{2}$ ⁠. Putting $x = 2y$ in $y^2 = 2x$ ⁠: $y^2 = 4y$ ⁠, so $y = 0$ or $y = 4$ ⁠. They meet at $(0, 0)$ and $(8, 4)$ ⁠.

Why?

\begin{aligned} y^2 &= 2(2y) = 4y \\ y^2 - 4y &= 0 \\ y(y - 4) &= 0 \end{aligned}

$y^2 = 4y$ gives $y(y - 4) = 0$ ⁠. Watch out: do not divide by $y$ ⁠, or the root $y = 0$ is lost.

Then $x = 2y$ gives $x = 0$ and $x = 8$ ⁠.

The region (shaded) between the parabola

y^2 = 2x

and the line

x = 2y

⁠, from

(0, 0)

(8, 4)

⁠.

\text{Area} = \int_0^8\left(\sqrt{2x} - \frac{x}{2}\right)dx

Why?

The region is above the x-axis, so use the upper branch $y = \sqrt{2x}$ of the parabola. Between 0 and 8 it is above the line: at $x = 2$ ⁠, $\sqrt{4} = 2$ is above $\frac22 = 1$ ⁠.

\begin{aligned} &= \left[\sqrt2\cdot\frac23x^{\frac32} - \frac{x^2}{4}\right]_0^8 \\ &= \frac{2\sqrt2}{3}\cdot 16\sqrt2 - 16 \\ &= \frac{64}{3} - 16 \end{aligned}

Why?

Write $\sqrt{2x} = \sqrt2\,x^{\frac12}$ and use the power rule:

\int x^{\frac12}\,dx = \frac23x^{\frac32}

At $x = 8$ ⁠: $8^{\frac32} = \left(\sqrt8\right)^3 = \left(2\sqrt2\right)^3 = 16\sqrt2$ ⁠, and $\frac{8^2}{4} = 16$ ⁠.

Then $\frac{2\sqrt2}{3}\cdot16\sqrt2 = \frac{64}{3}$ ⁠, since $\sqrt2\cdot\sqrt2 = 2$ ⁠.

\begin{aligned} \sqrt2\cdot\frac23 &= \frac{2\sqrt2}{3} \\[4pt] \sqrt8 &= \sqrt{4 \times 2} = 2\sqrt2 \\ (2\sqrt2)^3 &= 2^3(\sqrt2)^3 = 8 \times 2\sqrt2 \\ &= 16\sqrt2 \\[4pt] \frac{2\sqrt2}{3}\cdot 16\sqrt2 &= \frac{2 \times 16 \times 2}{3} = \frac{64}{3} \end{aligned}

At $x = 0$ both terms are 0.

Last, $\frac{64}{3} - 16$ ⁠. Write 16 with denominator 3, then subtract the tops:

\begin{aligned} 16 &= \frac{16 \times 3}{3} = \frac{48}{3} \\ \frac{64}{3} - \frac{48}{3} &= \frac{64 - 48}{3} = \frac{16}{3} \end{aligned}

$\therefore$ Required area $= \frac{16}{3}$ sq. units

For Visually Impaired

(a) $25x^2 + 16y^2 = 400$ is the ellipse $\frac{x^2}{16} + \frac{y^2}{25} = 1$ ⁠. In the first quadrant,

y = \frac54\sqrt{4^2 - x^2}

Why?

Divide the equation by 400: $\frac{25}{400} = \frac{1}{16}$ and $\frac{16}{400} = \frac{1}{25}$ ⁠.

Solve for $y$ ⁠:

\begin{aligned} \frac{y^2}{25} &= 1 - \frac{x^2}{16} \\ y^2 &= 25\left(1 - \frac{x^2}{16}\right) \\ &= 25\cdot\frac{16 - x^2}{16} \\ &= \frac{25}{16}\left(16 - x^2\right) \\[4pt] y &= \sqrt{\frac{25}{16}}\sqrt{16 - x^2} \\ &= \frac54\sqrt{4^2 - x^2} \end{aligned}

In the first quadrant $y \ge 0$ ⁠, so take the positive root.

The ellipse is symmetric about both axes, so

Why?

The four quadrants hold equal pieces of the ellipse. Find the first-quadrant area ( $x$ from 0 to 4) and multiply by 4.

\begin{aligned} \text{Area} &= 4\int_0^4\frac54\sqrt{4^2 - x^2}\,dx \\ &= 5\int_0^4\sqrt{4^2 - x^2}\,dx \\ &= 5\bigg[\frac{x}{2}\sqrt{16 - x^2} \\ &\qquad + 8\sin^{-1}\frac{x}{4}\bigg]_0^4 \\ &= 5\left(0 + 8 \cdot \frac{\pi}{2}\right) - 0 \end{aligned}

Why?

$4 \times \frac54 = 5$ ⁠. Then the standard integral of $\sqrt{a^2 - x^2}$ with $a = 4$ ⁠, so $\frac{a^2}{2} = 8$ ⁠.

At $x = 4$ ⁠: the root is 0 and $\sin^{-1}1 = \frac{\pi}{2}$ ⁠. So the area is $5 \times 4\pi = 20\pi$ ⁠.

\begin{aligned} \frac{4}{2}\sqrt{16 - 4^2} &= 2 \times \sqrt{0} = 0 \\ 8\sin^{-1}\frac{4}{4} &= 8 \times \frac{\pi}{2} = 4\pi \\ 5(0 + 4\pi) &= 20\pi \end{aligned}

At $x = 0$ ⁠: $0 + 8\sin^{-1}0 = 0$ ⁠.

Check: an ellipse with semi-axes $a$ and $b$ has area $\pi ab = \pi \times 4 \times 5$ ⁠.

$\therefore$ Required area $= 20\pi$ sq. units

(b) $y^2 = 8x$ is a parabola. In the first quadrant $y = \sqrt{8x}$ ⁠, so the area above the x-axis, below the parabola, between $x = 2$ and $x = 5$ is

Why?

The region is bounded by the curve above, the x-axis below and the two vertical lines at the sides, so the area is $\int_2^5 y\,dx$ ⁠.

\begin{aligned} \text{Area} &= \int_2^5\sqrt{8x}\,dx \\ &= \left[2\sqrt2\cdot\frac23x^{\frac32}\right]_2^5 \\ &= \frac{4\sqrt2}{3}\left(5\sqrt5 - 2\sqrt2\right) \end{aligned}

Why?

$\sqrt{8x} = 2\sqrt2\,x^{\frac12}$ ⁠, and $\int x^{\frac12}\,dx = \frac23x^{\frac32}$ ⁠.

$5^{\frac32} = 5\sqrt5$ and $2^{\frac32} = 2\sqrt2$ ⁠.

Multiply out: $\sqrt2 \cdot 5\sqrt5 = 5\sqrt{10}$ and $\sqrt2 \cdot 2\sqrt2 = 4$ ⁠.

\begin{aligned} \sqrt{8x} &= \sqrt8\,\sqrt x = 2\sqrt2\,x^{\frac12} \\[4pt] 2\sqrt2\cdot\frac23 &= \frac{4\sqrt2}{3} \\[4pt] \text{top} - \text{bottom} &= \frac{4\sqrt2}{3}\left(5^{\frac32} - 2^{\frac32}\right) \\ &= \frac{4\sqrt2}{3}\left(5\sqrt5 - 2\sqrt2\right) \\[4pt] \sqrt2\left(5\sqrt5 - 2\sqrt2\right) &= 5\sqrt{10} - 4 \end{aligned}

So the area is $\frac43\left(5\sqrt{10} - 4\right)$ ⁠.

$\therefore$ Required area $= \dfrac43\left(5\sqrt{10} - 4\right)$ sq. units

Where this comes from

Chapter 8, Application of Integrals: Area between a curve and a line.

Question type: (a) A curve and a line + Circles (combined); (b) A curve and a line

How to spot it: click to see

(a) Type 7. A curve and a line: top minus bottom

Spot it: the region is shut in by a curve and a line, like a parabola and $y = x$ ⁠, sometimes written as a set with $\le$ signs. Your book has no worked question of this kind, but CBSE sample papers ask it.

Here: Read the set: under the semicircle $y = \sqrt{4 - x^2}$ ⁠, on the far side of the line $x + y = 2$ ⁠. They meet at $(0, 2)$ and $(2, 0)$ ⁠; the circle is on top, so integrate $\sqrt{4 - x^2} - (2 - x)$ ⁠.

(a) 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: The circle part, $\int_0^2\sqrt{4 - x^2}\,dx$ ⁠, is one quarter of radius 2: the square-root card gives $\pi$ ⁠, and the line part takes off 2.

(b) Type 7. A curve and a line: top minus bottom

Here: $y^2 = 2x$ and $x = 2y$ meet at $(0, 0)$ and $(8, 4)$ ⁠. From 0 to 8 the parabola $\sqrt{2x}$ is above the line $\frac x2$ ⁠, so integrate $\sqrt{2x} - \frac x2$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 7, Integrals.

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Application of Integrals in past papers: question 8 of 17 · see them all

Q34: Area between the circle y=4−x2y = \sqrt{4 - x^2}y=4−x2​ and x+y=2x + y = 2x+y=2⁠; or between y2=2xy^2 = 2xy2=2x and x−2y=0x - 2y = 0x−2y=0

Q34: Area between the circle $y = \sqrt{4 - x^2}$ and $x + y = 2$ ⁠; or between $y^2 = 2x$ and $x - 2y = 0$