CBSE Sample Paper 2020-21 · Part B: Section IV · 3 marks

Q34: Area between $x^2 + y^2 = 4$ ⁠, $y = \sqrt3x$ and the x-axis, or the area of $x^2 + 9y^2 = 36$

Sample 2020-21Ch 8 · Application of Integrals Ch 7 · Integrals

The question

(a) Find the area of the region bounded by the curves $x^2 + y^2 = 4$ ⁠, $y = \sqrt3x$ and $x$ -axis in the first quadrant

(b) Find the area of the ellipse $x^2 + 9y^2 = 36$ using integration

Exam answer

(a)

Solving $y = \sqrt3x$ and $x^2 + y^2 = 4$ ⁠, we get $x^2 + 3x^2 = 4$ $\Rightarrow x^2 = 1$ $\Rightarrow x = 1$ (first quadrant). They meet at $(1, \sqrt3)$ ⁠.

Why?

Substitute $y = \sqrt3x$ into the circle: $y^2 = (\sqrt3x)^2 = 3x^2$ ⁠. Then $4x^2 = 4$ ⁠, so $x = \pm 1$ ⁠. In the first quadrant $x = 1$ ⁠, and $y = \sqrt3 \times 1 = \sqrt3$ ⁠.

\begin{aligned} x^2 + (\sqrt3x)^2 &= 4 \\ x^2 + 3x^2 &= 4 \\ 4x^2 &= 4 \\ x^2 &= 1 \quad (\div 4) \\ x &= \pm 1 \end{aligned}

The region (shaded) bounded by the line

y = \sqrt3x

⁠, the circle and the x-axis.

\begin{aligned} \text{Area} &= \sqrt3\int_0^1 x\,dx \\ &\quad + \int_1^2\sqrt{2^2 - x^2}\,dx \end{aligned}

Why?

The top edge of the region changes at $x = 1$ ⁠. From $x = 0$ to $1$ it is the line $y = \sqrt3x$ ⁠; from $x = 1$ to $2$ it is the circle, $y = \sqrt{4 - x^2}$ (upper half). So the area is split into two integrals of $y\,dx$ ⁠.

The circle meets the x-axis at $x = 2$ ⁠, its radius, and $4 = 2^2$ ⁠.

\begin{aligned} &= \frac{\sqrt3}{2}\left[x^2\right]_0^1 + \bigg[\frac{x}{2}\sqrt{2^2 - x^2} \\ &\qquad + 2\sin^{-1}\left(\frac{x}{2}\right)\bigg]_1^2 \end{aligned}

Why?

\begin{aligned} \sqrt3\int_0^1 x\,dx &= \sqrt3\left[\frac{x^2}{2}\right]_0^1 \\ &= \frac{\sqrt3}{2}\left[x^2\right]_0^1 \end{aligned}

$\int x\,dx = \frac{x^2}{2}$ ⁠, which puts the $\frac12$ in front. For the circle part, the standard integral is

\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}

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

\begin{aligned} &= \frac{\sqrt3}{2} + \left[2 \times \frac{\pi}{2} - \frac{\sqrt3}{2} - 2 \times \frac{\pi}{6}\right] \\ &= \pi - \frac{\pi}{3} = \frac{2\pi}{3} \end{aligned}

Why?

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

At $x = 1$ ⁠: $\frac12\sqrt{4 - 1} = \frac{\sqrt3}{2}$ and $\sin^{-1}\frac12 = \frac{\pi}{6}$ ⁠.

The two $\frac{\sqrt3}{2}$ terms cancel, leaving $\pi - \frac{\pi}{3}$ ⁠.

Each piece, one at a time:

\begin{aligned} \text{Line part} &= \frac{\sqrt3}{2}\left(1^2 - 0^2\right) = \frac{\sqrt3}{2} \\[4pt] \text{At } x = 2&: \frac22\sqrt{4 - 4} + 2\sin^{-1}1 \\ &= 0 + 2 \times \frac{\pi}{2} = \pi \\[4pt] \text{At } x = 1&: \frac12\sqrt{4 - 1} + 2\sin^{-1}\frac12 \\ &= \frac{\sqrt3}{2} + 2 \times \frac{\pi}{6} \\ &= \frac{\sqrt3}{2} + \frac{\pi}{3} \end{aligned}

Circle part = (value at 2) minus (value at 1). Then add the line part:

\begin{aligned} &\frac{\sqrt3}{2} + \pi - \left(\frac{\sqrt3}{2} + \frac{\pi}{3}\right) \\ &= \frac{\sqrt3}{2} + \pi - \frac{\sqrt3}{2} - \frac{\pi}{3} \\ &= \pi - \frac{\pi}{3} = \frac{3\pi}{3} - \frac{\pi}{3} \\ &= \frac{2\pi}{3} \end{aligned}

$\therefore$ Required area $= \frac{2\pi}{3}$ sq. units

(b)

$x^2 + 9y^2 = 36$ is the ellipse $\frac{x^2}{36} + \frac{y^2}{4} = 1$ ⁠. In the first quadrant, $y = \frac13\sqrt{6^2 - x^2}$ ⁠, $0 \le x \le 6$ ⁠.

Why?

Dividing $x^2 + 9y^2 = 36$ by 36 gives the standard form, with $a = 6$ on the x-axis and $b = 2$ on the y-axis.

\begin{aligned} \frac{x^2}{36} + \frac{9y^2}{36} &= \frac{36}{36} \\ \frac{x^2}{36} + \frac{y^2}{4} &= 1 \quad \left(\tfrac{9}{36} = \tfrac14\right) \\ a^2 = 36,\ b^2 &= 4 \\ a = 6,\ b &= 2 \end{aligned}

For $y$ ⁠: $9y^2 = 36 - x^2$ ⁠, so $y = \frac13\sqrt{36 - x^2}$ ⁠, taking $y \ge 0$ in the first quadrant.

\begin{aligned} 9y^2 &= 36 - x^2 \\ y^2 &= \frac{36 - x^2}{9} \\ y &= \frac{\sqrt{36 - x^2}}{\sqrt9} = \frac13\sqrt{36 - x^2} \end{aligned}

The ellipse

x^2 + 9y^2 = 36

⁠, with vertices

(\pm 6, 0)

and

(0, \pm 2)

⁠.

The ellipse is symmetric about both axes, so

Why?

The two axes cut the ellipse into four equal quarters. So the whole area is 4 times the area in the first quadrant, which is $\int_0^6 y\,dx$ ⁠.

\begin{aligned} \text{Area} &= 4 \times \frac13\int_0^6\sqrt{6^2 - x^2}\,dx \end{aligned}

Why?

Use the standard integral

\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}

with $a = 6$ ⁠, so $\frac{a^2}{2} = 18$ ⁠. Also $4 \times \frac13 = \frac43$ ⁠.

\begin{aligned} &= \frac43\bigg[\frac{x}{2}\sqrt{6^2 - x^2} \\ &\qquad + 18\sin^{-1}\left(\frac{x}{6}\right)\bigg]_0^6 \\ &= \frac43\left[18 \times \frac{\pi}{2} - 0\right] \\ &= \frac43 \times 9\pi \end{aligned}

Why?

At $x = 6$ ⁠: $\sqrt{36 - 36} = 0$ and $\sin^{-1}1 = \frac{\pi}{2}$ ⁠. At $x = 0$ ⁠: both terms are 0, since $\sin^{-1}0 = 0$ ⁠.

\begin{aligned} \text{At } 6&: \frac62\sqrt{36 - 36} + 18\sin^{-1}\frac66 \\ &= 3(0) + 18\sin^{-1}1 \\ &= 18 \times \frac{\pi}{2} = 9\pi \\[4pt] \text{At } 0&: 0 + 18\sin^{-1}0 = 0 \\[4pt] \frac43 \times 9\pi &= \frac{4 \times 9\pi}{3} = \frac{36\pi}{3} \\ &= 12\pi \end{aligned}

Check: the area of an ellipse with semi-axes $a$ and $b$ is $\pi ab = \pi \times 6 \times 2 = 12\pi$ ⁠.

$\therefore$ Area of the ellipse $= 12\pi$ sq. units

Where this comes from

Chapter 8, Application of Integrals: Area of a circle sector; area of an ellipse.

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

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: The line $y = \sqrt3x$ meets the circle at $(1, \sqrt3)$ ⁠. With the $x$ -axis below, the top edge changes there: the line from 0 to 1, then the circle from 1 to 2.

(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 piece is $\int_1^2\sqrt{4 - x^2}\,dx$ ⁠, with radius 2 and the square-root card; here it is not a full quarter, as it starts at $x = 1$ ⁠.

(b) Type 4. Ellipses: read off $a$ and $b$ ⁠, then times four

Spot it: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ⁠, with numbers under $x^2$ and $y^2$ ⁠.

Here: Divide by 36: $\frac{x^2}{36} + \frac{y^2}{4} = 1$ ⁠, so $a = 6$ ⁠, $b = 2$ ⁠, and the top edge is $\frac13\sqrt{36 - x^2}$ ⁠. Four quarters give $12\pi = \pi ab$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 7, Integrals.

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

Q34: Area between x2+y2=4x^2 + y^2 = 4x2+y2=4⁠, y=3xy = \sqrt3xy=3​x and the x-axis, or the area of x2+9y2=36x^2 + 9y^2 = 36x2+9y2=36

Q34: Area between $x^2 + y^2 = 4$ ⁠, $y = \sqrt3x$ and the x-axis, or the area of $x^2 + 9y^2 = 36$