Question

The area in the positive quadrant is enclosed by the circle $x^2+y^2=4$, the line $x=y\sqrt{3}$ and the $x$-axis is

• A. $\frac{\mathrm{\pi }}{2}$
• B. $\frac{\mathrm{\pi }}{4}$
• C. $\frac{\mathrm{\pi }}{3}$
• D. $\mathrm{\pi }$

Answer 1

The correct option is C

$\frac{\mathrm{\pi }}{3}$

Explanation for correct answer:

Step 1:Representing the curve

Given the equation of a circle is

$x^2+y^2=4\to \left(i\right)$

The radius of this circle is $2$ units.

Also.

$$\begin{gathered} x=y\sqrt{3} \\ \Rightarrow y=\frac{x}{\sqrt{3}}\to \left(ii\right) \end{gathered}$$

Put equation $\left(ii\right)$ in equation $\left(i\right)$, we get,

$$\begin{gathered} x^2+{\left(\frac{x}{\sqrt{3}}\right)}^2=4 \\ \Rightarrow \frac{3x^2+x^2}{3}=4 \\ \Rightarrow 4x^2=12 \\ \Rightarrow x^2=\frac{12}{4} \\ \Rightarrow x^2=3 \\ \Rightarrow x=\pm \sqrt{3} \end{gathered}$$

Put $x=\sqrt{3}$ in equation $\left(ii\right)$ we get

$$\begin{gathered} \Rightarrow y=\frac{\sqrt{3}}{\sqrt{3}} \\ y=1 \end{gathered}$$

Therefore the point of intersection in the positive quadrant is $\left(\sqrt{3},1\right)$

Step 2:Determining the area in the positive quadrant

The area of a shaded region is

$\begin{array}{l}A={\int }_0^{\sqrt{3}}\frac{x}{\sqrt{3}}dx+{\int }_{\sqrt{3}}^2\sqrt{4-x^2}dx\\ =\frac{1}{\sqrt{3}}{\left[\frac{x^2}{2}\right]}_0^{\sqrt{3}}+{\left[\frac{1}{2}x\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}x\right]}_{\sqrt{3}}^2\\ =\frac{3}{2\sqrt{3}}-0+\left(\frac{1}{2}\times 2\sqrt{4-2^2}\right)+\frac{4}{2}{\sin }^{-1}2-\left(\frac{1}{2}\times \sqrt{3}\times \sqrt{4-{\sqrt{3}}^2}\right)-\frac{4}{2}{\sin }^{-1}\sqrt{3}\\ =\frac{\sqrt{3}}{2}+2\times \frac{\pi }{2}-\frac{\sqrt{3}}{2}-2\frac{\pi }{3}\\ =\pi -\frac{2\pi }{3}\\ =\frac{\pi }{3}\end{array}$

Therefore the area of the shadow region is equal to $\frac{\mathrm{\pi }}{3}$ square units.

Hence, the correct answer is option (C).