Question

Prove that the area in the first quadrant enclosed by the x-axis, the line x = $\sqrt{3}y$ and the circle x² + y² = 4 is π/3.

Answer 1

$x^2+y^2=4$ represents a circle with centre O(0,0) and radius 2 , cutting x axis at A(2,0) and A'(-2,0)

$x=\sqrt{3}y$ represents a straight line passing through O(0,0)

Solving the two equations we get

$$\begin{gathered} x^2+y^2=4\mathrm{and}x=\sqrt{3}y \\ \Rightarrow {\left(\sqrt{3}y\right)}^2+y^2=4 \\ \Rightarrow 4y^2=4 \\ \Rightarrow y=\pm 1 \\ \Rightarrow x=\pm \sqrt{3} \\ B\left(\sqrt{3},1\right)\mathrm{and}B\text{'}\left(-\sqrt{3},-1\right)\mathrm{are}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{circle}\mathrm{and}\mathrm{straight}\mathrm{line} \\ \mathrm{Shaded}\mathrm{area}\left(OBQAO\right)=\mathrm{area}\left(OBPO\right)+\mathrm{area}\left(BAPB\right) \\ =\frac{1}{\sqrt{3}}{\int }_0^{\sqrt{3}}xdx+{\int }_{\sqrt{3}}^2\sqrt{4-x^2}dx \\ =\frac{1}{\sqrt{3}}{\left[\frac{x^2}{2}\right]}_0^{\sqrt{3}}+{\left[\frac{1}{2}x\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\left(\frac{x}{2}\right)\right]}_{\sqrt{3}}^2 \\ =\frac{\sqrt{3}}{2}+0-\frac{\sqrt{3}}{2}+2\left({\sin }^{-1}1-{\sin }^{-1}\frac{\sqrt{3}}{2}\right) \\ =\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}+2\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{3}\right) \\ =\frac{\mathrm{\pi }}{3}\mathrm{sq}\mathrm{units} \\ \mathrm{Area}\mathrm{bound}\mathrm{by}\mathrm{the}\mathrm{circle}\mathrm{and}\mathrm{straight}\mathrm{line}\mathrm{above}x\mathrm{axis}=\frac{\mathrm{\pi }}{3}\mathrm{sq}\mathrm{units} \end{gathered}$$