Question

Find the area of the region in the first quadrant enclosed by x-axis, the line y = $\sqrt{3}x$ and the circle x² + y² = 16.

Answer 1

$x^2+y^2=16$ represents a circle with centre O(0,0) and cutting the x axis at A(4,0)

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

Point of intersection is obtained by solving the two equations

$$\begin{gathered} x^2+y^2=16\mathrm{and}y=\sqrt{3}x \\ \Rightarrow x^2+{\left(\sqrt{3}x\right)}^2=16 \\ \Rightarrow 4x^2=16 \\ \Rightarrow x=\pm 2 \\ \Rightarrow y=\pm 2\sqrt{3} \\ B\left(2,2\sqrt{3}\right)\mathrm{and}B\text{'}\left(-2,-2\sqrt{3}\right)\mathrm{are}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{circle}\mathrm{and}\mathrm{straight}\mathrm{line} \\ \mathrm{Shaded}\mathrm{area}\left(OBQAO\right)=\mathrm{area}\left(OBPO\right)+\mathrm{area}\left(PBQAP\right) \\ ={\int }_0^2\sqrt{3}xdx+{\int }_2^4\sqrt{16-x^2}dx \\ =\sqrt{3}{\left[\frac{x^2}{2}\right]}_0^2+{\left[\frac{1}{2}x\sqrt{16-x^2}+\frac{16}{2}{\sin }^{-1}\left(\frac{x}{4}\right)\right]}_2^4 \\ =2\sqrt{3}+8\times \frac{\mathrm{\pi }}{2}-2\sqrt{3}-8\times \frac{\mathrm{\pi }}{6} \\ =2\sqrt{3}+4\mathrm{\pi }-2\sqrt{3}-\frac{4\mathrm{\pi }}{3} \\ =\frac{8\mathrm{\pi }}{3}\mathrm{sq}\mathrm{units} \\ \mathrm{Area}\mathrm{bound}\mathrm{by}\mathrm{the}\mathrm{circle}\mathrm{and}\mathrm{straight}\mathrm{line}\mathrm{above}\mathrm{x}\mathrm{axis}=2\sqrt{3}+\left(-2\sqrt{3}+8\times \frac{2\mathrm{\pi }}{6}\right)=\frac{8\mathrm{\pi }}{3}\mathrm{sq}\mathrm{units} \end{gathered}$$