Question

Find the area of the region bounded by the curve y = $\sqrt{1-x^2}$, line y = x and the positive x-axis.

Answer 1

$$\begin{gathered} y=\sqrt{1-x^2} \\ \Rightarrow y^2=1-x^2 \\ \Rightarrow x^2+y^2=1 \end{gathered}$$
Hence, $y=\sqrt{1-x^2}$ represents the upper half of the circle x² + y² = 1 a circle with centre O(0, 0) and radius 1 unit.
y = x represents equation of a straight line passing through O(0, 0)
Point of intersection is obtained by solving two equations

$$\begin{gathered} y=x \\ y=\sqrt{1-x^2} \\ \Rightarrow x=\sqrt{1-x^2} \\ \Rightarrow x^2=1-x^2 \\ \Rightarrow 2x^2=1 \\ \Rightarrow x=\pm \frac{1}{\sqrt{2}} \\ \Rightarrow y=\pm \frac{1}{\sqrt{2}} \\ \mathrm{D}\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\mathrm{and}\mathrm{D}\text{'}\left(\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}}\right)\mathrm{are}\mathrm{two}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{between}\mathrm{the}\mathrm{circle}\mathrm{and}\mathrm{the}\mathrm{straight}\mathrm{line} \\ \mathrm{And}D\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\mathrm{is}\mathrm{the}\mathrm{intersection}\mathrm{point}\mathrm{of}\mathit{}y=\sqrt{1-x^2}\mathrm{and}y=x. \\ \mathrm{Required}\mathrm{area}=\mathrm{Shaded}\mathrm{area}\left(\mathit{O}\mathit{D}\mathit{A}\mathit{E}\mathit{O}\right) \\ =\mathrm{Area}\left(\mathit{O}\mathit{D}\mathit{E}\mathit{O}\right)\hspace{0.17em}+\mathrm{area}\left(\mathit{}\mathit{E}\mathit{D}\mathit{A}\mathit{E}\right).....\left(1\right) \\ \mathrm{Now},\mathrm{area}\left(\mathit{O}\mathit{D}\mathit{E}\mathit{O}\right)={\int }_0^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}xdx \\ ={\left[\frac{x^2}{2}\right]}_0^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.} \\ =\frac{1}{2}{\left(\frac{1}{\sqrt{2}}\right)}^2 \\ =\frac{1}{4}\mathrm{sq}\mathrm{units}.....\left(2\right) \\ \mathrm{Area}\left(\mathit{E}\mathit{D}\mathit{A}\mathit{E}\right)={\int }_{\frac{1}{\sqrt{2}}}^1\sqrt{1-x^2}dx \\ ={\left[\frac{1}{2}x\sqrt{1-x^2}+\times \frac{1}{2}\times {\sin }^{-1}\left(\frac{x}{1}\right)\right]}_{\frac{1}{\sqrt{2}}}^1 \\ =0+\frac{1}{2}{\sin }^{-1}\left(1\right)-\frac{1}{2}\times \frac{1}{\sqrt{2}}\times \sqrt{1-{\left(\frac{1}{\sqrt{2}}\right)}^2}-\frac{1}{2}{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right) \\ =\frac{1}{2}\times \frac{\mathrm{\pi }}{2}-\frac{1}{4}-\frac{1}{2}\times \frac{\mathrm{\pi }}{4}\left\{\mathrm{using},{\sin }^{-1}\left(1\right)=\frac{\mathrm{\pi }}{2}\mathrm{and}{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\mathrm{\pi }}{4}\right\} \\ =\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{8}-\frac{1}{4} \\ =\frac{\mathrm{\pi }}{8}-\frac{1}{4}\mathrm{sq}\mathrm{units}.....\left(3\right) \\ \mathrm{From}\left(1\right),\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ \mathrm{Area}\left(\mathit{O}\mathit{D}\mathit{A}\mathit{E}\mathit{O}\right)=\frac{1}{4}+\frac{\mathrm{\pi }}{8}-\frac{1}{4}=\frac{\mathrm{\pi }}{8}\mathrm{sq}.\mathrm{units} \end{gathered}$$