Question

Draw a rough sketch of the graph of the function y = 2$\sqrt{1-x^2}$, x ∈ [0, 1] and evaluate the area enclosed between the curve and the x-axis.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ y=2\sqrt{1-x^2} \\ \Rightarrow \frac{y}{2}=\sqrt{1-x^2} \\ \Rightarrow \frac{y^2}{4}=1-x^2 \\ \Rightarrow x^2+\frac{y^2}{4}=1 \\ \Rightarrow \frac{x^2}{1}+\frac{y^2}{4}=1 \\ \mathrm{Since}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{equation}\frac{x^2}{1}+\frac{y^2}{4}=1,\mathrm{all}\mathrm{the}\mathrm{powers}\mathrm{of}\mathrm{both}x\mathrm{and}y\mathrm{are}\mathrm{even},\mathrm{the}\mathrm{curve}\mathrm{is}\mathrm{symmetrical}\mathrm{about}\mathrm{both}\mathrm{the}\mathrm{axes}. \\ \therefore \mathrm{Required}\mathrm{area}=\mathrm{area}\mathrm{enclosed}\mathrm{by}\mathrm{ellipse}\mathrm{and}\mathit{}x-\mathrm{axis}\mathrm{in}\mathrm{first}\mathrm{quadrant} \\ (1,0),(-1,0)\mathrm{are}\mathrm{the}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{curve}\mathrm{and}x-\mathrm{axis} \\ (0,2),(0,-2)\mathrm{are}\mathrm{the}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{curve}\mathrm{and}y-\mathrm{axis} \\ \mathrm{Slicing}\mathrm{the}\mathrm{area}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant}\mathrm{into}\mathrm{vertical}\mathrm{stripes}\mathrm{of}\mathrm{height}=\left|y\right|\mathrm{and}\mathrm{width}=dx \\ \therefore \mathrm{Area}\mathrm{of}\mathrm{approximating}\mathrm{rectangle}=\left|y\right|dx \\ \mathrm{Approximating}\mathrm{rectangle}\mathrm{can}\mathrm{move}\mathrm{between}x=0\mathrm{and}\mathit{}x=1 \\ A=\mathrm{Area}\mathrm{of}\mathrm{enclosed}\mathrm{curve}\mathrm{above}x-\mathrm{axis}={\int }_0^1\left|y\right|dx \\ \Rightarrow A={\int }_0^{1}2\sqrt{1-x^2}dx \\ =2{\int }_0^1\sqrt{1-x^2}dx \\ =2{\left[\frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x\right]}_0^1 \\ =2\left\{\frac{1}{2}{\sin }^{-1}1\right\} \\ =2\left\{\frac{1}{2}\left(\frac{\pi }{2}-0\right)\right\} \\ \Rightarrow A=\frac{\pi }{2}\mathrm{sq}.\mathrm{units} \end{gathered}$$