Question

Sketch the region {(x, y) : 9x² + 4y² = 36} and find the area of the region enclosed by it, using integration.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ 9x^2+4y^2=36.....\left(1\right) \\ \Rightarrow 4y^2=36-9x^2 \\ \Rightarrow y^2=\frac{9}{4}\left(4-x^2\right) \\ \Rightarrow y=\frac{3}{2}\sqrt{\left(4-x^2\right)}.....\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{we}\mathrm{get} \\ \Rightarrow \frac{x^2}{4}+\frac{y^2}{9}=1 \\ \mathrm{Since}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{equation}\frac{x^2}{4}+\frac{y^2}{9}=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{Area}\mathrm{enclosed}\mathrm{by}\mathrm{the}\mathrm{curve}\mathrm{and}\mathrm{above}\mathrm{x}\mathrm{axis}=4\times \mathrm{area}\mathrm{enclosed}\mathrm{by}\mathrm{ellipse}\mathrm{and}\mathit{}x-\mathrm{axis}\mathrm{in}\mathrm{first}\mathrm{quadrant} \\ (2,0),(-2,0)\mathrm{are}\mathrm{the}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{curve}\mathrm{and}x-\mathrm{axis} \\ (0,3),(0,-3)\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}x=2 \\ A=\mathrm{Area}\mathrm{of}\mathrm{enclosed}\mathrm{curve}=4{\int }_0^2\left|y\right|dx \\ \Rightarrow A=4{\int }_0^2\frac{3}{2}\sqrt{4-x^2}dx\left[\mathrm{From}\left(2\right)\right] \\ =4\times \frac{3}{2}{\int }_0^2\sqrt{4-x^2}dx \\ =6{\int }_0^2\sqrt{2^2-x^2}dx \\ =6{\left[\frac{x}{2}\sqrt{2^2-x^2}+\frac{1}{2}{2}^2{\sin }^{-1}\frac{x}{2}\right]}_0^2 \\ =6\left\{0+\frac{1}{2}4{\sin }^{-1}1\right\} \\ =6\left\{\frac{1}{2}\times 4\left(\frac{\pi }{2}\right)\right\} \\ \Rightarrow A=6\pi \mathrm{sq}\mathrm{units} \end{gathered}$$