Question

Find the area enclosed by the curve x = 3cost, y = 2sint. [NCERT EXEMPLAR]

Answer 1

The given curve x = 3cost, y = 2sint represents the parametric equation of the ellipse.

Eliminating the parameter t, we get

$\frac{x^2}{9}+\frac{y^2}{4}={\cos }^{2}t+{\sin }^{2}t=1$

This represents the Cartesian equation of the ellipse with centre (0, 0). The coordinates of the vertices are $\left(\pm 3,0\right)$ and $\left(0,\pm 2\right)$.



∴ Required area = Area of the shaded region

= 4 × Area of the region OABO

$$\begin{gathered} =4\times {\int }_0^3{y}_{\mathrm{Ellipse}}dx \\ =4{\int }_0^3\sqrt{4\left(1-\frac{x^2}{9}\right)}dx \\ =4\times \frac{2}{3}{\int }_0^3\sqrt{9-x^2}dx \\ =\frac{8}{3}{\overline{)\left(\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}{\sin }^{-1}\frac{x}{3}\right)}}_0^3 \\ =\frac{8}{3}\left[\left(0+\frac{9}{2}{\sin }^{-1}1\right)-\left(0+0\right)\right] \\ =\frac{8}{3}\times \frac{9}{2}\times \frac{\mathrm{\pi }}{2} \\ =6\mathrm{\pi }\mathrm{square}\mathrm{units} \end{gathered}$$