Question

Find the area enclosed by the curve x = 3cost, y = 2sin t.

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 $$(\pm 3,0)$$ and $$(0,\pm 2)$$

 

∴ Required area = Area of the shaded region
                          = 4 × Area of the region OABO

$$=4\times {\int }_0^3{y}_{\text{ Ellipse }}dx$$

$$=4{\int }_0^3\sqrt{4(1-\frac{x^2}{9})}dx$$
$$=4\times \frac{2}{3}{\int }_0^3\sqrt{9-x^2}dx$$
$$={\frac{8}{3}(\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}{\sin }^{-1}\frac{x}{3})|}_0^3$$
$$=\frac{8}{3}[(0+\frac{9}{2}{\sin }^{-1}1)-(0+0)]$$
$$=\frac{8}{3}\times \frac{9}{2}\times \frac{\pi }{2}$$
$$=6\pi \text{ square units }$$