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 \[\left( \pm 3, 0 \right)\] and \[\left( 0, \pm 2 \right)\]

 

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

\[= 4 \times \int_0^3 y_{\text{ Ellipse }} dx\]

\[ = 4 \int_0^3 \sqrt{4\left( 1 - \frac{x^2}{9} \right)}dx\]
\[ = 4 \times \frac{2}{3} \int_0^3 \sqrt{9 - x^2}dx\]
\[ = \left.\frac{8}{3} \left( \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2} \sin^{- 1} \frac{x}{3} \right)\right|_0^3 \]
\[ = \frac{8}{3}\left[ \left( 0 + \frac{9}{2} \sin^{- 1} 1 \right) - \left( 0 + 0 \right) \right]\]
\[ = \frac{8}{3} \times \frac{9}{2} \times \frac{\pi}{2}\]
\[ = 6\pi \text{ square units }\]