Question

Find the area enclosed by the curve $x=3\cos t,y=2\sin t$.

Answer 1

R.E.F image

Solution :-

Given $x=3cost$ $y=2sint$

$\Rightarrow cost=\frac{x}{3}$ and $sint=\frac{y}{2}$

Now $co{s}^{2}t+si{n}^2=1$

$\Rightarrow \frac{x^2}{a}+\frac{y^2}{4}=1$___(1) [From $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1]$

we have to find area of shaded portion

$\Rightarrow$ Area = 4 Area of OAB

$=4{\int }_0^{3}ydx$

From (i) $y^2=\frac{4}{9}(9-x^2)$

$\therefore$ Area $=4{\int }_0^3\frac{2}{3}\sqrt{9-x^2}dx$

$=\frac{8}{3}{\int }_0^3\sqrt{9-x^2}dx$

$=\frac{8}{3}{\int }_0^3\sqrt{3^2-x^2}dx$

$=\frac{8}{3}{|\frac{x}{2}\sqrt{3^2-x^2}+\frac{9}{2}si{n}^{-1}\frac{x}{3}|}_0^3$

$=\frac{8}{3}\left[\frac{9}{2}si{n}^{-1}\right(sin\frac{\pi }{2}\left)\right]$

$=6\pi$ squnits