Find the area enclosed by the curve `x=3cos t ,y=2sint`

Explanation

Given `x=3 cos t quad y=2 sin t`


`Rightarrow cos {t}=frac{{x}}{3}` and `sin {t}=frac{{y}}{2}`


Now `cos ^{2} {t}+sin ^{2}=1`


`Rightarrow frac{x^{2}}{a}+frac{y^{2}}{4}=1 quad` (1) `[.` From `frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}=1` ]


`Rightarrow` Area `=4` Area of `{OAB}`


`=4 int_{0}^{3} y d x`


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


`therefore` Area `=4 int_{0}^{3} frac{2}{3} sqrt{9-{x}^{2}} {dx}`


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