Question
Find the area enclosed by the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
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Answer
$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
$ = 4\int_0^a {\frac{b}{a}\sqrt {{a^2} - {x^2}} dx} $
$=\frac{4b}{a} \left[ {\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}(\frac{x}{a})} \right]_0^a$
$=\frac{4b}{a} \left[ {(\frac{a}{2}×0+\frac{a^2}{2}sin^{-1}1)-0} \right]$
$= \frac{4b}{a}\frac{a^2}{2}\frac{π}{2}$
$ = \pi ab$ sq. units
$ = 4\int_0^a {\frac{b}{a}\sqrt {{a^2} - {x^2}} dx} $
$=\frac{4b}{a} \left[ {\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}(\frac{x}{a})} \right]_0^a$
$=\frac{4b}{a} \left[ {(\frac{a}{2}×0+\frac{a^2}{2}sin^{-1}1)-0} \right]$
$= \frac{4b}{a}\frac{a^2}{2}\frac{π}{2}$
$ = \pi ab$ sq. units
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