Find the area enclosed by the ellipse frac{x^ | vedclass

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(A) From the figure,the area of the region $ABA'B'A$ bounded by the ellipse is given by:Area $= 4 	imes (	ext{Area of the region } AOBA 	ext{ in th

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$\pi a b$

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(A) From the figure,the area of the region $ABA'B'A$ bounded by the ellipse is given by:Area $= 4 	imes (	ext{Area of the region } AOBA 	ext{ in the first quadrant bounded by the curve, } x	ext{-axis and the ordinates } x = 0, x = a)$(as the ellipse is symmetrical about both $x$-axis and $y$-axis)Area $= 4 \int_{0}^{a} y \, dx$ (taking vertical strips)Now,$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ gives $y = \pm \frac{b}{a} \sqrt{a^{2}-x^{2}}$. Since the region $AOBA$ lies in the first quadrant,$y$ is taken as positive.So,the required area is:Area $= 4 \int_{0}^{a} \frac{b}{a} \sqrt{a^{2}-x^{2}} \, dx$$= \frac{4b}{a} \left[ \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left( \frac{x}{a} ight) ight]_{0}^{a}$$= \frac{4b}{a} \left[ \left( \frac{a}{2} 	imes 0 + \frac{a^{2}}{2} \sin^{-1}(1) ight) - (0 + 0) ight]$$= \frac{4b}{a} \left( \frac{a^{2}}{2} 	imes \frac{\pi}{2} ight) = \pi ab$.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

Find the area enclosed by the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.

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