Find the area of the smaller region bounded b | vedclass

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(A) The area of the smaller region bounded by the ellipse,$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$ and the line,$\frac{x}{a}+\frac{y}{b}=1,$ is re

Vedclass · Accepted Answer

$\frac{ab}{4}(\pi-2)$

Vedclass · Answer

(A) The area of the smaller region bounded by the ellipse,$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$ and the line,$\frac{x}{a}+\frac{y}{b}=1,$ is represented by the shaded region $BCAB$ as:Area $BCAB = 	ext{Area}(OBCAO) - 	ext{Area}(OBAO)$$= \int_{0}^{a} b \sqrt{1-\frac{x^{2}}{a^{2}}} \, dx - \int_{0}^{a} b\left(1-\frac{x}{a}ight) \, dx$$= \frac{b}{a} \int_{0}^{a} \sqrt{a^{2}-x^{2}} \, dx - \frac{b}{a} \int_{0}^{a}(a-x) \, dx$$= \frac{b}{a} \left[ \left\{ \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \frac{x}{a} ight\}_{0}^{a} - \left\{ ax - \frac{x^{2}}{2} ight\}_{0}^{a} ight]$$= \frac{b}{a} \left[ \left\{ \frac{a^{2}}{2} \left(\frac{\pi}{2}ight) ight\} - \left\{ a^{2} - \frac{a^{2}}{2} ight\} ight]$$= \frac{b}{a} \left[ \frac{a^{2}\pi}{4} - \frac{a^{2}}{2} ight]$$= \frac{ba^{2}}{2a} \left[ \frac{\pi}{2} - 1 ight]$$= \frac{ab}{2} \left[ \frac{\pi-2}{2} ight] = \frac{ab}{4}(\pi-2) 	ext{ sq. units.}$

Find the area of the smaller region bounded by the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and the line $\frac{x}{a}+\frac{y}{b}=1$.

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