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Question Number 118073 by bemath last updated on 15/Oct/20

$$\mathrm{Evaluate}\underset{0}{\stackrel{a}{\int }}b\sqrt{1-\frac{{\mathrm{x}}^2}{a^2}}dx$$

Answered by john santu last updated on 15/Oct/20

$$I=\underset{0}{\stackrel{a}{\int }}\frac{b}{a}\sqrt{a^2-x^2}dx;theintegralrepresentsthearea$$$$ofaquartercirclehavingradiusa$$$$I=\frac{b}{a}\times \frac{1}{4}\pi a^2=\frac{\pi ab}{4}$$

Commented by bemath last updated on 15/Oct/20

$$\mathrm{gave}\mathrm{kudos}$$

Commented by MJS_new last updated on 15/Oct/20

$$\mathrm{right}\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{a}\mathrm{circle}\mathrm{but}\mathrm{an}\mathrm{ellipse}$$

Commented by john santu last updated on 16/Oct/20

$$nosir.itsacircle.Ify=\sqrt{a^2-x^2}$$$$\Rightarrow y^2+x^2=a^2$$

Commented by MJS_new last updated on 16/Oct/20

$$\mathrm{yes}.\mathrm{but}y=\frac{b}{a}\sqrt{a^2-x^2}\mathrm{is}\mathrm{an}\mathrm{ellipse}$$

Answered by 1549442205PVT last updated on 15/Oct/20

$$\mathrm{Evaluate}\underset{0}{\stackrel{a}{\int }}b\sqrt{1-\frac{{\mathrm{x}}^2}{a^2}}dx$$$$\mathrm{Put}\mathrm{x}=\mathrm{acos}\phi (\phi \in [0,\pi ])\Rightarrow \mathrm{dx}=-\mathrm{asin}\phi \mathrm{d}\phi$$$$\mathrm{I}=-{\int }_{\pi /2}^0{\mathrm{absin}}^2\phi \mathrm{d}\phi ={\int }_0^{\pi /2}\mathrm{ab}\frac{1-\cos 2\phi }{2}\mathrm{d}\phi$$$$=(\frac{\mathrm{ab}\phi }{2}-\frac{\mathrm{ab}}{4}\sin 2\phi ){\mid }_0^{\pi /2}=\frac{\mathrm{ab}\pi }{4}$$

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