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Question Number 105815 by I want to learn more last updated on 31/Jul/20

Answered by john santu last updated on 01/Aug/20

$$\frac{dx}{d\theta }=1+\sin \theta ;\frac{dy}{d\theta }=\sin \theta$$$$\underset{0}{\stackrel{2\pi }{\int }}\sqrt{\sin {}^2\theta +{(1+\sin \theta )}^2}d\theta$$$$\underset{0}{\stackrel{2\pi }{\int }}\sqrt{2\sin {}^2\theta +2\sin \theta +1}d\theta$$$$\sqrt{2}\underset{0}{\stackrel{2\pi }{\int }}\sqrt{\sin {}^2\theta +\sin \theta +\frac{1}{2}}d\theta$$$$\sqrt{2}\underset{0}{\stackrel{2\pi }{\int }}\sqrt{{(\sin \theta +\frac{1}{2})}^2+\frac{1}{4}}d\theta$$$$$$

Commented by bemath last updated on 01/Aug/20

$$let\sin \theta +\frac{1}{2}=t\Rightarrow dx=\frac{dt}{\cos \theta }$$$$\cos \theta =\sqrt{4-{(2t-1)}^2}=\sqrt{3-4t^2+4t}$$$$I=\sqrt{2}\underset{1/2}{\stackrel{1/2}{\int }}\sqrt{t^2+\frac{1}{4}}\sqrt{3-4t^2+4t}dt=0$$

Commented by I want to learn more last updated on 01/Aug/20

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