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Question Number 12577 by tawa last updated on 26/Apr/17

$$\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{generated}\mathrm{when}\mathrm{the}\mathrm{curve}\mathrm{x}=\mathrm{a}(\theta -\sin \theta ),(1-\cos \theta )$$$$\theta =0,\theta =\pi \mathrm{rotates}\mathrm{about}\mathrm{x}-\mathrm{axis}\mathrm{through}2\pi \mathrm{radian}.$$$$\mathrm{Note}:1-\cos \theta =2{\sin }^2\left(\frac{\theta }{2}\right)$$

Answered by mrW1 last updated on 26/Apr/17

$$x=a(\theta -\sin \theta )$$$$y=1-\cos \theta$$$$A=2\pi {\int }_{x_1}^{x_2}ydx=2\pi a\int (1-\cos \theta )(1-\cos \theta )d\theta$$$$=2\pi a\int {4\sin }^4\frac{\theta }{2}d\theta$$$$=16\pi a{\int }_0^{\pi }{\sin }^4\frac{\theta }{2}d\frac{\theta }{2}$$$$=16\pi a{\int }_0^{\frac{\pi }{2}}{\sin }^{4}tdt$$$$=16\pi a{\left[\frac{\sin 4t-8\sin 2t+12t}{32}\right]}_0^{\frac{\pi }{2}}$$$$=16\pi a\frac{12}{32}\times \frac{\pi }{2}=3a{\pi }^2$$

Commented by tawa last updated on 26/Apr/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by ajfour last updated on 26/Apr/17

$$Areagenerated=2\pi {\int }_0^{x_1}ydx$$$$A=2\pi {\int }_0^{\pi }a{(1-\cos \theta )}^{2}d\theta ...\left(i\right)$$$$=2\pi {\underset{0}{\int }}^{\pi }a{(1+\cos \theta )}^{2}d\theta ...\left(ii\right)$$$$\left(i\right)+\left(ii\right)gives:$$$$2A=2\pi a{\int }_0^{\pi }(2+2\cos {}^2\theta )d\theta$$$$A=\pi a{\int }_0^{\pi }(3+\cos 2\theta )d\theta$$$$=\pi a{[3\theta +\frac{\sin 2\theta }{2}]}_0^{\pi }$$$$A=3{\mathit{\pi }}^2\mathit{a}.$$

Commented by tawa last updated on 26/Apr/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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