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Question Number 23283 by ajfour last updated on 28/Oct/17

$$Computetheareaofaloopof$$$$thecurve\mathit{\rho }=\mathit{a}\sin 2\mathit{\theta };andeven$$$$sketchthecurve,please.$$

Answered by mrW1 last updated on 28/Oct/17

$$\mathrm{a}\mathrm{small}\mathrm{loop}\mathrm{for}\theta \mathrm{from}0\mathrm{to}\frac{\pi }{2}:$$$${\mathrm{A}}_1=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}{\rho }^2\mathrm{d}\theta$$$$=\frac{{\mathrm{a}}^2}{2}{\int }_0^{\frac{\pi }{2}}{\sin }^{2}2\theta \mathrm{d}\theta$$$$=\frac{{\mathrm{a}}^2}{4}{\int }_0^{\frac{\pi }{2}}(1-\cos 4\theta )\mathrm{d}\theta$$$$=\frac{{\mathrm{a}}^2}{4}{[\theta -\frac{\sin 4\theta }{4}]}_0^{\frac{\pi }{2}}$$$$=\frac{\pi {\mathrm{a}}^2}{8}$$$$\mathrm{for}\mathrm{a}\mathrm{total}\mathrm{loop}\mathrm{from}0\mathrm{to}2\pi ,$$$$\mathrm{A}=4\times \frac{\pi {\mathrm{a}}^2}{8}=\frac{\pi {\mathrm{a}}^2}{2}$$

Commented by mrW1 last updated on 28/Oct/17

Commented by ajfour last updated on 28/Oct/17

$$thankyouverymuchSir,right$$$$answer,nicequestion;ithought$$$$ofsharingitwithyou.$$$$i^{\prime }vecreatedacaromboardquestion,$$$$pleaseviewit,sir.$$

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