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Question Number 208693 by Frix last updated on 21/Jun/24

$$\mathrm{Calculate}\mathrm{the}\mathrm{area}\mathrm{enclosed}\mathrm{by}\mathrm{the}\mathrm{curve}$$$${(\frac{1}{x}-2)}^2+{(\frac{1}{y}-2)}^2=1$$

Answered by mr W last updated on 21/Jun/24

$$\frac{1}{x}-2=u\Rightarrow x=\frac{1}{2+u}$$$$\frac{1}{y}-2=v\Rightarrow y=\frac{1}{2+v}$$$$u^2+v^2=1$$$$u=r\cos \theta$$$$v=r\sin \theta$$$$dA=dxdy=\frac{dudv}{{(2+u)}^2{(2+v)}^2}$$$$dudv=rdrd\theta$$$$A={\int }_0^{2\pi }{\int }_0^1\frac{rdrd\theta }{{(2+r\cos \theta )}^2{(2+r\sin \theta )}^2}$$$$\approx 0.296131$$

Commented by mr W last updated on 21/Jun/24

Commented by mr W last updated on 21/Jun/24

Commented by Frix last updated on 21/Jun/24

$$\mathrm{Yes}.$$$$\mathrm{I}\mathrm{got}\mathrm{the}\mathrm{exact}\mathrm{result}\mathrm{following}\mathrm{a}\mathrm{different}$$$$\mathrm{path}:,\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{easy}...$$$$A=\frac{8\sqrt{3}}{147}\pi$$

Answered by mr W last updated on 22/Jun/24

$$\{\begin{array}{l}x=\frac{1}{2+\cos t}\\ y=\frac{1}{2+\sin t}\end{array}$$$$dA=\frac{ydx-xdy}{2}=\frac{({yx}^{\prime }-{xy}^{\prime })dt}{2}$$$$dA=\frac{1}{2}\times \frac{2(\sin t+\cos t)+1}{{(2+\sin t)}^2{(2+\cos t)}^2}\times dt$$$$A=\frac{1}{2}\mid {\int }_0^{2\pi }\frac{2(\sin t+\cos t)+1}{{(2+\sin t)}^2{(2+\cos t)}^2}dt\mid$$$$=\frac{8\pi }{49\sqrt{3}}\approx 0.296131$$

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