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Question Number 208693 by Frix last updated on 21/Jun/24
Calculate the area enclosed by the curve ((1/x)−2)^2 +((1/y)−2)^2 =1
$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{area}\:\mathrm{enclosed}\:\mathrm{by}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\left(\frac{\mathrm{1}}{{x}}−\mathrm{2}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{y}}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{1} \\ $$
Answered by mr W last updated on 21/Jun/24
(1/x)−2=u ⇒x=(1/(2+u)) (1/y)−2=v ⇒y=(1/(2+v)) u^2 +v^2 =1 u=r cos θ v=r sin θ dA=dxdy=((dudv)/((2+u)^2 (2+v)^2 )) dudv=rdrdθ A=∫_0 ^(2π) ∫_0 ^1 ((rdrdθ)/((2+r cos θ)^2 (2+r sin θ)^2 )) ≈0.296131
$$\frac{\mathrm{1}}{{x}}−\mathrm{2}={u}\:\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}+{u}} \\ $$$$\frac{\mathrm{1}}{{y}}−\mathrm{2}={v}\:\Rightarrow{y}=\frac{\mathrm{1}}{\mathrm{2}+{v}} \\ $$$${u}^{\mathrm{2}} +{v}^{\mathrm{2}} =\mathrm{1} \\ $$$${u}={r}\:\mathrm{cos}\:\theta \\ $$$${v}={r}\:\mathrm{sin}\:\theta \\ $$$${dA}={dxdy}=\frac{{dudv}}{\left(\mathrm{2}+{u}\right)^{\mathrm{2}} \left(\mathrm{2}+{v}\right)^{\mathrm{2}} } \\ $$$${dudv}={rdrd}\theta \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{rdrd}\theta}{\left(\mathrm{2}+{r}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} \left(\mathrm{2}+{r}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\approx\mathrm{0}.\mathrm{296131} \\ $$
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
Yes. I got the exact result following a different path:, but it′s not easy... A=((8(√3))/(147))π
$$\mathrm{Yes}. \\ $$$$\mathrm{I}\:\mathrm{got}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{result}\:\mathrm{following}\:\mathrm{a}\:\mathrm{different} \\ $$$$\mathrm{path}:,\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{easy}… \\ $$$${A}=\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{147}}\pi \\ $$
Answered by mr W last updated on 22/Jun/24
{ ((x=(1/(2+cos t)))),((y=(1/(2+sin t)))) :} dA=((ydx−xdy)/2)=(((yx′−xy′)dt)/2) dA=(1/2)×((2(sin t+cos t)+1)/((2+sin t)^2 (2+cos t)^2 ))×dt A=(1/2)∣∫_0 ^(2π) ((2(sin t+cos t)+1)/((2+sin t)^2 (2+cos t)^2 ))dt∣ =((8π)/(49(√3)))≈0.296131
$$\begin{cases}{{x}=\frac{\mathrm{1}}{\mathrm{2}+\mathrm{cos}\:{t}}}\\{{y}=\frac{\mathrm{1}}{\mathrm{2}+\mathrm{sin}\:{t}}}\end{cases} \\ $$$${dA}=\frac{{ydx}−{xdy}}{\mathrm{2}}=\frac{\left({yx}'−{xy}'\right){dt}}{\mathrm{2}} \\ $$$${dA}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}\left(\mathrm{sin}\:{t}+\mathrm{cos}\:{t}\right)+\mathrm{1}}{\left(\mathrm{2}+\mathrm{sin}\:{t}\right)^{\mathrm{2}} \left(\mathrm{2}+\mathrm{cos}\:{t}\right)^{\mathrm{2}} }×{dt} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\mid\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{2}\left(\mathrm{sin}\:{t}+\mathrm{cos}\:{t}\right)+\mathrm{1}}{\left(\mathrm{2}+\mathrm{sin}\:{t}\right)^{\mathrm{2}} \left(\mathrm{2}+\mathrm{cos}\:{t}\right)^{\mathrm{2}} }{dt}\mid \\ $$$$\:\:\:\:=\frac{\mathrm{8}\pi}{\mathrm{49}\sqrt{\mathrm{3}}}\approx\mathrm{0}.\mathrm{296131} \\ $$

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