Calculate-the-area-enclosed-by-the-curve-1-x-2-2-1-y-2-2-1-

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

Calculate the area enclosed by the 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 θ

v = r \sin θ

d A = d x d y = \frac{d u d v}{{(2 + u)}^{2} {(2 + v)}^{2}}

d u d v = r d r d θ

A = \int_{0}^{2 π} \int_{0}^{1} \frac{r d r d θ}{{(2 + r \cos θ)}^{2} {(2 + r \sin θ)}^{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

Yes .

I got the exact result following a different

path :, but {it}^{'} s not easy \dots

A = \frac{8 \sqrt{3}}{147} π

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

{\begin{cases} x = \frac{1}{2 + \cos t} \\ y = \frac{1}{2 + \sin t} \end{cases}

d A = \frac{y d x - x d y}{2} = \frac{({y x}^{'} - {x y}^{'}) d t}{2}

d A = \frac{1}{2} \times \frac{2 (\sin t + \cos t) + 1}{{(2 + \sin t)}^{2} {(2 + \cos t)}^{2}} \times d t

A = \frac{1}{2} ∣ \int_{0}^{2 π} \frac{2 (\sin t + \cos t) + 1}{{(2 + \sin t)}^{2} {(2 + \cos t)}^{2}} d t ∣

= \frac{8 π}{49 \sqrt{3}} \approx 0.296131

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