Calculate the area of the green shaded portions


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Question Number 205161 by York12 last updated on 11/Mar/24

$Calculate the area of the green shaded portions$
Commented by York12 last updated on 11/Mar/24

	Answered by mr W last updated on 12/Mar/24

	Commented by mr W last updated on 13/Mar/24

	${(- \frac{1}{a} + \frac{1}{b} + \frac{1}{r_{n}} + \frac{1}{r_{n + 1}})}^{2} = 2 (\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{r_{n}^{2}} + \frac{1}{r_{n + 1}^{2}})$ ${(- \frac{1}{a} + \frac{1}{b})}^{2} + \frac{1}{r_{n}^{2}} + \frac{1}{r_{n + 1}^{2}} + 2 (- \frac{1}{a} + \frac{1}{b}) (\frac{1}{r_{n}} + \frac{1}{r_{n +}}) + \frac{2}{r_{n} r_{n + 1}} = 2 (\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{r_{n}^{2}} + \frac{1}{r_{n + 1}^{2}})$ $\frac{1}{r_{n + 1}^{2}} - 2 (\frac{1}{b} - \frac{1}{a} + \frac{1}{r_{n}}) \frac{1}{r_{n + 1}} + \frac{1}{r_{n + 1}^{2}} - 2 (\frac{1}{b} - \frac{1}{a}) \frac{1}{r_{n}} + {(\frac{1}{b} - \frac{1}{a})}^{2} + \frac{4}{a b} = 0$ $\frac{1}{r_{n + 1}^{2}} - 2 (\frac{1}{r_{n}} + \frac{1}{b} - \frac{1}{a}) \frac{1}{r_{n + 1}} + {(\frac{1}{r_{n}} - \frac{1}{b} + \frac{1}{a})}^{2} + \frac{4}{a b} = 0$ $\Rightarrow \frac{1}{r_{n + 1}} = \frac{1}{r_{n}} + \frac{1}{b} - \frac{1}{a} + 2 \sqrt{\frac{1}{r_{n}} (\frac{1}{b} - \frac{1}{a}) - \frac{1}{a b}}$ $\Rightarrow \frac{1}{r_{n - 1}} = \frac{1}{r_{n}} + \frac{1}{b} - \frac{1}{a} - 2 \sqrt{\frac{1}{r_{n}} (\frac{1}{b} - \frac{1}{a}) - \frac{1}{a b}}$ $\Rightarrow \frac{1}{r_{n + 1}} + \frac{1}{r_{n - 1}} = 2 (\frac{1}{r_{n}} + \frac{1}{b} - \frac{1}{a})$ $s a y k_{n} = \frac{1}{r_{n}}, α = \frac{1}{a}, β = \frac{1}{b}$ $\Rightarrow k_{n + 1} - 2 k_{n} + k_{n - 1} + 2 (α - β) = 0$ $k_{n} = A + B n + (β - α) n^{2}$ $i n c u r r e n t e x a m p l e :$ $a = 2, b = 1, r_{0} = a - b = 1$ $k_{0} = A = \frac{1}{r_{0}} = \frac{1}{a - b}$ $k_{1} = A + B + (β - α)$ $k_{- 1} = A - B + (β - α) = k_{1}$ $\Rightarrow B = 0$ $\Rightarrow k_{n} = \frac{1}{a - b} + (\frac{1}{b} - \frac{1}{a}) n^{2} = (\frac{a - b}{a b}) [\frac{a b}{{(a - b)}^{2}} + n^{2}]$ $r_{n} = \frac{a b}{(a - b) [\frac{a b}{{(a - b)}^{2}} + n^{2}]} = \frac{a b}{(a - b) (λ^{2} + n^{2})}$ $w i t h λ = \frac{\sqrt{a b}}{a - b}$ $S = π r_{0}^{2} + 2 π \underset{n = 1}{\sum^{\infty}} r_{n}^{2}$ $S = π {(a - b)}^{2} + \frac{2 a^{2} b^{2} π}{{(a - b)}^{2}} \underset{n = 1}{\sum^{\infty}} \frac{1}{{(λ^{2} + n^{2})}^{2}}$ $w e h a v e (s e e Q 205173)$ $\underset{n = 1}{\sum^{\infty}} \frac{1}{{(λ^{2} + n^{2})}^{2}} = \frac{π^{2}}{4 λ^{2} \sinh^{2} (λ π)} + \frac{π}{4 λ^{3} \tanh (λ π)} - \frac{1}{2 λ^{4}}$ $S = π {(a - b)}^{2} + \frac{2 a^{2} b^{2} π}{{(a - b)}^{2}} [\frac{π^{2}}{4 λ^{2} \sinh^{2} (λ π)} + \frac{π}{4 λ^{3} \tanh (λ π)} - \frac{1}{2 λ^{4}}]$ $\Rightarrow S = \frac{(a - b) \sqrt{a b} π^{2}}{2} [\frac{λ π}{\sinh^{2} (λ π)} + \frac{1}{\tanh (λ π)}]$ $w i t h a = 2, b = 1, λ = \sqrt{2}$ $S = \frac{π^{2}}{\sqrt{2}} [\frac{\sqrt{2} π}{\sinh^{2} (\sqrt{2} π)} + \frac{1}{\tanh (\sqrt{2} π)}]$ $= \frac{π^{2} (e^{\sqrt{2} π} + e^{- \sqrt{2} π})}{\sqrt{2} (e^{\sqrt{2} π} - e^{- \sqrt{2} π})} + \frac{4 π^{3}}{e^{2 \sqrt{2} π} + e^{- 2 \sqrt{2} π} - 2}$ $\approx 6.997 958 338$
	Commented by mr W last updated on 12/Mar/24

	Commented by York12 last updated on 12/Mar/24

	$thank you$
	Commented by York12 last updated on 12/Mar/24

	$very good solution$
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