Question Number 1985 by 123456 last updated on 27/Oct/15
![f:[a,b]→R ∫_a ^b fdx=∫_a ^b (√(1+((df/dx))^2 ))dx does π∫_a ^b f^2 dx=2π∫_a ^b f(√(1+((df/dx))^2 ))dx ?](https://www.tinkutara.com/question/Q1985.png)
$${f}:\left[{a},{b}\right]\rightarrow\mathbb{R} \\ $$$$\underset{{a}} {\overset{{b}} {\int}}{fdx}=\underset{{a}} {\overset{{b}} {\int}}\sqrt{\mathrm{1}+\left(\frac{{df}}{{dx}}\right)^{\mathrm{2}} }{dx} \\ $$$$\mathrm{does} \\ $$$$\pi\underset{{a}} {\overset{{b}} {\int}}{f}^{\mathrm{2}} {dx}=\mathrm{2}\pi\underset{{a}} {\overset{{b}} {\int}}{f}\sqrt{\mathrm{1}+\left(\frac{{df}}{{dx}}\right)^{\mathrm{2}} }{dx}\:\:? \\ $$
Commented by Yozzi last updated on 28/Oct/15
![∫_a ^b fdx=∫_a ^b (√(1+((df/dx))^2 ))dx (area = arc length for x∈[a,b]) ⇒∫_a ^b (f−(√(1+((df/dx))^2 )))dx=0.....(1) If b≠a, (1) is true iff f−(√(1+((df/dx))^2 ))=0. ⇒f^2 =1+((df/dx))^2 ⇒f^2 −1=((df/dx))^2 ⇒(df/dx)=±(√(f^2 −1)) (∣f∣>1) ⇒∫(df/( (√(f^2 −1))))=±∫dx Let u=cosh^(−1) (f/a) a≠0 coshu=f/a Differentiating wrt f ⇒sinhu×u^′ =1/a u^′ =(1/(asinhu))=(1/(a(√(cosh^2 u−1)))) u^′ =(1/(a(√((f^2 /a^2 )−1))))=(1/(a×(1/a)(√(f^2 −a^2 ))))=(1/( (√(f^2 −a^2 )))) Let a=1, (du/df)=(1/( (√(f^2 −1))))⇒u=∫(df/( (√(f^2 −1)))) ∴ cosh^(−1) f=±x+C f=cosh(C±x) C,x∈R⇒ f:[a,b]→R Let f=cosh(C+x). ∴(df/dx)=sinh(C+x) ⇒ ((df/dx))^2 +1=sinh^2 (x+C)+1=cosh^2 (x+C) ⇒(√(1+((df/dx))^2 ))=∣cosh(x+C)∣ Now, cosh(x+C)≥1>0 ∀x,C∈R. ∴ ∣cosh(x+C)∣=cosh(x+C) ⇒(√(1+((df/dx))^2 ))=cosh(x+C) Let I=2π∫_a ^b f(√(1+((df/dx))^2 ))dx=2π∫_a ^b cosh(x+C)×cosh(x+C)dx I=2π∫_a ^b cosh^2 (x+C)dx Now, let Q=π∫_a ^b f^2 dx=π∫_a ^b cosh^2 (x+C)dx ∴ I=2Q⇒Q≠I if f=cosh(x+C). We deduce I≠Q also if f=cosh(C−x).](https://www.tinkutara.com/question/Q1986.png)
$$\int_{{a}} ^{{b}} {fdx}=\int_{{a}} ^{{b}} \sqrt{\mathrm{1}+\left(\frac{{df}}{{dx}}\right)^{\mathrm{2}} }{dx} \\ $$$$\left({area}\:=\:{arc}\:{length}\:{for}\:{x}\in\left[{a},{b}\right]\right) \\ $$$$\Rightarrow\underset{{a}} {\overset{{b}} {\int}}\left({f}−\sqrt{\mathrm{1}+\left(\frac{{df}}{{dx}}\right)^{\mathrm{2}} }\right){dx}=\mathrm{0}…..\left(\mathrm{1}\right) \\ $$$${If}\:{b}\neq{a},\:\left(\mathrm{1}\right)\:{is}\:{true}\:{iff}\: \\ $$$$\:\:\:\:\:{f}−\sqrt{\mathrm{1}+\left(\frac{{df}}{{dx}}\right)^{\mathrm{2}} }=\mathrm{0}. \\ $$$$\Rightarrow{f}^{\mathrm{2}} =\mathrm{1}+\left(\frac{{df}}{{dx}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{f}^{\mathrm{2}} −\mathrm{1}=\left(\frac{{df}}{{dx}}\right)^{\mathrm{2}} \: \\ $$$$\Rightarrow\frac{{df}}{{dx}}=\pm\sqrt{{f}^{\mathrm{2}} −\mathrm{1}}\:\:\:\left(\mid{f}\mid>\mathrm{1}\right) \\ $$$$\Rightarrow\int\frac{{df}}{\:\sqrt{{f}^{\mathrm{2}} −\mathrm{1}}}=\pm\int{dx} \\ $$$${Let}\:{u}={cosh}^{−\mathrm{1}} \left({f}/{a}\right)\:\:\:{a}\neq\mathrm{0} \\ $$$${coshu}={f}/{a} \\ $$$${Differentiating}\:{wrt}\:{f} \\ $$$$\Rightarrow{sinhu}×{u}^{'} =\mathrm{1}/{a} \\ $$$${u}^{'} =\frac{\mathrm{1}}{{asinhu}}=\frac{\mathrm{1}}{{a}\sqrt{{cosh}^{\mathrm{2}} {u}−\mathrm{1}}} \\ $$$${u}^{'} =\frac{\mathrm{1}}{{a}\sqrt{\frac{{f}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\mathrm{1}}}=\frac{\mathrm{1}}{{a}×\frac{\mathrm{1}}{{a}}\sqrt{{f}^{\mathrm{2}} −{a}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\:\sqrt{{f}^{\mathrm{2}} −{a}^{\mathrm{2}} }} \\ $$$${Let}\:{a}=\mathrm{1},\:\frac{{du}}{{df}}=\frac{\mathrm{1}}{\:\sqrt{{f}^{\mathrm{2}} −\mathrm{1}}}\Rightarrow{u}=\int\frac{{df}}{\:\sqrt{{f}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\therefore\:{cosh}^{−\mathrm{1}} {f}=\pm{x}+{C} \\ $$$${f}={cosh}\left({C}\pm{x}\right)\:{C},{x}\in\mathbb{R}\Rightarrow\:{f}:\left[{a},{b}\right]\rightarrow\mathbb{R} \\ $$$$ \\ $$$${Let}\:{f}={cosh}\left({C}+{x}\right).\:\therefore\frac{{df}}{{dx}}={sinh}\left({C}+{x}\right) \\ $$$$\Rightarrow\:\left(\frac{{df}}{{dx}}\overset{\mathrm{2}} {\right)}+\mathrm{1}={sinh}^{\mathrm{2}} \left({x}+{C}\right)+\mathrm{1}={cosh}^{\mathrm{2}} \left({x}+{C}\right) \\ $$$$\Rightarrow\sqrt{\mathrm{1}+\left(\frac{{df}}{{dx}}\right)^{\mathrm{2}} }=\mid{cosh}\left({x}+{C}\right)\mid \\ $$$${Now},\:{cosh}\left({x}+{C}\right)\geqslant\mathrm{1}>\mathrm{0}\:\forall{x},{C}\in\mathbb{R}. \\ $$$$\therefore\:\mid{cosh}\left({x}+{C}\right)\mid={cosh}\left({x}+{C}\right) \\ $$$$\Rightarrow\sqrt{\mathrm{1}+\left(\frac{{df}}{{dx}}\right)^{\mathrm{2}} \:}={cosh}\left({x}+{C}\right) \\ $$$${Let}\:{I}=\mathrm{2}\pi\int_{{a}} ^{{b}} {f}\sqrt{\mathrm{1}+\left(\frac{{df}}{{dx}}\right)^{\mathrm{2}} }{dx}=\mathrm{2}\pi\int_{{a}} ^{{b}} {cosh}\left({x}+{C}\right)×{cosh}\left({x}+{C}\right){dx} \\ $$$${I}=\mathrm{2}\pi\underset{{a}} {\overset{{b}} {\int}}{cosh}^{\mathrm{2}} \left({x}+{C}\right){dx} \\ $$$${Now},\:{let}\:{Q}=\pi\int_{{a}} ^{{b}} {f}^{\mathrm{2}} {dx}=\pi\int_{{a}} ^{{b}} {cosh}^{\mathrm{2}} \left({x}+{C}\right){dx} \\ $$$$\therefore\:{I}=\mathrm{2}{Q}\Rightarrow{Q}\neq{I}\:{if}\:{f}={cosh}\left({x}+{C}\right). \\ $$$${We}\:{deduce}\:{I}\neq{Q}\:{also}\:{if}\:{f}={cosh}\left({C}−{x}\right). \\ $$