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So I want to find the length of the function $y = a - 2\sqrt{ax} + x$ in the interval $(0, a)$, assuming $a > 0$.

I found the derivative: $ y' = 1 - \frac {\sqrt{a}}{\sqrt{x}} $

Then using the formula $$l = \int_{0}^{a}\sqrt{1+(y'(x))^2} dx$$ I get the integral: $$l =\int_{0}^{a}\sqrt{2-2\sqrt{\frac{a}{x}}+\frac{a}{x}} dx$$

Which is improper, because at the lower bound $0$ $\sqrt{\frac{a}{x}}$ and $\frac{a}{x}$ approach infinity.

$$ l = \int_{0}^{a}\sqrt{2x-2\sqrt{ax}+a}*x^{-1/2} dx = \frac{1}{\sqrt{a}}\int_{0}^{a}\sqrt{2x-2\sqrt{ax}+a} d2\sqrt{ax}+a$$ With substitution as suggested by the comments: $y=\sqrt{\frac{a}{x}}$, $dx=-\frac{a}{y^2}dy$: (I think I made a mistake, because this one doesn't converge) $$l = -a*\int_{\infty}^{1}\sqrt{2-2y+y^2}*\frac{1}{y^2}dy=a*\int_{1}^{\infty}\sqrt{2-2y+y^2}*\frac{1}{y^2}dy$$

How can I solve this? And is $l = \sqrt{2}a$, because I found that as an answer to a similar question but without integration?

Average_C_Enjoyer
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1 Answers1

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Substitute $t=\sqrt{\frac xa}$ \begin{align} l =&\int_{0}^{a}\sqrt{2-2\sqrt{\frac{a}{x}}+\frac{a}{x}} \ {dx} =2a \int_0^1 \sqrt{2t^2-2t+1}\ dt\\ =& \ a \int_0^1 \frac{\sqrt{2t^2-2t+1}}{t-\frac12}\ d\bigg[\left(t-\frac12\right)^2\ \bigg]\\ \overset{ibp}= &\ a+\frac a2\int_0^1 \frac1{\sqrt{2t^2-2t+1}}\ dt = a+ \frac a{\sqrt2}\sinh^{-1}(1) \end{align}

Quanto
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  • How do you go from line 2 to 3? Don't you just multiply by $\frac{2t-1}{2t-1}$ and where does the 2 in front of the integral go? I think on the 4th line you fixed it since it's supposed to be 1/2 instead of 1 [2a(1/2+1/4...)] instead of [a(1 +1/4...)]. – Average_C_Enjoyer Mar 31 '24 at 16:21
  • @Average_C_Enjoyer - u r right. just fixed it – Quanto Mar 31 '24 at 19:06