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The region $D$ is formed by the area under a curve of length $\ln(1+\sqrt{2})$ that connected points $(\ln(1+\sqrt{2}), \sqrt{2})$ and $(\sqrt{2}, 2)$. What is the maximum volume of the solid of revolution obtained by rotating the region $D$ around the $x$-axis?

$$ \begin{align*} & I(f)=\int\limits_{\ln(1+\sqrt{2})}^{\sqrt{2}}\pi f^2\ dx+\lambda\left(\ \int\limits_{\ln(1+\sqrt{2})}^{\sqrt{2}}\sqrt{1+f'^2}\ dx-\ln(1+\sqrt{2})\right)\\\\ & (1+f'^2)(f^2-c^2)=\lambda^2\\\\ & ds=\frac{c_2}{c_1-f^2}\ dx\\ \end{align*} $$ I'm stuck on solving this differential equation. I'm not sure if it can be solved with parametric equations or some "trick".

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  • Here is a related question and answers. The solution involves elliptic curves, and might not have a closed form. I attempted using the Euler-Lagrange equation on functional $$F(x,f,f')=\pi f^2 +\lambda\sqrt{1+f'^2},,$$ which yields $$2\pi f-\frac{\partial}{\partial x}\left(\frac{\lambda f'}{\sqrt{1+f'^2}}\right)=0$$ but did not go further. – cjferes Sep 07 '23 at 18:55

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