length of $\sqrt{x} + \sqrt{y} = 1$ ? on $[0,1]$

Question

how to calcuate length of $\sqrt{x} + \sqrt{y} = 1$ ?

I put $x = \cos^4{t} , \ y = \sin^4{t}$ but failed.

Do you have Any good idea form here?

Answer 1

$$ \sqrt{x}+\sqrt{y}=1\implies y'=-\frac{\sqrt{y}}{\sqrt{x}} $$ Substituting $x\mapsto u^2$ and $2u-1\mapsto\tan(\phi)$ gives $$ \begin{align} \int_0^1\sqrt{1+y'^2}\,\mathrm{d}x &=\int_0^1\sqrt{1+\frac yx}\,\mathrm{d}x\\ &=\int_0^1\sqrt{1+\frac{1+x-2\sqrt{x}}x}\,\mathrm{d}x\\ &=2\int_0^1\sqrt{2u^2-2u+1}\,\mathrm{d}u\\ &=\sqrt2\int_0^1\sqrt{\left(2u-1\right)^2+1}\,\mathrm{d}u\\ &=\frac1{\sqrt2}\int_{-\pi/4}^{\pi/4}\sec^3(\phi)\,\mathrm{d}\phi\\ &=\frac1{\sqrt2}\int_{-\pi/4}^{\pi/4}\frac{\mathrm{d}\sin(\phi)}{\left(1-\sin^2(\phi)\right)^2}\\ &=\frac1{4\sqrt2}\int_{-\pi/4}^{\pi/4}\left(\frac{\mathrm{d}\sin(\phi)}{(1-\sin(\phi))^2}+\frac{\mathrm{d}\sin(\phi)}{1-\sin(\phi)}+\frac{\mathrm{d}\sin(\phi)}{(1+\sin(\phi))^2}+\frac{\mathrm{d}\sin(\phi)}{1+\sin(\phi)}\right)\\ &=\frac1{4\sqrt2}\left[\frac{2\sin(\phi)}{\cos^2(\phi)}+2\log\left(\frac{1+\sin(\phi)}{\cos(\phi)}\right)\right]_{-\pi/4}^{\pi/4}\\ &=\frac1{2\sqrt2}\left[\tan(\phi)\sec(\phi)+\log(\tan(\phi)+\sec(\phi))\vphantom{\frac12}\right]_{-\pi/4}^{\pi/4}\\ &=\frac1{2\sqrt2}\left[2\sqrt2+2\log(\sqrt2+1)\right]\\ &=1+\frac{\log(\sqrt2+1)}{\sqrt2} \end{align} $$