<s>[INST] Arc length of an implicit function Find the arc length of $$y=\frac{x^4}{8} + \frac{1}{4y^2}$$ for the region between $~x=1~$ to $~x=2~$.  
I have tried implicitly deriving it to find $~\frac{dy}{dx}~$, but since $~\frac{dy}{dx}~$ is dependent on two variables, you cant integrate through the normal definition of $$\int_1^2 \sqrt{1+(dy/dx)^2}dx$$
 [/INST] Why not to look at $x$ as a function of $y$ and then integrate along $y$ ?
For the range of interest, we have
$$x=\frac{\left(8 y^3-2\right)^{1/4}}{\sqrt{y}}$$
$$x'=\frac{2 y^3+1}{y^{3/2} \left(8 y^3-2\right)^{3/4}}$$ and we face the problem of
$$L=\int_a^b \sqrt{1+\frac{\left(2 y^3+1\right)^2}{y^3 \left(8 y^3-2\right)^{3/2}}}\,dy$$ $a$ being the solution of $y=\frac{1}{8} + \frac{1}{4y^2}$ and  $b$ the solution of $y=2 + \frac{1}{4y^2}$ that is to say
$$a=\frac{1}{24} \left(1+\sqrt[3]{1729-24 \sqrt{5190}}+\sqrt[3]{1729+24
   \sqrt{5190}}\right)\approx 0.674504$$
$$b=\frac{1}{12} \left(8+\sqrt[3]{728-24 \sqrt{465}}+2 \sqrt[3]{91+3 \sqrt{465}}\right)\approx 2.05897$$ and now numerical integration.
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