solve $L=\int_0^4 \sqrt{1+\frac{9}{4}y^4+\frac{9}{2}y^2}dy$

Question

I need help with this excercise.

Find the arc length of the function $$x=\frac{1}{2}(y^2+2)^{3/2}$$ from $y=0$ to $y=4$.

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$$L=\int_a^b \sqrt{1+(\frac{dx}{dy})^2}dy$$

Now, $$x=\frac{1}{2}(y^2+2)^{3/2}$$ $$\frac{dx}{dy}=\frac{3}{2}y\sqrt{y^2+2}$$

Then,

$$L=\int_0^4 \sqrt{1+(\frac{3}{2}y\sqrt{y^2+2})^2}dy$$ $$L=\int_0^4 \sqrt{1+\frac{9}{4}y^4+\frac{9}{2}y^2}dy$$

How to solve this integral?

Answer 1

We can instead take the parametrization

$y=\sqrt{2}\tan(a), x=\sqrt{2}\sec^3({a})$

$y'^2=2\sec^4(a), x'^2=18\sec^6(a)\tan^2(a)$

We then need to compute

$\int_{0}^{\arctan(2\sqrt{2})} \sqrt{18\sec^6(a)\tan^2(a)+2\sec^4(a)}da$

Making the substiution $u=\tan{a}$ transfomrs the integral to

$\int_{0}^{2\sqrt{2}} \sqrt{18u^2+20}du$

Now take $u=\sqrt{\frac{20}{18}}\sinh{v}$

then we're done hope you can continue from here