Arc Length of $x^{\frac{2}{3}}+y^{\frac23}=1$ How do I find the arch length of 
$$x^{\frac{2}{3}}+y^{\frac23}=1$$
The hint given was "4x the arc in first quadrant"
I think I am supposed to use the formula: 
$$L=\int^b_a \sqrt{1+(f'(x))^2} dx$$

I tried plotting the equation in a graphing utility like https://www.desmos.com/calculator, which results in an error. 
So I tried expanding the term : 
$$y=\sqrt{(1-x^{\frac{2}{3}})^3} = \sqrt{1-2x^{\frac{2}{3}}-x^2}$$
Then how do I proceed? Complete the square? Doesn't look likes its in an appropriate form? 
 A: Note that the $x^{2/3}+y^{2/3}=1$ can be parametrized by $x(t)=\cos^3 t$ and $y(t)=\sin^3 t$, where $0\leq t\leq 2\pi$. Now you can apply the formula (see here)
$$s=\int_0^{2\pi}\sqrt{\left(\frac{dx(t)}{dt}\right)^2+\left(\frac{dy(t)}{dt}\right)^2}dt$$
to find the arclength.
Note added: This is how you can finish the problem: since $x(t)=\cos^3 t$ and $y(t)=\sin^3 t$, we have $\displaystyle\frac{dx(t)}{dt}=-3\cos^2t \sin t$ and $\displaystyle\frac{dy(t)}{dt}=3\sin^2t \cos t$. Substitute it into above formula, we have 
$$s=\int_0^{2\pi}\sqrt{9\cos^2 t\sin^2 t}dt=4\int_0^{\pi/2}3\cos t\sin tdt=6\int_0^{\pi/2}\sin (2t)dt=-3\cos 2t\Big|_0^{\pi/2}=6.$$
A: The error you found is probably due to trying to take a negative number to the $2/3$rds power. If we assume that $x^{2/3} := (x^{2})^{1/3}$ then the solution set of $x^{2/3}+y^{2/3}=1$ is symmetric about the $x$ axis (since if $(x,y)$ is an element of the solution set, then so to is $(x,-y)$) and symmetric about the $y$-axis (since if $(x,y)$ is an element of the solution set, then so too is $(-x,y)$.) These symmetries imply the hint. 
So if $L$ is the arclength of the entire solution set, then
\begin{aligned}
L=4\int_{0}^{1} \sqrt{1+f'(x)^{2}} \, dx
\end{aligned}
From here, it's simply aggravation and annoyance:
$$
y^{2/3}=1-x^{2/3} \Longrightarrow y=f(x)=(1-x^{2/3})^{3/2}
$$
Then we compute the derivative:
$$
f'(x) = \frac{3}{2}(1-x^{2/3})^{1/2}\cdot\left(-\frac{2}{3}x^{-1/3}\right)
=-\frac{\sqrt{1-x^{2/3}}}{x^{1/3}}
$$
and its square
$$
f'(x)^{2} = \frac{1-x^{2/3}}{x^{2/3}}=-1+x^{-2/3}
$$
so that our integrand is
$$
\sqrt{1+f'(x)^{2}}= x^{-1/3}
$$
Then $4\int_{0}^{1}x^{-1/3} \, dx = 6$.
A: $$(y')^2=\frac{y^{2/3}}{x^{2/3}} ~\text {and }~ y^{2/3}=1-x^{2/3}$$
Hence :
$$L=\int\limits_{a}^{b} \sqrt{1+\frac{y^{2/3}}{x^{2/3}}} \,dx$$
$$L=\int\limits_{a}^{b} \sqrt{\frac{1}{x^{2/3}}} \,dx$$
$$L=\int\limits_{a}^{b} x^{-1/3} \,dx$$
