Calculate the length of the closed curve $x^{2/3} + y^{2/3} = 4$ I realise that this function forms a closed curve, and the range of both $x$ and $y$ are: $-8 \leq x, y \leq 8$. 
I began by differentiating the function implicitly, arriving at a expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$:
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x} &: \frac{2}{3}x^{-\frac{1}{3}} + \frac{2}{3}y^{-\frac{1}{3}}\cdot\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \\
&\implies x^{-\frac{1}{3}} + y^{-\frac{1}{3}}\cdot\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \\
&\implies \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{y^{-\frac{1}{3}}}{y^{-\frac{1}{3}}} = -\left(\frac{y}{x}\right)^{\frac{1}{3}}
\end{align}
Once that was done, I applied the formula for the arc length of a curve:
\begin{align}
&\int_{a}^{b}{\sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}}\\
&=\int_{a}^{b}{\sqrt{1 + \left(\frac{y}{x}\right)^\frac{2}{3}}}\\
&=\int_{a}^{b}{\sqrt{1 + \frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}}}\\
&\vdots\\
&=2\int_{a}^{b}{\frac{1}{\sqrt[\leftroot{4}\uproot{1}3]{x}}} \\
&=3\left[x^\frac{2}{3}\right]_{a}^{b}
\end{align}
Now, I'd like to know what the bounds are. 
As mentioned earlier, the function has a range (and domain) of $[-8, 8]$, but inserting them gives a definite integral with a result of zero, which is clearly incorrect. Using the bound $[0, 8]$ gives a result of $12$ which is more believable. However, I plotted the graph in Desmos, and I think the answer is definitely off by a large margin.
I believe (this was by sheer visual inspection) the length is approximately—but not equal to—the circumference of a circle with radius $8$, so I should be getting an answer closer to $16\pi \approx 50.$ I notice that $12 \times 4 = 48$; is the answer as simple as that? 
Is there a more rigorous method of choosing the bounds of the integral?

With parametric equations you have $$x=8cos^3 \theta, y=8\sin ^3 \theta $$
Then you integrate  $$L=4 \int _{0}^{\pi/2} \sqrt {(\frac {dx}{d\theta})^2 +(\frac {dx}{d\theta})^2} d\theta $$
The derivatives are straight forward and you can finish it.
