# 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.

I would compute the arc length for the first quadrant. You can multiply by $$4$$ to get the total because of symmetry. That means your square root is unambiguous-you can use the plus sign as you have, so the range is $$0$$ to $$8$$. As you say, that gives $$12$$ so the total circumference is $$48$$.

• Thank you for the errata—I've edited the original post. Nov 12, 2018 at 3:05

$$f(x) = (4-x^{2/3})^{3/2}$$

and

$$g(x) = -(4-x^{2/3})^{3/2}$$

To get the total arc length, you integrate the arc length for each of them, and add them together. This gives you:

$$\int_{-8}^8 \sqrt{1 + (f^\prime(x))^2}dx + \int_{-8}^8 \sqrt{1 + (g^\prime(x))^2}dx$$

In your case, this simplifies to:

$$2\int_{-8}^8 \sqrt{1 + (f^\prime(x))^2}dx$$

This integral should not produce 0. You need to be careful when evaluating it, since $$\sqrt{z^2} = |z|$$, not $$z$$. Or, you can notice that in this case, the curve is symmetrical when reflected in the y axis, so the integral becomes:

$$4\int_{0}^8 \sqrt{1 + (f^\prime(x))^2}dx$$

And then you don't have to worry about negative numbers.