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 $48$.

  • $\begingroup$ Thank you for the errata—I've edited the original post. $\endgroup$ – SRSR333 Nov 12 '18 at 3:05

Your curve is really made of two functions:

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


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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.