# Length of an Astroid

I am having trouble with this question:

Calculate the length of the astroid of $x^{\frac23}+y^{\frac23}=1$. s = ?

I approached it by doing the following:

1. setting $x^{\frac23}=1$ because then I can find $x=\sqrt{1}=1$
2. Then I set $$s = 4\int_0^1 \sqrt{1 + \left(\frac{dy}{dx}\right)^2}dx$$
3. Then I did implicit differentiation to get

$$\frac23 x^{-\frac13} + \frac23 y^{-\frac13}\left(\frac{dy}{dx}\right) = 0,\\ \frac{dy}{dx} = -\left(\frac{y}{x}\right)^{\frac13}.$$

I stopped there... I appreciate the help. Thank you!

• Can you write $y$ as a function of $x$, i.e., $y=f(x)$? – David H Apr 12 '14 at 22:14

## 3 Answers

$$ds^2=dx^2+dy^2$$ $$ds=\sqrt{dx^2+dy^2}=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$ $$s=\int \sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$

You had the right approach, but the problem was with the implicit differentiation. Since you have $y$ in there, it becomes a problem when integrating.

Instead, isolate $y$ from the original and take the derivative of that: $$x^{\frac23}+y^{\frac23}=1$$ $$y^{\frac23}=1-x^{\frac23}$$ $$y=\left(1-x^{\frac23}\right)^\frac32$$

Now when you find $\frac{dy}{dx}$, it will only be in terms of $x$. So take the derivative of that: $$\frac{dy}{dx}=\frac32 \left(1-x^{\frac23}\right)^\frac12 \left(-\frac23x^{-\frac13}\right)$$

Get rid of the fractions:

$$\frac{dy}{dx}=\left(1-x^{\frac23}\right)^\frac12 \left(-x^{-\frac13}\right)$$

Now plug it in:

$$s=4\int_0^1 \sqrt{1+\left(\left(1-x^{\frac23}\right)^\frac12 \left(-x^{-\frac13}\right)\right)^2}dx=4\int_0^1 \sqrt{1+\left(\left(1-x^{\frac23}\right)^\frac12 \left(-x^{-\frac13}\right)\left(1-x^{\frac23}\right)^\frac12 \left(-x^{-\frac13}\right)\right)}dx=4\int_0^1 \sqrt{1+\left(\left(1-x^{\frac23}\right) \left(x^{-\frac23}\right)\right)}dx=4\int_0^1 \sqrt{1+\left(x^{-\frac23}-1\right)}dx=4\int_0^1 \sqrt{x^{-\frac23}}dx=4\int_0^1 x^{-\frac13}dx$$

So easy to integrate now!

$$s=4\int_0^1 x^{-\frac13}dx=4\left(\frac32x^\frac23\Biggr|_0^1\right)=4\left(\frac32\right)=\therefore 6$$

A potentially easier way to do this is to parametrize the astroid by taking advantage of the trig identity $\cos^2(\theta)+\sin^2(\theta) = 1$.

Take $x = \cos^3(t)$ and $y = \sin^3(t)$ where $0\leq t \leq 2\pi$.

By symmetry, we can simply find the arclength of $1/4$th the astroid and multiply by $4$ at the end. The bounds on our integration will be $0 \leq t \leq \pi/2$.

Now simply use the formula for arclength:

$$\int_0^{\pi/2} \sqrt{\left( \frac{dx}{dt}\right)^2 + \left( \frac{dy}{dt} \right)^2} \ dt$$

• Thank you very much! This was very helpful! – Vanessa Vitiello May 8 '14 at 22:40

You can also continue using your formula for s.

$$s=4\int_0^1\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$

When you fill in $$\frac{dy}{dx} = -\frac{y^\frac{1}{3}}{x^\frac{1}{3}}$$ in $$s$$, you get

$$s=4\int_0^1 \sqrt{1+\left(\frac{y^\frac{1}{3}}{x^\frac{1}{3}}\right)^2}dx$$

$$s=4\int_0^1 \sqrt{\frac{x^\frac{2}{3}+y^\frac{2}{3}}{x^\frac{2}{3}}}dx$$

$$s=4\int_0^1 \frac{1}{x^\frac{1}{3}}\sqrt{x^\frac{2}{3}+{y^\frac{2}{3}}}dx$$

But the original formula says that $$x^\frac{2}{3}+y^\frac{2}{3}=1$$, so

$$s=4\int_0^1 \frac{1}{x^\frac{1}{3}}dx=4\int_0^1 x^\frac{-1}{3}dx=4\left(\frac{3}{2}\right)=6$$

• Great answer and welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. – dantopa Jan 24 '19 at 15:55