# Arc length of $y^3 = x^2$

I am trying to find the arc length of $y^3 = x^2$ and I am suppose to use two formulas, one for in terms of x and one for in terms of y.

At first I need to find (0,0) (1,1) and I start with in terms of x

$$y\prime = \frac{2}{3}x^\frac{-1}{3}$$

$$\left(\frac{2}{3}x^\frac{-1}{3}\right)^2 = \frac{4}{9x^\frac{2}{3}}$$

$$\int_0^1 \sqrt{1 + \left( \frac{2}{3}x^\frac{-1}{3}\right)^2}\, dx$$

$$\int_0^1 \sqrt{1 + \frac{4}{9x^\frac{2}{3}}}\, dx$$

This is undefined at 0 so it is an improper function. I do not think I can continue.

Now in terms of y.

$$x = y^\frac{3}{2}$$

$$x \prime = \frac{3\sqrt{y}}{2}$$ $$\left(\frac{3\sqrt{y}}{2}\right)^2 = \frac{9y}{4}$$

$$\int_0^1 \sqrt{1+ \frac{9y}{4}}\, dy$$

I have no idea how to factor that out, I have tried many ways but I can not get it.

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For the last integral, let $u=1+\frac{9y}{4}$. – André Nicolas Jun 7 '12 at 0:55

$$\int_0^1\sqrt{1 + {9y\over 4}} = {4\over 9}\int_1^{13/4} \sqrt{y}\,dy = {8\over 27}\left({13\sqrt{13}\over 8} - 1\right) = {13\sqrt{13} - 8\over 27}$$

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For the last integral, let $u=1+\frac{9y}{4}$. Then $du=\frac{9}{4}\,dy$, so $dy=\frac{4}{9}\,du$. The indefinite integral is $\int \frac{9}{4}u^{1/2}\,du$, and it's almost over.

The first integral is a bit more scary. I would rewrite your $\sqrt{1 + \frac{4}{9x^{2/3}}}$ as $\sqrt{\frac{9x^{2/3} + 4}{9x^{2/3}}}$ and then pull the denominator out to conclude that we are interested in the indefinite integral $$\int \frac{1}{3x^{1/3}}\sqrt{9x^{2/3}+4}\,\,dx$$ Now let $u=9x^{2/3}+4$. Then $du=6x^{-1/3}\,dx=\frac{6}{x^{1/3}}\,dx$. Then the $\frac{1}{3x^{1/3}}\,dx$ term is just $\frac{1}{18}du$.

You are right in viewing the above integral as an improper integral, since there is the appearance of trouble at $0$. In principle, we should integrate from $\epsilon$ to $1$, and find the limit as $\epsilon$ approaches $0$ from the right. In practice, the substitution $u=x^{2/3}$ makes the problem disappear.

Remark: There is a fair bit of difference in appearance of difficulty between the two approaches. But in fact they are just a substitution away from each other. However, the appropriate substitution may not be obvious.

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Neater is to do it parametrically, setting $x=t^2$ and $y=t^3$, and using the formula $$s = \int_0^1\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}dt\\ =\int_0^1t\sqrt{9t^2+4}dt\,.$$ Now you set $u=9t^2+4$, getting $\int_4^{13}u^{1/2}du/18$, evaluating to $(13^{3/2}-8)/27$.

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I have no idea what parametrically means and I do not think we are intended to use this since we have not learned it in class. – user138246 Jun 7 '12 at 12:24

For $$\int_0^1 \sqrt{1 + \frac{9y}{4}} dy$$ you could try making the substitution $y = \frac{4}{9}\sinh^2 \theta$ where $\theta$ goes from $0$ to $\sinh^{-1} \frac{3}{2}$. This substitution is likely to be nice since $1 + \sinh^2 \theta = \cosh^2 \theta$.

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I didn't downvote, but see André Nicolas' note above. Using $\sinh$ would make things harder then they need to be. – Argon Jun 7 '12 at 1:08
@Argon, yep, I saw that (and I've upvoted both André's and ncmathsadist's answer, because they are much cleaner), but I don't think this answer will give the wrong result, so I'm not going delete it. (Thanks for the comment btw, much nicer than an anonymous downvote.) – huon Jun 7 '12 at 1:25