# Arc-Length of Curves

I want to calculate the arc length of a full cycle(? by this I mean from $$t=0$$ to $$t=2\pi$$) of a cardioid. The parameterization of a cardioid is given by: $$\gamma(t)=(2\cos(t)-\cos(2t),2\sin(t)-\sin(2t)) \\ |\gamma'| = \sqrt{8-8\cos(t)}$$

And thus, the arc length of a full cycle/rotation is: $$\int^{2\pi}_0 \sqrt{8-8\cos(t)} dt.$$

Which is non-negative. That is why I don't think we need to take the absolute value of the function. The issue, why am I getting $$0$$ as the value of the integral? Furthermore, to tackle this, because of symmetry, I tried to take $$\int^\pi_0 |\gamma'|dt$$ instead, then multiply it by 2. However, the result of this integral is negative. What? A non-negative function everywhere has a negative integral?

Edit: $$u = 8+8cos(t) \Rightarrow \frac{du}{dt} = -8sin(t), \\ \int^{2\pi}_0 \sqrt{8-8cos(t)}dt = \int^{2\pi}_0 \sqrt{8-8cos(t)} \frac{\sqrt{8+8cos(t)}}{\sqrt{8+8cos(t)}}dt= \\ =\int^{a'}_{a''}\frac{|8sin(t)|}{\sqrt{u}}\cdot\frac{1}{-8sin(t)}du$$ Now, I think I realize my mistake. I should separate this into 2, from $$0$$ to $$\pi$$, and from $$\pi$$ to $$2\pi$$. The final result would be

$$\int^0_{16} -\frac{1}{\sqrt{u}}du + \int^{16}_0 \frac{1}{\sqrt{u}}du$$

Note that $$a':= u = 8 + 8cos(\pi)=0, a'':= 8 + 8cos(0) = 16$$ for the first integral, and $$a':= u = 8 + 8cos(2\pi)=16, a'':= 8 + 8cos(pi) = 0$$

So, the final answer will be 16.

• Since we don't know how you got $0$ as the value of the integral, how can we answer your question why you got that answer? I suggest you show us how you actually calculated that, so that we can answer your question. Otherwise, it's not clear what you are actually asking. Jun 8, 2021 at 22:47
• The cycloid is "tricky" because the integrand you get after substitution (to deal with the square-root) changes sign between quadrants: that's why your definite integral came to zero. Integrate only from $\ 0 \$ to $\ \pi/2 \$ and then multiply by the appropriate factor.
– user882145
Jun 8, 2021 at 22:58
• I've edited @LeeMosher Jun 8, 2021 at 23:44
• @boojum: That looks like a good answer. Jun 8, 2021 at 23:51
• @AyamGorengPedes Your result should be correct now; the arc length of a full cycle of a cycloid of radius $\ R \$ is $\ 8R \ \ .$ I know about this particular arc-length integral because I've had students come to me frantically with the same question you had -- it is not a "stupid" mistake. You learn from experience to watch for integrands that change sign over an interval/region of integration. Stewart's textbook has this arc-length calculation as a brief example and, until maybe the most recent editions, left the "trip-hazard" concerning cancelation between quadrants unremarked.
– user882145
Jun 9, 2021 at 0:11