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.

  • 3
    $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Jun 8, 2021 at 22:47
  • 5
    $\begingroup$ 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. $\endgroup$
    – user882145
    Jun 8, 2021 at 22:58
  • $\begingroup$ I've edited @LeeMosher $\endgroup$ Jun 8, 2021 at 23:44
  • 1
    $\begingroup$ @boojum: That looks like a good answer. $\endgroup$
    – Lee Mosher
    Jun 8, 2021 at 23:51
  • 3
    $\begingroup$ @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. $\endgroup$
    – user882145
    Jun 9, 2021 at 0:11


You must log in to answer this question.

Browse other questions tagged .