0
$\begingroup$

I don't understand why these two problems are solved differently here the first one $fig(1)$ and 2nd one $fig(2)$. Why did we take the limit $\displaystyle \lim_{r\to0^+}\int_r^\pi \sqrt{2-2cost}\space dt$ in the arc length problem but not in the Surface are problem, Although the cycloid is not differentiable at $\theta=0$ and $\theta=\pi$

enter image description here

$fig(1)$

enter image description here

$fig(2)$

$\endgroup$
4
  • 2
    $\begingroup$ Welcome to Math.SE! <> The images are small and a little blurry, but taking a limit as $r \to 0^+$ in the arc length integral was unnecessary because the arc length element is computed from a parametrization. <> Re: ...the cycloid is not differentiable...: The given parametrization is smooth everywhere, and can be used to calculate arc length and surface area. It's true that the cycloid has a vertical tangent where it touches the axis. That necessitates an improper integral if we use the arc length formula for a graph $y = f(x)$. $\endgroup$
    – Andrew D. Hwang
    Commented Jun 12, 2022 at 12:11
  • 1
    $\begingroup$ Can you get higher resolution images than that? Anyway, your question is why $\int_0^\pi\sqrt{1-\cos t}\mathrm{d}t$ needed more care than $\int_0^\pi(3-2\cos^2\frac{\theta}{2})|\sin\frac{\theta}{2}|\mathrm{d}\theta$, with the former changed to $\lim_{r\to0^+}\int_r^\pi\sqrt{1-\cos t}\mathrm{d}t$. I don't think it did, since an integrand doesn't have to be differentiable at $0$ to be integrable on $[0,\,\pi]$. $\endgroup$
    – J.G.
    Commented Jun 12, 2022 at 12:15
  • $\begingroup$ @AndrewD.Hwang So we only care about the smoothness of what's inside the integral, right? $\endgroup$
    – user1062191
    Commented Jun 12, 2022 at 12:31
  • 1
    $\begingroup$ Essentially "yes": For a proper integral it's sufficient that the integrand be continuous. We'd only need to take a limit as $r \to 0^+$ if the integrand were unbounded in every neighborhood of $0$, e.g., $\int_0^1 \frac{dx}{\sqrt{x}}$. $\endgroup$
    – Andrew D. Hwang
    Commented Jun 12, 2022 at 18:06

0

Reset to default

You must log in to answer this question.