I am tasked with finding the length of the circle $x^2+y^2=4$ on the hyperboloid $x^2+y^2-z^2-1=0$.

The hyperboloid can be parametrized as $$x(u,v)=(\sinh(u)\sin(v), \sinh(u)\cos(v), \cosh(u)).$$

In particular, we have $$x_u=(\cosh(u)\sin(v),\cosh(u)\cos(v),\sinh(u))$$ $$x_v=(\sinh(u)\cos(v),-\sinh(u)\sin(v),0).$$

The circle of radius 2 can be parametrized as $$c(t)=(2\cos t,2\sin t).$$

The next order of business is to compute the coefficients of the first fundamental form on $c(t)$.

$$E=\langle x_u,x_u \rangle=\cosh(2\cos(t))^2\sin(2\sin(t))^2+\cosh(2\cos(t))^2\cos(2\sin(t))^2+\sinh(2\cos(t))^2$$ $$F=\langle x_u,x_v\rangle=0$$ $$G=\langle x_u,x_v\rangle=\sinh(2\cos(t))^2\cos(2\sin(t))^2+\sinh(2\cos(t))^2\sin(2\sin(t))^2.$$

Now, we can compute the differential of arc length $$ds=\sqrt{Edu^2+Gdv^2}=\sqrt{(\cosh(2\cos(t))^2\sin(2\sin(t))^2+\cosh(2\cos(t))^2\cos(2\sin(t))^2+\sinh(2\cos(t))^2)\sin(t)^2+(\sinh(2\cos(t))^2\cos(2\sin(t))^2+\sinh(2\cos(t))^2\sin(2\sin(t))^2)\cos(t)^2}$$

Now, we simply integrate the differential on the interval $[0,2\pi]$.


The issue is I have no idea how to integrate the differential.

Is the work up to the integral correct? Also, how can I integrate $ds$ if so? I've tried using Maple, but it cannot integrate such a function.

Thank you.

  • $\begingroup$ The first equation is that of a cylinder, not a circle. This cylinder intersects the hyperboloid in two disjoint curves. Are you meant to find the total length of both or just one? Also, this computation seems overly complicated. It’s easy to see that the intersection curves are planar, and those planes are orthogonal to the $z$-axis. That should simplify things tremendously. $\endgroup$
    – amd
    May 13, 2019 at 19:58
  • $\begingroup$ Over and over, I read a question and totally misunderstand; maybe that’s the case for this one, too. But: you’re telling me that the curve is given by $x^2+y^2=4$, $z=\pm\sqrt3$. I’ll take the plus sign, to guarantee a single curve. Isn’t your curve a circle of radius $2$? $\endgroup$
    – Lubin
    May 13, 2019 at 20:04
  • $\begingroup$ The problem is to compute the length of the curve $x^2+y^2=4$ on the surface $x^2+y^2-z^2-1=0$. I don't think I am supposed to just find the length of the intersection of the associated cylinder with the hyperboloid. I think I am supposed to do something similar to Example 3.2.1 in this link: web.mit.edu/hyperbook/Patrikalakis-Maekawa-Cho/node28.html. Edit: I realize there is a bit of notational abuse in the problem statement. Sorry for any confusion that may have caused. $\endgroup$
    – Karambwan
    May 13, 2019 at 20:11
  • $\begingroup$ You made several mistakes in the computation of the fundamental form. $\endgroup$
    – user65203
    May 13, 2019 at 20:33

1 Answer 1


Computing correctly the fundamental form,

$$\begin{cases}E=\cosh^2(u)+\sinh^2(u), \\F=0, \\G=\sinh^2(u).\end{cases}$$

Then the path is $\sinh(u)=2$, so that $du=0$.



  • $\begingroup$ I had substituted the coordinates of the curve into the first fundamental form. What is the justification for writing the path, the circle of radius 2, as $\cosh(u)=2$? $\endgroup$
    – Karambwan
    May 13, 2019 at 21:03
  • $\begingroup$ @Saru In the end, this is exactly twice the circumference of a circle with radius $2$. $\endgroup$
    – amd
    May 13, 2019 at 21:09
  • $\begingroup$ @amd: my bad, $4\pi$. $\endgroup$
    – user65203
    May 14, 2019 at 6:12
  • $\begingroup$ @saru: my bad again, $x^2+y^2=4=\sinh^2(u)$. $\endgroup$
    – user65203
    May 14, 2019 at 6:13
  • $\begingroup$ Could you explain that bit in detail? I'm not quite seeing where $x^2+y^2=4=\sinh^2(u)$ comes from. $\endgroup$
    – Karambwan
    May 14, 2019 at 15:35

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