I was given the task to calculate the length of a curve c with constant $r$ and $\theta$ and $\varphi = t$ in polar coordinates $$f = r(\sin(\varphi) \sin(\theta), \cos(\varphi) \sin(\theta), \cos(\theta))^T$$ using the following formula involving the metric of $f^{-1}$: $$L = \int_0^{2\pi} \sqrt{(<c^\prime,c^\prime>_{c(t)})}dt$$ Where $<>$ is the metric induced by the euclidian metric.

First I tried to calculate the metric and the the Integral but I got stuck and dont know how to actually continue solving the integral: First I calculated the basis vectors of the map $f^{-1}$
$$\partial_r = \frac{\partial}{\partial r}f = (\sin(\varphi) \sin(\theta), \cos(\varphi) \sin(\theta), \cos(\theta))$$

$$\partial_\varphi = r(\cos(\varphi)\sin(\theta), -\sin(\varphi) \sin(\theta), 0)$$

$$\partial_\theta = r(\sin(\varphi) \cos(\theta), \cos(\varphi) \cos(\theta), -\sin(\theta))$$

Now we can calculate the euclidian product of those vectors to get our metric components:

$$g_{1,1} = (\sin(\varphi) \sin(\theta))^2+(\cos(\varphi) \sin(\theta))^2+\cos(\theta)^2 = \sin(\theta)^2+\cos(\theta)^2 = 1$$

$$g_{1,2} = g_{2,1} = \sin(\varphi) \sin(\theta)r\cos(\varphi)\sin(\theta) - \cos(\varphi) \sin(\theta)r\sin(\varphi) \sin(\theta) = 0$$

$$g_{1,3} = g_{3,1} = \sin(\varphi) \sin(\theta) r \sin(\varphi) \cos(\theta) + \cos(\varphi) \sin(\theta)r\cos(\varphi) \cos(\theta) - \sin(\theta)r\cos(\theta) = r\sin(\theta)\cos(\theta) - \sin(\theta)r\cos(\theta) = 0$$

$$g_{2,2} = r^2((\cos(\varphi)\sin(\theta))^2 + (\sin(\varphi) \sin(\theta))^2 = r^2 \sin(\theta)^2$$

$$g_{2,3} = g_{3,2} = r^2(\cos(\varphi)\sin(\theta)\sin(\varphi) \cos(\theta) - \sin(\varphi) \sin(\theta)\cos(\varphi) \cos(\theta) ) = 0 $$

$$g_{3,3} = r^2((\sin(\varphi) \cos(\theta))^2 + (\cos(\varphi) \cos(\theta))^2 + \sin(\theta)^2 ) =r^2(cos(\theta)^2+\sin(\theta)^2) =r^2 $$ We then have: $$G = \begin{pmatrix} 1 & 0 & 0 \\ 0 & r^2 \sin(\theta)^2 & 0 \\ 0 & 0 & r^2 \end{pmatrix}$$ Our curve is given as: $$c (t) = r_0(\sin(t)\sin(\theta_0), \cos(t) \sin(\theta_0), \cos(\theta_0))$$ I then try to calculate $<c^\prime,c^\prime>_{c(t)}$ which is equal to

$$c^{\prime T} G(r_0,t,\theta_0)*c^\prime$$

$$c^\prime = r_0(\cos(t)\sin(\theta_0), -\sin(t) \sin(\theta_0), 0)$$

$$G(r_0,t,\theta_0)*c^\prime = r_0(\cos(t)\sin(\theta_0), -\sin(t) \sin(\theta_0) r_0^2 \sin(\theta_0)^2, 0)$$

$$\Rightarrow c^{\prime T} G(r_0,t,\theta_0)*c^\prime = (r_0\cos(t)\sin(\theta_0))^2+(\sin(t) r_0^2 \sin(\theta_0)^2)^2$$

And now I dont know how to actually continue integrating this

  • $\begingroup$ This answer is difficult to follow as it is currently written. Could you add in some justification to your working to make it clearer $\endgroup$
    – FD_bfa
    Commented May 10, 2023 at 19:47
  • $\begingroup$ I tried to make it a bit more clearer. Hope its understandable now $\endgroup$
    – soph6626
    Commented May 10, 2023 at 19:56

1 Answer 1


You mixed the metric for the spherical coordinates with the curve expressed in cartesian coordinates from its values in spherical ones (e.g. $r\sin(\varphi) \sin(\theta)$ is the $x$ coordinate). The curve in spherical coordinates is this:

$c(t)=(r_0,t,\theta_0)$, then

$c'(t)=(0,1,0)$ and

$c^{\prime T} G(r_0,t,\theta_0)*c^\prime =r_0^2 \sin(\theta_0)^2$, so

$L = \displaystyle\int_0^{2\pi}\sqrt{(c^{\prime T}G(r_0,t,\theta_0)*c^\prime )}dt=\int_0^{2\pi} \sqrt{r_0^2 \sin(\theta_0)^2}dt=2\pi r_0\sin(\theta_0)$

  • 1
    $\begingroup$ Thank you! Yeah that makes sense, the task was worded in a confusing way so I assumed I had to do it like that. $\endgroup$
    – soph6626
    Commented May 10, 2023 at 20:32

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