Length of curve in 3D spherical coordinate

Question

let $r$ be the magnitude of a vector in 3D with Spherical co-ordinate $(r,\theta,\phi)$ and cartesian coordinates is $(x,y,z)$, whose angle with $z$ axis is $\phi$ and projection of the vector makes angle $\theta$ with $x$ axis. We know the the relations $$x=r\sin\theta\cos\phi$$ $$y=r\sin\theta\sin\phi$$ $$z=r\cos\phi$$

Now $\gamma(t):[0,1]\to \mathbb{R}^3$ be a curve $\gamma(t)=(x(t),y(t),z(t))$ Now could you confirm me that lenghth of the curve is given by the formulae $$l_{\gamma}=\int_{0}^{1}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2}?$$ if yes how did we come to this formulae? please help

and from the above relation of cartesian and spherical co ordinates I calculated that the lenght is given by $l_{\gamma}=\int_{a}^{b}\sqrt{(\frac{dr}{dt})^2+r^2(\frac{d\theta}{dt})^2}$, am I right? and what wil be the upper and lower limit of tis integration?

Answer 1

When $0<t_1-t_0\ll 1$ one has $$\eqalign{|\gamma(t_1)-\gamma(t_0)|^2&= (x(t_1)-x(t_0))^2+(y(t_1)-y(t_0))^2+(z(t_1)-z(t_0))^2 \cr &\doteq \bigl(\dot x^2(t_0)+\dot y^2(t_0)+\dot z^2(t_0)\bigr)(t_1-t_0)^2\cr}$$ and therefore $$|\gamma(t_1)-\gamma(t_0)|\doteq \sqrt{\dot x^2(t_0)+\dot y^2(t_0)+\dot z^2(t_0)}\ (t_1-t_0)\ .$$ Introducing a partition $0=t_0<t_1<\ldots<t_N=1$ of $[0,1]$ we obtain $$\eqalign{L(\gamma)&\doteq\sum_{k=1}^N|\gamma(t_k)-\gamma(t_{k-1})|\cr &\doteq\sum_{k=1}^N\sqrt{\dot x^2(t_{k-1})+\dot y^2(t_{k-1})+\dot z^2(t_{k-1})}\ (t_k-t_{k-1})\cr &\doteq \int_0^1\sqrt{\dot x^2(t)+\dot y^2(t)+\dot z^2(t)}\ dt\ .\cr}$$ A real proof, using proper assumptions, as well as $\epsilon$ and $\delta$, would make this argument precise.

Using spherical coordinates $(r,\phi,\theta)$, where $\phi=\arg(x,y)$ and $\theta=\arg(\sqrt{x^2+y^2},z)$ one has $$x=r\cos\phi\cos\theta,\quad y=r\sin\phi\cos\theta,\quad z=r\sin\theta\ .$$ When $r$, $\phi$ and $\theta$ are given as functions of $t$ it follows by the chain rule that $$\dot x=\cos\phi\cos\theta\ \dot r-r\sin\phi\cos\theta\ \dot\phi-r\cos\phi\sin\theta\ \dot\theta\ ,$$ and similarly for the other two. Now compute $\dot x^2+\dot y^2+\dot z^2$ in terms of the spherical variables and simplify. You should get $$ds=\sqrt{\dot r^2 + r^2\cos^2\theta\ \dot\phi^2+ r^2\ \dot\theta^2}\ dt\ .$$

Answer 2

To the confirmation of the formula for the length of the curve:

The length of the curve

$$ \gamma:t\mapsto\vec x(t)= \left( \begin{array}{c} x^1(t)\\ x^2(t)\\ x^3(t) \end{array} \right)\ ; \quad t\in[0,1] $$

can be approximated by the length of the polygonal path that goes from $\vec x_0=\vec x(0)$ passing through $\vec x_i=\vec x(t_i),\ i=1,2,...n-1\ $ to $\vec x_n=\vec x(1)$, where $t_0=0<t_1<t_2<\ ...<t_n=1$ is a partition of the interval $[0,1]$.

The length of a segment from $\vec x_{i-1}$ to $\vec x_i$ is by Pythagoras $$ \sqrt{\sum_k(x_i^k-x_{i-1}^k)^2}\ ;\quad k\in\{1,2,3\}\ . $$ (The main diagonal of a cube is the square root of the sum of the squares of its sides).

From Rolle's theorem we know that $x_i^k-x_{i-1}^k=(t_i-t_{i-1})\frac{\mathrm dx^k}{\mathrm dt}_{|\xi_{i,k}}$. If we take for every component $k$ a common $\xi_i$ we will make a small error that will go to zero with finer and finer partitions of our interval, because $\sqrt{\sum_k{(\frac{\mathrm dx^k}{\mathrm dt}_{|\xi_{i,k}})^2}}$ is uniformly continuous in its arguments.

Our approximation of the polygon path becomes then

$$ \sum_{i=1}^n(t_i-t_{i-1})\sqrt{\sum_k{(\frac{\mathrm dx^k}{\mathrm dt}_{|\xi_{i}})^2}}\quad, $$

which is the Riemann sum of the integral

$$ l(0,1,\gamma)=\int\limits_0^1\sqrt{\sum_k(\frac{\mathrm dx^k}{\mathrm dt})^2}\,\mathrm dt \quad. $$

Source: Skript Dragon Chapter 12 Integration, Wegintegral (in German)