# Chris' Math Blog

## Arc Length in Different Coordinate Systems

This post will deal with converting the arc length formulas in two and three dimensions from rectangular coordinates to polar (2-D), cylindrical (3-D) and spherical (3-D) coordinates.

Two Dimensions

In $\mathbb{R}^2$, we define the arc length of a function $y=f(x)$ over the interval $a\leq x\leq b$ to be

$L=\displaystyle\int_a^b \sqrt{1+\left(\frac{\,dy}{\,dx}\right)^2}\,dx$

If we define a parametric function $x=x(t)$ and $y=y(t)$, we observe that $\dfrac{\,dy}{\,dx}=\dfrac{\,dy/\,dt}{\,dx/\,dt}$.  Substituting this into the arc length formula yields

$\displaystyle\int_a^b\sqrt{1+\left(\frac{\,dy/\,dt}{\,dx/\,dt}\right)^2}\,dx$

Getting a common denominator gives us

\begin{aligned}\displaystyle\int_a^b\sqrt{\frac{(\,dx/\,dt)^2+(\,dy/\,dt)^2}{(\,dx/\,dt)^2}}\,dt&=\int_a^b\sqrt{\left(\frac{\,dx}{\,dt}\right)^2+\left(\frac{\,dy}{\,dt}\right)^2}\frac{\,dt}{\,dx}\,dx\\ &=\int_a^b\sqrt{\left(\frac{\,dx}{\,dt}\right)^2+\left(\frac{\,dy}{\,dt}\right)^2}\,dt\end{aligned}

In polar coordinates, we know that $x=r\cos\theta$ and $y=r\sin\theta$.   If $r=r(t)$ and $\theta=\theta(t)$, then we observe that

$\dfrac{\,dy}{\,dt}=r\cos\theta\dfrac{\,d\theta}{\,dt}+\sin\theta\dfrac{\,dr}{\,dt}\text{ and }\dfrac{\,dx}{\,dt}=\cos\theta\dfrac{\,dr}{\,dt}-r\sin\theta\dfrac{\,d\theta}{\,dt}$

Therefore,

\begin{aligned}L &=\displaystyle\int_{\alpha}^{\beta}\sqrt{\left(\cos\theta\frac{\,dr}{\,dt}-r\sin\theta\frac{\,d\theta}{\,dt}\right)^2+\left(r\cos\theta\frac{\,d\theta}{\,dt}+\sin\theta\frac{\,dr}{\,dt}\right)^2}\,dt\\ \implies L&=\displaystyle\int_{\alpha}^{\beta}\sqrt{\left(\frac{\,dr}{\,dt}\right)^2+r^2\left(\frac{\,d\theta}{\,dt}\right)^2}\,dt\end{aligned}

\begin{aligned}\implies L &=\displaystyle\int_{\alpha}^{\beta}\sqrt{\left(\frac{\,d\theta}{\,dt}\right)^2\left[r^2+\left(\frac{\,dr}{\,d\theta}\right)^2\right]}\,dt\\\implies L &=\displaystyle\int_{\alpha}^{\beta}\sqrt{r^2+\left(\frac{\,dr}{\,d\theta}\right)^2}\frac{\,d\theta}{\,dt}\,dt\end{aligned}

\begin{aligned}\implies L&=\displaystyle\int_{\alpha}^{\beta}\sqrt{r^2+\left(\frac{\,dr}{\,d\theta}\right)^2}\,d\theta\end{aligned}

Three Dimensions

Let us now consider two additional coordinate systems in $\mathbb{R}^3$:  the cylindrical and spherical coordinate system.

The cylindrical coordinate system is a 3-D version of the polar coordinate system in 2-D with an extra component for $z$.  We can slightly modify our arc length equation in polar to make it apply to the cylindrical coordinate system given that $r=r(t)$, $\theta=\theta(t)$.  We start from this step:

$\displaystyle\int_{a}^{b}\sqrt{\left(\frac{\,dr}{\,dt}\right)^2+\left(\frac{\,d\theta}{\,dt}\right)^2}\,dt$

From rectangular coordinates, the arc length of a parameterized function is

$\displaystyle\int_a^b\sqrt{\left(\frac{\,dx}{\,dt}\right)^2+\left(\frac{\,dy}{\,dt}\right)^2+\left(\frac{\,dz}{\,dt}\right)^2}\,dt$.

So from this, it follows that in cylindrical coordinates, we have arc length defined to be

$\displaystyle\int_a^b\sqrt{\left(\frac{\,dr}{\,dt}\right)^2+r^2\left(\frac{\,d\theta}{\,dt}\right)^2+\left(\frac{\,dz}{\,dt}\right)^2}\,dt$

To define arc length in the spherical coordinate system, we need to know first how to convert points from spherical to cylindrical coordinates (and then spherical to rectangular).   I leave without proof that $r=\rho\sin\phi$, $\theta=\theta$, and $z=\rho\cos\phi$.  Since we know that $x=r\cos\theta$, $y=r\sin\theta$, and $z=z$ goes from cylindrical to rectangular, it follows that $x=\rho\sin\phi\cos\theta$, $y=\rho\sin\phi\sin\theta$, and $z=\rho\cos\phi$ takes points in spherical coordinates and converts them into points from the rectangular coordinate system.

Again we consider the arc length formula for a parametric curve in the rectangular coordinate system:

$\displaystyle\int_a^b\sqrt{\left(\frac{\,dx}{\,dt}\right)^2+\left(\frac{\,dy}{\,dt}\right)^2+\left(\frac{\,dz}{\,dt}\right)^2}\,dt$

If we let $\rho=\rho(t)$, $\phi=\phi(t)$ and $\theta=\theta(t)$, we now can find expressions for $\frac{\,dx}{\,dt}$, $\frac{\,dy}{\,dt}$ and $\frac{\,dz}{\,dt}$.  I leave the calculations of these three to the reader, but I will, for the sake of space, write the values to each one.

Since $x=\rho\sin\phi\cos\theta$, we have

$\dfrac{\,dx}{\,dt}=\sin\phi\cos\theta\dfrac{\,d\rho}{\,dt}-\rho\sin\phi\sin\theta\dfrac{\,d\theta}{\,dt}+\rho\cos\phi\cos\theta\dfrac{\,d\phi}{\,dt}$

Since $y=\rho\sin\phi\sin\theta$, we have

$\dfrac{\,dy}{\,dt}=\sin\phi\sin\theta\dfrac{\,d\rho}{\,dt}+\rho\sin\phi\cos\theta\dfrac{\,d\theta}{\,dt}+\rho\cos\phi\sin\theta\dfrac{\,phi}{\,dt}$

Since $z=\rho\cos\phi$, we have

$\dfrac{\,dz}{\,dt}=\cos\phi\dfrac{\,d\rho}{\,dt}-\rho\sin\phi\dfrac{\,d\phi}{\,dt}$

When you plug this into the rectangular arc length formula and after a ton of painful simplifying (which I have left out to spare your sanity), we see that the arc length formula in spherical coordinates is:

$\displaystyle\int_a^b\sqrt{\left(\frac{\,d\rho}{\,dt}\right)^2+\rho^2\sin^2\phi\left(\frac{\,d\theta}{\,dt}\right)^2+\rho^2\left(\frac{\,d\phi}{\,dt}\right)^2}\,dt$