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 $latex \mathbb{R}^2$, we define the arc length of a function $latex y=f(x)$ over the interval $latex a\leq x\leq b$ to be
$latex L=\displaystyle\int_a^b \sqrt{1+\left(\frac{\,dy}{\,dx}\right)^2}\,dx$
If we define a parametric function $latex x=x(t)$ and $latex y=y(t)$, we observe that $latex \dfrac{\,dy}{\,dx}=\dfrac{\,dy/\,dt}{\,dx/\,dt}$. Substituting this into the arc length formula yields
$latex \displaystyle\int_a^b\sqrt{1+\left(\frac{\,dy/\,dt}{\,dx/\,dt}\right)^2}\,dx$
Getting a common denominator gives us
$latex \begin{aligned}\displaystyle\int_a^b\sqrt{\frac{(\,dx/\,dt)^2+(\,dy/\,dt)^2}{(\,dx/\,dt)^2}}\,dt&=\int_a^b\sqrt{\left(\frac{\,dx}{\,dt}\right)^2+\left(\frac{\,dy}{\,dt}\right)^2}\frac{\,dt}{\,dx}\,dx\\ &=\int_a^b\sqrt{\left(\frac{\,dx}{\,dt}\right)^2+\left(\frac{\,dy}{\,dt}\right)^2}\,dt\end{aligned}$
In polar coordinates, we know that $latex x=r\cos\theta$ and $latex y=r\sin\theta$. If $latex r=r(t)$ and $latex \theta=\theta(t)$, then we observe that
$latex \dfrac{\,dy}{\,dt}=r\cos\theta\dfrac{\,d\theta}{\,dt}+\sin\theta\dfrac{\,dr}{\,dt}\text{ and }\dfrac{\,dx}{\,dt}=\cos\theta\dfrac{\,dr}{\,dt}-r\sin\theta\dfrac{\,d\theta}{\,dt}$
Therefore,
$latex \begin{aligned}L &=\displaystyle\int_{\alpha}^{\beta}\sqrt{\left(\cos\theta\frac{\,dr}{\,dt}-r\sin\theta\frac{\,d\theta}{\,dt}\right)^2+\left(r\cos\theta\frac{\,d\theta}{\,dt}+\sin\theta\frac{\,dr}{\,dt}\right)^2}\,dt\\ \implies L&=\displaystyle\int_{\alpha}^{\beta}\sqrt{\left(\frac{\,dr}{\,dt}\right)^2+r^2\left(\frac{\,d\theta}{\,dt}\right)^2}\,dt\end{aligned}$
$latex \begin{aligned}\implies L &=\displaystyle\int_{\alpha}^{\beta}\sqrt{\left(\frac{\,d\theta}{\,dt}\right)^2\left[r^2+\left(\frac{\,dr}{\,d\theta}\right)^2\right]}\,dt\\\implies L &=\displaystyle\int_{\alpha}^{\beta}\sqrt{r^2+\left(\frac{\,dr}{\,d\theta}\right)^2}\frac{\,d\theta}{\,dt}\,dt\end{aligned}$
$latex \begin{aligned}\implies L&=\displaystyle\int_{\alpha}^{\beta}\sqrt{r^2+\left(\frac{\,dr}{\,d\theta}\right)^2}\,d\theta\end{aligned}$
Three Dimensions
Let us now consider two additional coordinate systems in $latex \mathbb{R}^3$: the cylindrical and spherical coordinate system.
The
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