Show that the path integral of $f(x,y)$ along a path given in polar coordinates by $r=r(\theta)$ where $\theta_1 ≤ \theta ≤ \theta_2$, is

$$\int_{\theta_1}^{\theta_2} f(r \cos \theta ,r \sin \theta) \sqrt {r^2+(\frac{dr}{d\theta})^2 } d\theta$$

I thought $x = r \cos \theta, y = r \sin \theta$

So $r(\theta)=(r\cos(\theta), r\sin(\theta))$, $\int_{\theta_1}^{\theta_2}f(r(\theta))||r'(\theta)||d\theta$

Then $\int_{\theta_1}^{\theta_2} f(r\cos\theta,r\sin\theta)\sqrt {r^2}d\theta$, ($\frac{dr}{d\theta})^2$ do not appear in the square root.

I cannot understand what happens. I think polar coordinates is something to do with it.

  • $\begingroup$ You have a bit troubles with your notation. You are missing that $r$ can depend on $\theta$. $\endgroup$ – Fabian Dec 1 '15 at 13:08
  • $\begingroup$ @Fabian I now thoroughly understand what I misunderstood!!!! $\endgroup$ – JAEMTO Dec 1 '15 at 14:01


1) The path integral of the scalar field $f(\bf{x})$ on a curve $C$ with parametric equation ${\bf{x}}={\bf{x}}(t)$ is defined as

$$I = \int\limits_C {f({\bf{x}}(t))\left\| {{{d{\bf{x}}} \over {dt}}(t)} \right\|dt} $$

2) In your example, we can find that

$$\eqalign{ & \theta \equiv t \cr & {\bf{x}} = r(\theta ){\bf{r}}(\theta ) \cr & {{d{\bf{x}}} \over {d\theta }} = {{dr} \over {d\theta }}(\theta ){\bf{r}}(\theta ) + r(\theta ){{d{\bf{r}}} \over {d\theta }}(\theta ){\mkern 1mu} \cr & {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\, = {{dr} \over {d\theta }}(\theta ){\bf{r}}(\theta ) + r(\theta ) \pmb{\theta} (\theta ) \cr & \left\| {{{d{\bf{x}}} \over {d\theta }}} \right\| = \sqrt {{r^2}(\theta ) + {{\left( {{{dr} \over {d\theta }}} \right)}^2}(\theta )} \cr} $$

  • $\begingroup$ What r(θ)θ(θ) means? $\endgroup$ – JAEMTO Dec 1 '15 at 13:59
  • $\begingroup$ @JAEMTO: Look at the answer again! I suceeded to write it in bold case! :) $\endgroup$ – H. R. Dec 1 '15 at 14:05
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    $\begingroup$ @JAEMTO: $\bf{r}$ and $\pmb{\theta}$ are unit vectors in polar coordinates. :) $\endgroup$ – H. R. Dec 1 '15 at 14:09
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    $\begingroup$ @JAEMTO: The bold case letters are vectors while the not bolds are just scalars. :) Can you understand this equation ${\bf{x}} = r(\theta ){\bf{r}}(\theta )$? $\endgroup$ – H. R. Dec 1 '15 at 14:12
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    $\begingroup$ @JAEMTO: I think you have some problems with the concept of parametric equations of a curve. In your example the parametric equations of the curve whose parameter is $\theta$ can be written in Cartesian coordinates as $${\bf{x}} = r(\theta )\cos (\theta ){\bf{i}} + r(\theta )\cos (\theta ){\bf{j}}$$ Or in polar coordinates as $${\bf{x}} = r(\theta ){\bf{r}}\left( \theta \right)$$ where $${\bf{r}}(\theta ) = \cos (\theta ){\bf{i}} + \sin (\theta ){\bf{j}}$$ $\endgroup$ – H. R. Dec 1 '15 at 14:24

First note that Polar Coordinates are orthogonal coordinates, so the unit vectors $\mathbf{\hat r}$ and $\mathbf{\hat \theta}$ are orthogonal.

Now, from a point $P=(r, \theta)$ a displacement given by an infinitesimal change in coordinates $(dr,d\theta)$ is a segment whose length can be found using the pythagorean formula. The displacement in the direction of $\mathbf{\hat r}$ is obviously $dr$ and the displacement in the direction of $\mathbf{\hat \theta}$ is $rd\theta$ ( as you can easily see thinking that $d\theta$ is an infinitesimal angle) , so we have:

$$ ds=\sqrt{dr^2 +r^2d\theta^2}=d\theta \sqrt{\left(\frac{dr}{d\theta}\right)^2 +r^2} $$


First equation is correct. It can be written also with respect to arc length.

$$\int_{\theta_1}^{\theta_2} f(r,\theta) \sqrt {r^2+(\frac{dr}{d\theta})^2 } d\theta$$ $$\int_{s_1}^{s_2} f(r,\theta)\, ds $$

$$\int_{x_1}^{x_2} g(x,y) \sqrt {1+(\frac{dy}{dx})^2 } dx $$ $$\int_{s_1}^{s_2} g(x,y) \, ds $$

  • $\begingroup$ What is the g then? You mean xy-coordinate? $\endgroup$ – JAEMTO Dec 1 '15 at 13:16
  • $\begingroup$ No harm using the same f label or g. $ f( r \cos \theta , \theta) = g( x, tan^{-1}\frac {y}{x}) $ . $\endgroup$ – Narasimham Dec 1 '15 at 13:22

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