4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

Hints

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

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @JAEMTO: $\bf{r}$ and $\pmb{\theta}$ are unit vectors in polar coordinates. :) $\endgroup$ – H. R. Dec 1 '15 at 14:09
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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
1
$\begingroup$

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:

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

$\endgroup$
1
$\begingroup$

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

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

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.