Evaluating A Path Integral In Polar Coordinates 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. 
 A: 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
$$\eqalign{
  & \theta  \equiv t  \cr 
  & {\bf{x}} = r(\theta ){\bf{r}}(\theta )  \cr 
  & {{d{\bf{x}}} \over {d\theta }} = {{dr} \over {d\theta }}(\theta ){\bf{r}}(\theta ) + r(\theta ){{d{\bf{r}}} \over {d\theta }}(\theta ){\mkern 1mu}   \cr 
  & {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\, = {{dr} \over {d\theta }}(\theta ){\bf{r}}(\theta ) + r(\theta ) \pmb{\theta} (\theta )  \cr 
  & \left\| {{{d{\bf{x}}} \over {d\theta }}} \right\| = \sqrt {{r^2}(\theta ) + {{\left( {{{dr} \over {d\theta }}} \right)}^2}(\theta )}  \cr} $$
A: 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:
$$
ds=\sqrt{dr^2 +r^2d\theta^2}=d\theta \sqrt{\left(\frac{dr}{d\theta}\right)^2 +r^2}
$$ 
A: First equation is correct. It can be written also with respect to arc length.
$$\int_{\theta_1}^{\theta_2} f(r,\theta) \sqrt {r^2+(\frac{dr}{d\theta})^2 } d\theta$$
$$\int_{s_1}^{s_2} f(r,\theta)\,  ds $$
$$\int_{x_1}^{x_2} g(x,y) \sqrt {1+(\frac{dy}{dx})^2 } dx $$
$$\int_{s_1}^{s_2} g(x,y)  \,  ds $$
