Integral in polar cooequationrdinate : what is $\int_\Gamma d\gamma $ for $\Gamma =\{(R,\varphi )\mid \varphi \in [0,2\pi]\}$? I was wondering if we can use polar equation to compute integral. For example, the equation of a circle of radius 2 is given by $r(\varphi )=2$. So $$\Gamma =\{\gamma (\varphi )=(2,\varphi )\mid \varphi \in [0,2\pi]\}.$$
Now, I was wondering that $$length(\Gamma )=\int_\Gamma d\gamma =\int_0^{2\pi} \left|\frac{d}{d\varphi }\gamma (\varphi )\right|d\varphi =\int_0^{2\pi}|(0,1)|d\varphi =2\pi,$$
and thus it doesn't work. Could someone tell me why it doesn't work ? And if it's possible to use polar equation (as we do with cartesian equation) to compute integral.
Other attempts
$$\frac{d}{d\varphi }\gamma (\varphi )=\frac{d}{d\varphi }(2e_r+\varphi e_\varphi )=2e_\varphi +e_\varphi -\varphi e_r=3e_\varphi -\varphi e_r=(-\varphi ,3).$$
So $$Length(\Gamma )=\int_0^{2\pi}\sqrt{\varphi ^2+9}d\varphi,$$
but it doesn't do $4\pi$.
 A: So, you want a path that runs (say, counter-clockwise) through the points with distance to the origin $2$. This would require a parameter, I'll use $t$. Then, you can write something like
$$\gamma(t)=(2\cos t,2\sin t),$$
in Cartesian coordinates.
In polar coordinates, $\gamma_\theta(t)=t$ and $\gamma_r(t)=2$. Now the length of the curve is given by
$$L_\gamma = \int_\gamma \mathrm d\gamma = \int_0^{2\pi}\left|\frac{\mathrm d}{\mathrm dt}\gamma(t)\right|\mathrm dt.$$
Now, to directly compute the integrand, namely the magnitude of the tangent vector of the curve, in polar coordinates, you need some care. We know that intuitively the outcome should be a vector of magnitude $2$, pointing counterclockwise, tangent to the circle. To obtain this, you need to differentiate between coordinates and vectors; in polar coordinates they cannot be taken to be equivalent.
In polar coordinates $(r, \theta)$, an infinitesimal displacement $(r,\theta)\to(r+\delta r,\theta+\delta \theta)$ should correspond to a vector $$\delta\vec v = [\delta r, r\cdot\delta\theta],$$ as can be verified by a diagram, or transforming into Cartesian coordinates and back. Therefore,
$$
\frac{\mathrm d}{\mathrm dt}\gamma(t) = [r',r\theta'] = [0, 2\cdot 1]=[0,2].
$$
Its magnitude would then be $2$.
A: Be careful, it's not true that the graph of a curve in polar is given by $(r(\varphi ),\varphi )$. In fact, if a curve has Polair coordinate $(r(\varphi ),\varphi )$, then it's graph is given by $$\{r(\varphi )e_r\mid \varphi \in I\}.$$ Here, $$\Gamma =\{(2,0)=2 e_r+0e_\varphi \mid \varphi \in [0,2\pi]\}.$$
Then $\frac{d}{d\varphi }\gamma (\varphi )=(0,-2)$ and thus $$\text{Length}(\Gamma )=\int_0^{2\pi}2d\varphi =4\pi.$$
