how do we interpret this integral from polar co-ordinates

Question

$$\text{Find } \int_C rdr$$

Where $C$ is any closed loop.

I feel that the answer is zero, i have no hard reasoning.

Here $r$ is the parameter from the polar coordinates.

Answer 1

There is a corollary of Stokes' theorem that would be useful to apply here:

$$\oint_{\partial\Sigma}f\,\mathrm{d}\mathbf{r}=\iint_{\Sigma}\mathrm{d}\mathbf{S}\times\nabla f$$

Using the formula for the gradient of a scalar field $f(\rho,\phi,z)$ in cylindrical coordinates, the gradient of $f(\rho)=\rho$ is

$$\nabla f(\rho)=\hat{\rho}\frac{\partial f}{\partial \rho}=\hat{\rho}$$

Assuming the closed loop lies in the $xy$-plane, the surface normal will be $\hat{n}=\hat{z}$. Also, suppose that the origin lies in the region bounded by the closed loop. Noting that $\hat{z}\times\hat{\rho}=\hat{\phi}$, the integral becomes:

$$\begin{align} \oint_{\partial\Sigma}\rho\,\mathrm{d}\mathbf{r}&=\iint_{\Sigma}\mathrm{d}\mathbf{S}\times\nabla \rho\\ &=\iint_{\Sigma}\mathrm{d}S\,\hat{\phi}\\ &=\hat{y}\iint_{\Sigma}\mathrm{d}S\,\cos{\phi} - \hat{x}\iint_{\Sigma}\mathrm{d}S\,\sin{\phi}\\ &=0 \end{align}$$

For any closed loop around the origin, the surface integrals will integrate $\phi$ over an interval of length $2\pi$, which is a full period for the trigonometric functions being integrated; hence the reason they vanish identically.

Answer 2

If you are certain that the integral you are interested in is $$ \int_{C} r\ \text{d}r$$ and not $$ \int_{C} r\ \text{d}{\bf{s}} \quad \Bigl(\text{or} \int_{C} r\ \text{d}\bf{r} \Bigr)$$ as I suggested in the comments then the answer to your question is obviously zero.

$$ \int_{C} r\ \text{d}r = \Bigl[\frac{1}{2}r^{2}\Bigr]_{\text{start point}}^{\text{end point}} = 0$$

since the start and end point on a closed loop are the same.

(Also, this closed loop we are imagining must actually just be a line at fixed angle $\theta$ moving first outwards, and then back inwards along the same straight line.)

(If you are actually interested in the second integral than you should use Stokes' Theorem, but be careful not to confuse $r$ from polar coordinates with the $\text{d}\bf{r}$ used to parametrise arclength.)