What is the difference between the line integrals $\oint_b\,ds$, $\oint_b\,dx$, and $\oint_b\,x\,ds$?

Question

Can someone explain to me what this means: Where c is the path of a circle with some radius b in clockwise direction.

$\oint_b \,ds$

$\oint_b \,dx$

$\oint_b \,xds$

Like I can't visualise this. Can someone help me?

Answer 1

• The first integral is the arclength of the curve. (Since your curve is a circle of radius $b$, this is just $2\pi b$.)
• The second integral is the line integral of the vector field $\langle x,y\rangle=\langle 1,0\rangle$ along the curve. By symmetry, this is zero.
• The third integral is the "mass" of the curve if the "mass" density per unit length is $x$. (Of course, $x$ won't really be a mass density, because it takes negative values. The point is that, in general, we interpret the integral $\int_c f\,ds$ as the integration of the density of some quantity per unit length.) Again by symmetry, the integral is zero.

Answer 2

Maybe directly computing them with polar coordinates might help.

Note that $ds=rd\theta=bd\theta$ is the arclength element.

$$\oint_b \,ds = \int_{\theta=0}^{2\pi} bd\theta=2\pi b$$

$$\oint_b \,dx = \int_{\theta=0}^{2\pi} d(b\cos\theta)= -b\int_{\theta=0}^{2\pi} \sin\theta d\theta = 0$$

$$\oint_b x\,ds = \int_{\theta=0}^{2\pi} b\cos\theta bd\theta= b^2\int_{\theta=0}^{2\pi} \cos\theta d\theta = 0$$