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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?

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    for the first one for example, Im taking the lenght of the path to the plane 1? How do I visualise a plane of 1? – 2017-02-28
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    A bit more complicated but https://en.wikipedia.org/wiki/Line_integral#/media/File:Line_integral_of_vector_field.gif – 2017-02-28
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    Do you mind elaborating? I don't really have a field in the first integral so I am confused. – 2017-02-28
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    Evaluating a line integral is like calculating work in physics. At every point you have a vector from a vector field and you are adding each contribution (scalar product) along the path. – 2017-02-28
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    Yeah but I don't really have a vector field right? – 2017-02-28
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    If the integrand function isn't there it just means that it's $1$. So you have a vector field that is uniformly $1$ everywhere. – 2017-02-28
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    You're summing up infinitesimal contributions of a vector field projected onto your contour. In the first case, the vector field is just 1 so you're just summing up differential line segments all along a closed curve and it'll just be the length of the curve. – 2017-02-28
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    Yeah ok that makes sense!! Do you have any ideas for the second and third integrals? I computed those to be 0 and not so sure why :( – 2017-02-28

2 Answers 2

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  • 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.
2

Maybe directly computing them with polar coordinates might help.

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

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

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

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