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I have seen expressions for 4 different line integrals:

$\int f\text{d}s $

$\int f\text{d}\textbf{s} $

$\int \textbf{F} \boldsymbol{\cdot} \text{d}\textbf{s}$

$\int \textbf{F} \times \text{d} \textbf{s}$

These give results that are scalar, vector, scalar, vector respetively.

Now the second integral I have only come across in one book, and it was only mentioned as a type of line integral. However I am not sure what it means. The first integral I understand as the area of the 'curtain' formed by the scalar field over a path; the third I understand as a summing of the components of the vector field that are parallel to the tangent to the line scaled by the magnitude of the length of the vector elements (which is one for parameterisation by arc length); the final one is the summing of the perpendicular components of the vector field instead.

However what does it mean to have a scalar field integrated over a vector?

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    $\begingroup$ I have edited your LaTeX formatting. Please verify if the new version correctly represents what you meant. $\endgroup$
    – CiaPan
    Commented Feb 28, 2017 at 11:45
  • $\begingroup$ @CiaPan Yes, thank you for editing this and apologies for bad formatting. I typed this on my phone so was unable to see what the post looked like as I was writing. $\endgroup$
    – Meep
    Commented Feb 28, 2017 at 11:59

1 Answer 1

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The second line integral is the summing of all the infinitesimal tangential vectors of a path which are scaled by the value of a scalar field $\,f$.

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If you are still confused, there is another way to understand it.

Suppose the second line integral is calculated in the $xy$-plane, so let $\,{\rm d}s=|\rm d\mathbf s|,$ $\mathbf n=(\rm d\mathbf s)/\rm |d\mathbf s|=(n_x\vec i+n_y\vec j),\,$ then we have

$$\int f\,{\rm d}\mathbf s\,=\,\int f\cdot{\mathbf n}\,{\rm d}s\,=\,\int f\cdot(n_x\vec i+n_y\vec j)\,{\rm d}s$$ $$=\,\left(\int fn_x\,{\rm d}s\right)\vec i+\left(\int fn_y\,{\rm d}s\right)\vec j$$

As you can see, the result is a vector with each component being the first type of line integral. More specifically, there is a 'curtain' formed by '$\,fn_x$' over a path along the $\vec i$-direction, and there is another 'curtain' formed by '$\,fn_y$' over the same path along the $\vec j$-direction.

Hence, in this way, the second line integral is a vector with opponents being areas of 'curtains' formed by different fields over the same path.


By the suggestion from comments, the following is a graph illustration of the component of $\,\int f{\rm d}{\mathbf s}\,$ in the $x$ direction.

Consider a very simple example in which $\,fn_x\equiv1\,$ over a path $\,S:x=y\,$ for $\,0\leq x\leq 1$ $\qquad$(graph is in the link below)

https://drive.google.com/open?id=0B8BsUx4gve5VdUxvWVRZLVdnbzg

Then construct the 'red rectangular curtain' from $\,S\,$ to $\,fn_x\,$, as showed in the graph below:

https://drive.google.com/open?id=0B8BsUx4gve5VWWg1TE1COVpPbU0

Now we have $$\int fn_x\,{\rm d}s=\text{area of the 'red curtain'}=\sqrt2$$

Then we transfer the 'curtain' onto the $x$ axis:

https://drive.google.com/open?id=0B8BsUx4gve5VdXh3YnZpa1ZtTTA

Since the height of the 'rectangular curtain' is $1$, so the width has the same value as the area, and we have the final graph:

https://drive.google.com/open?id=0B8BsUx4gve5VOElPM2R1TF9mam8

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  • $\begingroup$ +1 for the curtain explanation. Could you also post a picture or graph where the x component of the vector has curtain below it in x direction. $\endgroup$
    – Shashaank
    Commented Feb 28, 2017 at 14:42
  • $\begingroup$ Yes, graph is a good idea, but I may need some time to do it. $\endgroup$ Commented Feb 28, 2017 at 14:49
  • $\begingroup$ Yes off course .. $\endgroup$
    – Shashaank
    Commented Feb 28, 2017 at 14:51

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