In surface integrals the process is to use the dot product of the vector field with the unit normal vector first and multiply that with the area element. In the line integral the process is to use the dot product of the vector field with the length element. Why isn't the unit tangent vector used in the line integral computations while in the surface integral computations the unit normal vector is used?


2 Answers 2


The unit tangent is used. For a surface integral you have:

(field vector) $\cdot$ (unit normal) $\times$ ( scalar area element)

For a line integral you have:

(field vector) $\cdot$ (unit tangent) $\times$ (scalar length element)

often, however, the last product is combined into a vector length element. So the unit tangent is hidden in there.

  • $\begingroup$ when i read the vector field line integral it usually says $F \dot \mathbb{r'}(t)$ without the norm brackets. The norm brackets remind me to cancel out with the unit normal in the surface integral case- in the line integral case then do i just assume it's the numerator of the unit normal left over after cancellation like i usually do with the surface integral case? $\endgroup$ Jun 6 at 22:44
  • $\begingroup$ (unit tangent) x (scalar length) = $r'/|r'| \times |r'|$. $\endgroup$
    – B. Goddard
    Jun 7 at 11:12

In both a line and a surface integral, one must take into account the elementary line element or the elementary surface element associated with the choice of coordinates that one integrates by.

For a line integral however: $\int \mathbf{F}\cdot d\mathbf{r}$, the line element $d\mathbf{r}$ is normally represented as a vector, and in general cannot be "nicely" factored to a constant times the unit tangent vector $\hat{r}$ that points along the path. For example, in cylindrical coordinates we have:

$$d\mathbf{r} = d\rho\hat{\rho} + \rho d\varphi\hat{\varphi}+dz\hat{z}$$

Of course, if for example you are only integrating along a radial path, without varying the angle $\varphi$ or the height $z$ then your integral will reduce to:

$$\int F_{\rho} d\rho$$

Similarly, with a surface integral $\int \mathbf{F}\cdot d\mathbf{s}$ the surface element is in many standard coordinate systems constant for each independent integration variable. For example, again taking cylindrical coordinates, if we're integrating along a fixed radius $\rho$ we have $d\mathbf{s} = dS_{\rho}\hat{n} = \rho d\varphi dz \hat{n}$ so that in this case our surface integral can be written as:

$$\int_{z_i}^{z_f}\int_{\varphi_i}^{\varphi_f} \left(\mathbf{F}\cdot \hat{n}\right ) \rho d\varphi dz$$

However, probably the real answer to your question, is that the difference here is that the "surface normal vector" $d\mathbf{s}$ is usually not written explicitly as a vector, although apparently it is possible to do so (see this interesting post for more details). The only reason I know of for that, is that it usually just complicates things, such as being extra careful about the orientation, and hence signs, etc. Hence the preference is as you said, whenever possible, to express the integrand as a constant times a differential area element, multiplied by the dot product $\mathbf{F}\cdot\hat{n}$.


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