Why is it that a line integral of a vector field takes the dot product of the vector field with the tangent? This results in us taking the component of the vector field in the direction of the tangent of the curve we are integrating over. This can give us work done as a physical interpretation.

If this curve encloses an area, we can also use Green's theorem over that area instead of the line integral to get the same result. Green's theorem measures flux ALONG the boundary.

Now the analogue to that in a higher dimension is the surface integral of a vector field. However here the definition of the surface integral is taking the dot product of the vector field with the normal (more specifically the cross product of the first partial derivatives), with its interpretation as flux across the surface. Here, we can use the divergence theorem also if the surface encloses a volume. Divergence theorem measures the flux ACROSS the boundary. Why is there a difference?

So why is a surface integral measuring something across the boundary and the line integral something along?

EDIT: We notice the curl is actually a vector field itself.

Looking through some notes, if we take a vector field in two dimensions, we have components $$f = (P,Q,0)$$

Hence, curl is equal to $$curl f.k$$ which is just $$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$$ and an integral with this as its integrand is exactly Green's theorem.

The curl in this case is integrated over a flat surface on the xy plane and the normal to the curl is just the k component (standard basis of the z axis). By applying the definition of surface integrals on Stoke's theorem, we get Green's theorem. This shows that Green's theorem is just a special case of Stoke's theorem.

So the exact definition of the surface integral can be used to result in Green's theorem, and Green's theorem can be obtained directly by considering circulation over an area which links to our line integral, so there is definitely a link here that I can't seem to grasp.

EDIT 2: I would like to add the fact that I mention in my comments. The line integral and surface integral over scalar valued functions makes sense. One could say that the line integral uses a tangential component (the norm of our the derivative of our parametrisation), while the surface integral a normal (the norm of the cross product of the first partial derivatives), but this is only because of its geometrical interpretation. The tangent used in line integral approximates the line, while the norm of the cross product used in surface integrals approximate the surface using the area of a parallelogram - which is exactly the norm of the cross product between arbitrary vectors.

So why is the link in vector valued functions less obvious? What am I missing in my thinking?

  • $\begingroup$ The real question is rather why it's most common that a line integral uses the tangential component of a vector field, while it's most common that a surface integral uses the normal component of a vector field. It's possible for a line integral to use a normal component, and for a surface integral to use a tangential component. $\endgroup$ – md2perpe Jun 10 '17 at 8:07
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    $\begingroup$ Indeed that is really the reason behind this question. If we have a line integral over a scalar valued function, we use the norm of the tangent, while with the surface integral of a scalar valued function, it's the norm of the cross product of the first partial derivatives of the parametrisation. I guess you could say here the line integral uses a tangential component and the surface integral uses a normal component, but only because the surface integral uses that cross product for the area of a parallelogram, to approx the surface, much like we use tangents to approx the line $\endgroup$ – 14tim4 Jun 10 '17 at 8:13
  • $\begingroup$ So here it makes sense why the surface integral uses a normal component and the line integral using the tangential component. In terms of vector valued functions, I don't see why $\endgroup$ – 14tim4 Jun 10 '17 at 8:14
  • $\begingroup$ Maybe one could add, that the integrals are defined, such that they have an physical interpretation. $\endgroup$ – Fakemistake Jun 10 '17 at 9:32

You are mixing up two different things; the surface integral is not a generalization of the line integral.* This is easiest to see in two dimensions, where everything is an integral along curves and yet you will still find a difference.

For simplicity, let's consider a constant wind field blowing to the right, $\mathbf f(x,y)=(1,0)$. Also consider two curves, $A$ a horizontal line segment from $(0,0)$ to $(1,0)$, and $B$ a vertical line segment $B$ from $(0,0)$ to $(0,1)$. In 2D, given a vector field and a curve there are two different kinds of integral you can consider.

  1. Interpret the curve as a wire on which a bead is threaded. If you move the bead from one end to the other, how much does the wind help or hinder the motion of the bead? This is the usual line integral $\int \mathbf f\cdot\mathrm d\mathbf r$. It is large for curve $A$ and zero for curve $B$.

  2. Interpret the curve as a butterfly net being held stationary while the wind blows through it. How much air passes through it per unit time? This is the flux integral $\int \mathbf f\cdot\mathbf n\,\mathrm d\ell$, where $\mathbf n$ is the unit vector perpendicular to the curve tangent. It is zero for curve $A$ and large for curve $B$.

These two notions generalize to higher dimensions in different ways. The line integral remains an integral over a $1$-dimensional object, i.e. a curve. The flux integral becomes an integral over a over an $(n-1)$-dimensional object, i.e. a surface.

*In an abstract sense one could argue that they are both specializations of the same thing, but that will take us too far into the theory of differential forms.

  • $\begingroup$ Ohhh so my mistake was in the way I took the definitions in higher dimensions! Thank you so much, it makes sense! Is it true to say that a surface of an n dimensional function is a level curve of that ie something in n-1 dimensions and then the line stays as a line, a function say $$f(t) = (x_1, ..., x_n)$$ ? $\endgroup$ – 14tim4 Jun 15 '17 at 8:18
  • $\begingroup$ Also, is it correct, that if I find the integral of the curl of a vector field over an n-1 dimensional surface, this will be the same as the line integral over the boundary (of one dimension) of this n-1 dimensional surface, by direct application of Stoke's theorem? Now, say we have an integral of a vector field over another n-1 dimensional surface, which encloses some n dimensional volume, can I apply divergence theorem on this n dimensional volume to get the the integral over the vector field of the n-1 dimensional surface? Sorry, I'm not too sure what to call n dimensional objects. $\endgroup$ – 14tim4 Jun 15 '17 at 8:28
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    $\begingroup$ Re your first comment: Yes, a curve is the image of a function $f:[a,b]\to\mathbb R^n$, and one way to describe an $(n-1)$-dimensional surface in $\mathbb R^n$ is as the level set of a function $g:\mathbb R^n\to\mathbb R$. Such a surface is always closed, though, so you would have to impose additional conditions to restrict it to a boundary. $\endgroup$ – Rahul Jun 15 '17 at 23:57
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    $\begingroup$ Re your second comment: Divergence generalizes easily to higher dimensions, and the divergence theorem indeed holds. However, as far as I know, curl is only defined in three dimensions, so one cannot easily generalize the curl theorem this way. (By the way, the boundary of an $(n-1)$-dimensional surface is $(n-2)$-dimensional.) There actually is a powerful and elegant generalization called the generalized Stokes' theorem, but it requires some advanced mathematical machinery, not just vector calculus. $\endgroup$ – Rahul Jun 16 '17 at 0:05
  • $\begingroup$ Ah, thank you so much! $\endgroup$ – 14tim4 Jun 16 '17 at 0:07

If you really want to understand how Green's Theorem works and similar results such as Divergence Theorem and Stokes Theorem, I suggest you look into the Extended Fundamental Theorem of Calculus, which states that $$\int_C dw=\int_{\partial C} w$$ If you want an easy, intuitive approach, I suggest doing pages 63 and 64 this. Or if you want a slightly more rigorous version, I suggest reading through section 8, "Differential Geometry", of this

  • $\begingroup$ Thank you for the book! It was helpful, and it will be a great read during the summer! $\endgroup$ – 14tim4 Jun 18 '17 at 23:44
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    $\begingroup$ No problem. That's what I'm doing this summer too! $\endgroup$ – Isaac Browne Jun 18 '17 at 23:45
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    $\begingroup$ +1 just for the napkin link ;) $\endgroup$ – An old man in the sea. Jun 12 at 18:10

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