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Line/surface/volume integrals

  1. Jul 28, 2006 #1
    Lately I have been trying to learn the more intuitive/geometric meaning of integrals. I just have a copule of conceptual questions on different types of integrals and I would really appreciate it if someone would please help me.

    OK, I know this so far:
    [tex]\int_{I}f(x) dx[/tex]
    [tex]\int\int_{R}f(x,y) dxdy[/tex]
    [tex]\int\int\int_{R}f(x,yz) dxdydz[/tex]
    The first two can be interpreted as the area and the volume under the function f(x) and f(x,y), respectively, and the third one can be interpreted as the total mass given that f(x,y,z) represents the mass per unit volume.

    Now what about integrals such as line integrals ([tex]\int_{C}f(x) ds[/tex]), surface integrals ([tex]\int_{S}f(x,y) dS[/tex]) and volume integrals ([tex]\int_{V}f(x,y,z) dV$[/tex]? How do these differ from the other ones that I mentioned?
  2. jcsd
  3. Jul 28, 2006 #2
    The exact same thing but now you have a further constraint that it is on a curve, surface or volume.

    For example, the line integral would trace out the area between the curve and the function f(x).
  4. Jul 28, 2006 #3
    Really? So the surface integral would trace out the volume between the surface and the function f(x,y)?

    What about volume integrals ([tex]\int_{V}f(x,y,z) dV$[/tex]? They probably don't have any "geometric" meaning like the two above do they (since, you know, we can't really visualize a 4-dimension surface)?
  5. Jul 28, 2006 #4
    Well those are my interpretations of it anyway, I'm sure they are probaly wrong...

    Lets see what the experts say.
  6. Jul 28, 2006 #5


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    That's how I always think about it
  7. Jul 29, 2006 #6
    Sort of, but not exactly.

    Part of the confusion is probably the notation you're using. For example, you have an integral of ds with a function f(x). This might sound like a minor quibble, but it can cause confusion.

    Let's think back to what an integral means. You have the intuitive idea of it being the area or volume under a function, but that's not quite the mathematical meaning.

    All an integral is, is a summation of little pieces as the size of the little pieces goes to zero. For example, the 1-D integral is the limit of the riemann sum of rectangles with height f(x) and length dx; effectively the area. Similarly, the 2-D integral is the riemann sum of rectangular prisms with height f(x, y) and length dx, width dy. Etc for 3d integrals

    The basic premise is that you have some region under which you wish to integrate (we are considering definite, proper integrals only), and we chop it into little pieces. Little line segments (length dx) for 1D integrals, little rectangles for 2d integrals (area dx*dy) and so on, which all together form the region of integration (always a line in 1d, circles, rectangles, polygons, etc in 2D)

    Now we consider line and surface integrals. Now our region of integration is a curve or surface. Nothing has changed; all we merely do is chop up the curve or surface into little pieces, evaluate the function, and sum them up. For a curve, it is chopped up into infinitesimally small line segments ds; for a surface, infinitesimally small rectangles. (In this case, s is not a coordinate but rather the parameter to the parametric curve)

    For the analogy to areas and volumes, consider the constant function f(x) = 1. [tex]\int_{a}^{b}f(x) dx[/tex] merely finds the area under f(x) on the region from a to b, effectively finding the length of the region, a - b. Similarly, find [tex]\int_{a}^{b}\int_{c}^{d} f(x, y) dx dy[/tex] which is the volume under the rectangle a to b, c to d. Since f(x, y) = 1, this is the area of the region of integration, or (a - b)*(c - d)

    Now surface integrals. For some curve I, and f(s) = 1, [tex]\int_{I} f(s) d\vec{s}[/tex] finds the function f(s) evaluated for each infitesimally small line segment ds. This is the length of the curve I. See the analogy now? Similar for surface integrals, evaluating the constant function f(s, t) = 1 over a surface gives the surface area of the surface

    Does that make sense? :)
    Last edited: Jul 29, 2006
  8. Jul 30, 2006 #7
    This makes a lot of sense. This is a really sweet explanation, thank you jbusc!:smile:

    OK, I just have a little question on this part. So I see how you can get the length of the curve for line integrals and the area of a surface for surface integrals using f(s)=1 and f(s,t)=1, respectively, as your functions.

    Now, what about the following:

    * If instead of having f(s) and f(s,t) to be unity, you would have made them to be some arbitrary functions and then you would have gotten the area between your curve C and your function f(s) and, similarly, the voulme between your surface S and your function f(s,t), right?

    ** We can think of [tex]\int_{a}^{b} f(x) dx[/tex] and [tex]\int_{a}^{b}\int_{c}^{d} f(x,y) dxdy[/tex] as being a special case of [tex]\int_{I} f(s) ds[/tex] and [tex]\int\int_{D} f(s,t) dS[/tex], respectively, where our "curve" is some flat interval over the x-axis and where our "surface" is some flat rectangular region on the xy-plane, respectively. Am I correct?
    Last edited: Jul 31, 2006
  9. Jul 30, 2006 #8
    Hmm...yeah...sorta. I lied above, actually. Or rather told an incomplete truth. What I said only applies when you have parametrization based on arc length (in crude terms, the distance between f(s) and f(s + ds) = ds)

    The trick is, that s and t are arbitrary parameters, not coordinates. Suppose you have a function r(t) which describes the curve of the flight of a mosquito with respect to time. During it's flight the curve might stop, backtrack, stop, and start going forward again (as mosquitos have erratic flight). In this case the infinitsimal element dt is not related to distance traveled and so what I told you above about the line integral of unity along that path giving arc length is not true and it would give no result of length or area at all (however, it would give you total time elapsed)

    On the other hand, suppose you had a roller coaster. From the air the roller coaster looks like it traces out a closed 2D curve on the surface of the earth (call it C). We have a function, parameterized by arc-length along the ground, f(s) which gives the height of the roller coaster at a given distance s along the roller coaster (again, as measured along the ground, not along the track). If we wish to find the "area" under the roller coaster (perhaps to find the number of columns needed) the line integral over C of f(s)ds would indeed give the area.

    In general, in order to get the "area" as in the roller coaster, you have some conditions to meet. The function f(s) must be parameterized based on the arc length of the curve you're integrating over, not the arc length of the function; second, the function must give the distance to the curve in question.

    True but it might be considered a circular definition since line integrals are defined in terms of regular integrals, vis (thanks to wikipedia): [tex]\int_C f\ ds = \int_a^b f(\mathbf{r}(t)) |\mathbf{r}'(t)|\, dt[/tex] Issues such as this are probably covered better in complex analysis, where line integrals have supreme importance.
    Last edited: Jul 30, 2006
  10. Jul 31, 2006 #9
    Okay, the first thing to understand here is that here we are not concerned about dimensionality. In your first post you were asking about something of the form

    [tex]\int_C f(x) ds.[/tex]

    But here, our curve C is some arbitrary curve. We don't know if it's in [itex]\mathbb{R}[/itex], [itex]\mathbb{R}^2[/itex], [itex]\mathbb{R}^{15}[/itex], or perhaps some arbitrary manifold! All being a "curve" means is that it can be parameterized using only one parameter (ie. if C is in say, [itex]\mathbb{R}^3[/itex], then there's some parameterization [itex]x=f(t), y=g(t), z=h(t)[/itex] of C, where [itex]t[/itex] is just a real parameter running over some interval [itex]I[/itex]). This is the essential problem with just saying that line, surface and volume integrals mean the area/surface/volume between the function and the region in question (though this idea certainly has some merit if you're careful with it).

    Now, the integrand is just some arbitrary function defined on the same space that C is in (at least that's good enough for our purposes here, though not quite general). So again, if C is in, say, [itex]\mathbb{R}^3[/itex], then in general your integral will look something like

    [tex]\int_C f(x,y,z)d\vec{s}.[/tex]

    It's important to understand here that [itex]d\vec{s}[/itex] is not really a differential in the way that you usually think about one. Here [itex]d\vec{s}[/itex] is just a convenient way of writing the "arclength element" of C: if [itex]C[/itex] is parameterized in the way I described above (ie. [itex]x=f(t), y=g(t), z=h(t)[/itex], and assuming that C is smooth [read: x, y, z are differentiable fns of t]), then what we mean by [itex]d\vec{s}[/itex] is actually


    or in other words,

    [tex]|r^\prime(t)|dt = \big|\frac{d}{dt}<x,y,z>\big|dt,[/tex]

    as jbusc indicated. This is the same thing you derived for the arclength formula back in first year calculus.

    Next is to talk about how you should think about these. Let's go back to what jbusc was looking at. If (in the situation above), we have f(x,y,z) identically equal to unity, then our integral is

    [tex]\int_C f(x,y,z)d\vec{s} = \int_C d\vec{s}.[/tex]

    Given what I just explained about [itex]d\vec{s}[/itex], it should be pretty clear to you that this is just the arclength of the curve C.

    So if [itex]f(x,y,z)[/itex] is arbitrary, what happens? Well, the best way to think about it in my opinion is as a weighting function. If [itex]f(x,y,z)=1[/itex], then each infinitesimal arclength segment of C is valued equally in the integral and since [itex]d\vec{s}[/itex] is just the arclength element for C, you just get back the arclength when you take the sum. On the other hand, if [itex]f(x,y,z)[/itex] is some nonconstant function, then you are assigning different "weights" to different parts of C.

    Surface and volume integrals (and you can go further than that of course, to integrating over arbitrary 4-d hypersurfaces and so on) are basically just higher-dimensional analogs to that. The idea of the integrand as a "weighting function" works for them too.
    Last edited: Jul 31, 2006
  11. Jul 31, 2006 #10
    Also note that more generally line/surface/volume integrals do not have to be done wrt the arlength/area/volume element of the region you're integrating over. For example, if you feel like it, you can try to figure out how you'd evaluate something like

    [tex]\int_C f(x,y)dx + g(x,y)dy,[/tex]

    where [itex]C[/itex] is a curve in [itex]\mathbb{R}^2[/itex].

    Once you start learning about complex analysis and (especially) differential forms, these ideas will begin to seem much more natural :smile:
    Last edited: Jul 31, 2006
  12. Jul 31, 2006 #11
    If there's one thing to take away from this discussion, it's that an integral does not automatically mean area or volume. An integral can be completely unrelated to areas and volume - something they don't emphasize enough in intro calculus.

    The key idea: all an integral is, is a summation of little pieces. That's it. Nothing more. A summation of little pieces. Think of the [tex]\int[/tex] as nothing more than a big S for "summation"
  13. Jul 31, 2006 #12
    Well, for now at least (when you're dealing with definite integrals - obviously indefinite integrals are different!). Later on you may find that other interpretations are more useful.

    One thing to try to make sure about when you're doing mathematics is that you don't get "locked in" to one way of thinking about things :smile:
  14. Jul 31, 2006 #13
    Thank you very much jbusc and Data for this enlightning discussion. I'll make sure that I follow your advice(s) whenever I doing calculus. :biggrin:

    Keep up the good work! :cool:
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