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What the heck does a line integral mean?

  1. Apr 2, 2005 #1
    Okay, I've searched PF. I actually found a thread that confirmed some of my assumptions. I've searched the web. But I still want to know what the geometric interpretation of a line integral with respect to x (or y) is. The example that made me want to know was [tex] \int y^2 dx + x dy [/tex]; It was integrated over the curve which is actually a line segment that goes from (-5,-3) to (0,2). I got the answer -5/6 by converting both variables into one and using the Jacobian. What bothers me is I don't know what that -5/6 means! It has to mean something.
  2. jcsd
  3. Apr 2, 2005 #2


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    Preliminary note: A vector field [itex]\vec{F}(x,y,z)[/itex] is a function that associates a vector to every point in space. Exemple: [itex]\vec{F}(x,y,z)=x\vec{i}+cosy\vec{j}+e^{xyz}\vec{k}[/itex]

    Geometrically, the line integral means....

    Imagine a vector field [itex]\vec{F}(x,y,z)[/itex], and draw a path C in it. Now put an arbitrary number of points on your curve. Label them 0, 1, 2, 3, ... and draw a vector going from 0 to 1, then another going from 1 to 2, then another going from 2 to 3, etc. Call those [itex]\Delta \vec{r_0}[/itex], [itex]\Delta \vec{r_1}[/itex], [itex]\Delta \vec{r_2}[/itex], etc. respectively Then compute the scalar product between the value of [itex]\vec{F}[/itex] at point 0 and the vector [itex]\Delta \vec{r_0}[/itex], the sclalar product between the value of [itex]\vec{F}[/itex] at point 1 and the vector [itex]\Delta \vec{r_1}[/itex], etc and add all those numbers together. The value of the line integral of [itex]\vec{F}(x,y,z)[/itex] over the path C, noted

    [tex]\int_C \vec{F}\cdot d\vec{r}[/tex]

    is the sum you've just computed, only with the distance between the points 0,1,2,3, etc. taken to be infinitely small, and thus with and infinity of points on the curve C. It can be noted

    [tex]\lim_{max||\Delta \vec{r}|| \rightarrow 0} \sum_{i=0}^{n}\vec{F}(\vec{r}_i)\cdot \Delta \vec{r}_i[/tex]
  4. Apr 3, 2005 #3


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    A line integral is just an integral over a curve. It should be called a "curve integral". Imagine a function of two variables, say [itex] f(x,y)=x^2+y^2[/itex] over the x-y plane in the first quadrant and some f(x) in that plane, say a parabola [itex] x^2[/itex]. Draw a "curved" surface from the parabola up to f(x,y). The area of that surface between two points on the parabola is the line integral of f(x,y) along the line, curve, whatever, of [itex] x^2 [/itex] between the points a and b. Like:

    [tex] \int _{x^2} (x^2+y^2)ds [/tex]

    Note I didn't put the little circle on the integral sign. If the curve I was integrating over was a "closed path" like a circle or whatever, then the circle would indicate "line integral over a closed path".

    Now I have a question for you. What is a "surface integral"? Same dif right, just add another dimension.
  5. Apr 3, 2005 #4


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    Continue thinking of adding up rectangles. [itex]y^2 \, dx[/itex] means to add up rectangles whose heights are [itex]y^2[/itex] and whose bases are infinitessimal segments parallel to the x axis.
  6. Apr 3, 2005 #5


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    How about this one:

    Using the description above I gave for the line integral, suppose we have a function of two variables that's just a sheet above the x-y plane, you know, f(x,y)=5 say. Now I want to take the line integral of this function around the closed path of the circle [itex]x^2+y^2=1[/itex]. That is:

    [tex]\oint_{x^2+y^2=1} 5ds[/tex]

    Now, from what I said, you don't even need to calculate this. It's just [itex] 10\pi[/tex] right, the area of a cylinder of unit radius and length 5. Now, can you calculate the line integral and verify this?

    Think I better verify this for myself. Pretty sure though . . . now where did I put Swokowski . . .

    Ok, I'm sure.
    Last edited: Apr 3, 2005
  7. Apr 3, 2005 #6
    Yes, I verified that the surface area of your cylinder is 10pi via line integrals.

    Thanks guys! I stopped panicking after I figured out that the example I posted had to be related to a vector field. I understood how the cooresponding components of the vector field had to be multiplied by dx, dy, and dz in order to integrate properly. But I was confused when they tried to make us integrate a line integral w.r.t x or y before they introduced the integration over a vector field! I just couldn't see the purpose. Now I do.
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