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Vector line integral notation.

  1. Mar 10, 2013 #1
    Hey, I'm studying for a physics degree and have a general curiosity about vector calculus. Having learned about surface and line integrals for scalar functions in multivariable calculus I've been having some issues translating them into vector calculus. Though conceptually I haven't had much trouble yet I find myself struggling to interpret some notation.

    My main concern concerns have been with [itex]\vec{dl}[/itex] (I've sometimes seen it written [itex]\vec{dr}[/itex]) and [itex]\vec{ds}[/itex]. I've encountered [itex]dl[/itex] as a scalar when doing line integrals but not as a vector. After much searching I was able to discover that the vector [itex]\vec{ds}[/itex] is equal to [itex]ds\mathbf{\hat{n}}[/itex] where n is the unit vector normal to the surface. But I've still not been able to find such a definition for [itex]\vec{dl}[/itex]. I would appreciate if anyone could shed some light on what this actually is.
     
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  3. Mar 10, 2013 #2

    vela

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    Hmm, can you give an example of a line integral where you didn't have something like ##d\vec{l}## or ##d\vec{r}##?

    There's nothing too mysterious going on here: ##d\vec{r} = dx\,\hat{i} + dy\,\hat{j} + dz\,\hat{k}##. If you have a vector field ##\vec{F} = F_x\,\hat{i} + F_y\,\hat{j} + F_z\,\hat{k}##, then you get ##\vec{F}\cdot d\vec{r} = F_x\,dx + F_y\,dy + F_z\,dz##
     
  4. Mar 10, 2013 #3

    LCKurtz

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    Suppose you have a force field ##\vec F(x,y,z)## and a curve parameterized by ##\vec r(t) =\langle x(t),y(t),z(t)\rangle,\ a\le t \le b##. Since ##\frac {d\vec r}{dt}## is parallel to the curve, if you want to calculate the work done by the force moving along the curve you would calculate the integral$$
    W=\int_a^b \vec F(x(t),y(t),z(t))\cdot \frac{d\vec r}{dt}\, dt
    =\int_a^b \vec F(x(t),y(t),z(t))\cdot \langle \frac{dx}{dt},\frac {dy}{dt}\frac{dz}{dt}\rangle\, dt$$
    This is sometimes written in the differential form, using ##\frac{d\vec r}{dt}dt=d\vec r## and ##\frac{dx}{dt}dt = dx## etc, as ##\int_C\vec F \cdot d\vec r##. You can think of ##d\vec r = \langle dx, dy,dz\rangle##. Whatever notation you use, remember that it means ##\int_a^b\vec F \cdot \frac {d\vec r}{dt}\, dt##.
    For surface integrals, it is a good idea to use capital "##S##" as in ##d\vec S = \hat n\cdot dS## so as not to confuse arc length notation with surface area notation.
     
  5. Mar 10, 2013 #4


    One of the example questions I have is.
    Evaluate [itex]\oint \vec{a}\cdot \vec{dl} [/itex] around the circle [itex]x^{2} +y^{2}=b^{2}[/itex] for [itex]\vec{a}=\frac{\vec{r}}{r^{3}}[/itex]. Where r has its usual meaning is spherical polars.

    It's nothing complicated I was just never taught what [itex]\vec{dl}[/itex] actually is when represented as a vector, and can't find it anywhere in my notes.

    My first instinct was to assume that since [itex]\vec{ds}[/itex] is just [itex]ds\mathbf{\hat{n}}[/itex] then [itex]\vec{dl}[/itex] would just be a scalar line element (which in polars I figured would be [itex]rd\phi[/itex]) multiplied by a vector normal to the contour, ie [itex]\mathbf{\hat{r}}[/itex] but after I did the line integral I tried to verify it using stokes theorem only to find that the curl of a is zero. I'm almost certain this is because of the assumption I made on the definition of the vector dl.
     
  6. Mar 10, 2013 #5

    vela

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    You can think of it similarly to ##d\vec{S}## except that ##d\vec{l}## is tangent, not normal, to the contour and has magnitude ds. It's a little piece of the contour.
     
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