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Area element vector for parametric surface integrals

  1. Dec 30, 2013 #1
    When doing surface integrals of surfaces described parametrically, we use the area element dA = ndS = (rv x rw)dvdw

    Where dS is the surface area element and v and w are the parameters.

    I'm fine with the derivation of this (I think) but I don't understand why it's necessary to have n and dS together, as in not just make an expression for n and dS seperately.

    I think I'm misunderstanding something because I thought n = rv x rw because this is a vector perpendicular to rv and rw therefore perpendicular to the surface. But then if this were true, according to the equation for dA at the top that would make dS = dvdw, which isn't necessarily true e.g. if v and w are polars.

    Thanks
     
  2. jcsd
  3. Dec 30, 2013 #2

    HallsofIvy

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    I am not sure what question you are asking. It certainly is possible to "make an expression for n and dS separately"- that is why they are written as different symbols. "n" is the unit normal vector at any point and "dS" is the (scalar) differential of surface area. There is, however, little use for them separately which is why many people (myself included) prefer to write "[itex]d\vec{S}[/itex]" representing the "vector differential of area", a vector perpendicular to the surface whose "length" is the scalar differential of surface area.

    No, n is NOT "r_v x r_w" because n is the unit normal and r_v x r_w does not have unit length.
     
  4. Dec 30, 2013 #3

    vanhees71

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    Well, to put it a bit provocative, I've the opposite problem than BomboshMan. I never understood, why one should split the simple area-element vector as a product of the area element and the surface-normal unit vector, if I don't need this for some purpose. Usually you need invariant surface integrals of the form
    [tex]\int_{A} \mathrm{d}^2 \vec{F} \cdot \vec{V},[/tex]
    where [itex]\vec{V}[/itex] is a vector field anyway, and the area-element vector is indeed given by
    [tex]\mathrm{d}^2 \vec{F}=\mathrm{d} v \mathrm{d} w \frac{\partial \vec{r}}{\partial v} \times \frac{\partial \vec{r}}{\partial w}[/tex]
    The unit vector is, of course given by
    [tex]\vec{n}=\frac{\frac{\partial \vec{r}}{\partial v} \times \frac{\partial \vec{r}}{\partial w}}{\left | \frac{\partial \vec{r}}{\partial v} \times \frac{\partial \vec{r}}{\partial w} \right|}.[/tex]
     
  5. Dec 30, 2013 #4
    I like to make the follow comparation:
    [tex]\\ \int \vec{f}\cdot d\vec{r} \\ \iint \vec{f}\cdot d^2\vec{R}[/tex]
    where:
    [tex]\\ d\vec{r}=(dx,dy,dz) \\ d^2\vec{R}=(dydz,dzdx,dxdy)=d\vec{r} \wedge d\vec{r}[/tex]

    or:
    [tex]\\ \int \vec{f}\cdot \hat{t}ds \\ \iint \vec{f}\cdot \hat{n}d^2S[/tex]
    where:
    [tex]\\ \hat{t}=\frac{1}{\frac{dr}{dt}} \frac{d\vec{r}}{dt} \\ \\ \hat{n}= \frac{1}{\frac{d^2R}{dtds}} \frac{d^2\vec{R}}{dtds}[/tex]
    [tex]\\ds=dr=\sqrt{dx^2+dy^2+dz^2} \\d^2S=d^2R=\sqrt{(dydz)^2+(dzdx)^2+(dxdy)^2}[/tex]
    [tex]\\ds=\frac{dr}{dt}dt=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2}dt
    \\
    \\d^2S=\frac{d^2R}{dtds}=\sqrt{\left(\frac{dydz}{dtds}\right)^2+\left( \frac{dzdx}{dtds}\right)^2+\left(\frac{dxdy}{dtds}\right)^2}dtds[/tex]

    [tex]\frac{dxdy}{dtds}=\frac{\partial(x,y)}{\partial(t,s)}[/tex]
     
    Last edited by a moderator: Jan 2, 2014
  6. Dec 31, 2013 #5

    HallsofIvy

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    I agree completely with you, vanhees71. It makes much more sense to me to use [itex]d\vec{S}[/itex] rather than [itex]\vec{n}dS[/itex].
     
  7. Dec 31, 2013 #6
    me too!
     
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