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Green's Theorem in 3 dimensions problem

  1. May 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate: [tex] \int _C{xydx - yzdy + xzdz}[/tex]
    C: [tex]\vec{r}(t) = t\vec{i} + t^2\vec{j} + t^4\vec{k}[/tex]
    o <= t <= 1

    2. Relevant equations

    3. The attempt at a solution

    I understand that you cannot use Green's Theorem in 3 dimensions. How else can I go about solving this?
     
    Last edited by a moderator: May 12, 2009
  2. jcsd
  3. May 12, 2009 #2

    HallsofIvy

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    Re: Curve

    Just use the standard definition of a "line integral".

    On C, x= t, y= t2, and z= t4
    dx= dt, dy= 2tdt, and dz= 4t3.

    xydx- yzdy+ xzdz= (t)(t2)dt- (t2)(t4)(2tdt)+ (t)(t4)(4t3dt.

    Integrate that from t= 0 to t= 1.
     
  4. May 12, 2009 #3
    Re: Curve

    Ohhh snap. I remember talking about the fundamental theorem of line integrals now. Thanks!
     
  5. May 12, 2009 #4

    gabbagabbahey

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    Re: Curve

    [itex]x[/itex], [itex]y[/itex] and [itex]z[/itex] will all be functions of [itex]t[/itex] along the curve....You should be able to simply look at the parametric equation for your curve [itex]\vec{r}(t) = t\hat{i} + t^2\hat{j} + t^4\hat{k}[/itex] and read off what those functions are.:wink:....Then you can easily express [itex]dx[/itex], [itex]dy[/itex] and [itex]dz[/itex] in terms of [itex]dt[/itex] and you integral will simply be a single variable integration...

    EDIT: Halls beat me to it.
     
  6. May 12, 2009 #5
    Re: Curve

    Ok so I got:

    EDIT:
    [tex]\int _0^1{4t^4 dt} = \frac{4t^5}{5} => \frac{4}{5}[/tex] ?
     
    Last edited: May 12, 2009
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