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Homework Help: Line Integrals (I'm missing something here)

  1. Apr 16, 2005 #1
    Ok the question is:

    [tex]\vec F(x,y)=<3xy,8y^2>\,=3xy\,\hat i + 8y^2\,\hat j[/tex]
    [tex]C: y=8x^2[/tex]
    Where [tex]C[/tex] joins [tex](0,0)\,,\,(1,8)[/tex]

    Evaluate [tex]\int_{C} \vec F \cdot d\vec r [/tex]

    I'm unsure how to evaluate this. I really do not want just an answer, I want to know how to solve it.
    My first thought was to check if it is conservative then I can simply let [tex]\vec r(t) [/tex] be a line between the point. Or, (if conservative) I could use the fundamental theorem for line integrals. But:

    [tex]\frac{dP}{dy} \neq \frac{dQ}{dx}[/tex]

    ...so it is not conservative.

    My last idea is that I need to paramterize [tex]y=8x^2[/tex] and this is where I get kind of shaky as far as my abilities with vector calculus. (We just started this chapter). So any insight on this problem, or parameterizing functions would be awesome. Thanks in advance.
    Last edited: Apr 17, 2005
  2. jcsd
  3. Apr 16, 2005 #2

    I'm not familiar with the notation in your first equation. What does it mean?
  4. Apr 16, 2005 #3
    its just some vector notation, it is the same thing as:

    F(x,y) = 3*x*y (i) + 8*t^2 (j)

    where i and j are the unit vectors in the x and y directions
  5. Apr 16, 2005 #4


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    He's just using pointed brackets instead of rounded ones to signify vectors. Okay, my question is, if the vector field F is only a function of x and y, why does its definition have t in one of the components?
  6. Apr 17, 2005 #5
    Well that would be because I made an error!!
    hehe, sorry about that.
    It is fixed now.
  7. Apr 17, 2005 #6
    As far as I remember,

    [tex] \int{F(x)\bullet d\vec{r} = \int{F(y(x))y'(x)dx} [/tex]

    In case I translated it wrong, for us it was
    [tex] \int{F\bullet dr} = \int F(r(t))r'(t) dt [/tex]
  8. Apr 17, 2005 #7


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    Ok. You want to take the dot product of F along this curve with the vector dr. So you can just evaluate the integral of

    [tex]\int_0^1 \vec{F}(x, y = 8x^2)\cdot (dx\vec{i} + dy\vec{j})[/tex]

    taking care to express dy as a function of x and dx.

    What I have done here is I have set x to be the parameter.
  9. Apr 17, 2005 #8


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    Don't use that notation,please...It usually stands for inner product...

  10. Apr 17, 2005 #9
    Who are you talking about?
  11. Apr 17, 2005 #10


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    Wasn't it obvious i was refering to the vicious

    [tex] \vec{F}\left(x,y\right)=\left\langle 3xy,8y^{2}\right\rangle [/tex] ?

  12. Apr 17, 2005 #11
    the < > notation
  13. Apr 17, 2005 #12
    it's easier to write, and im lazy --- so naturally I am ok with your notation :D
  14. Apr 17, 2005 #13
    Ok wow, that was easy. Thanks guys.

    Yeah my book is all about using <,> notation to refer to vectors. I usually just use a matrix represntation but I didn't want to spend the time looking up the latex formatting. Anyways, thanks for the help.
  15. Apr 17, 2005 #14


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    Hmm,then your book doesn't treat vector and preHilbert spaces and uses the [itex] \cdot [/itex] notation for the (euclidean) scalar product.

  16. Apr 17, 2005 #15
    how big is your brain? seriously... does your neck hurt. :)
  17. Apr 17, 2005 #16


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    BTW,the latex formatting is

    [tex] \vec{F}=\left(F_{x},F_{y}\right) [/tex]

    Not too much to look for.

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