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Integrating vector functions

  1. Jul 31, 2010 #1
    1. The problem statement, all variables and given/known data
    See figure

    2. Relevant equations

    3. The attempt at a solution

    For an integral like this am I allowed to evaluate the cross product between the to vectors,

    [tex]\vec{u}[/tex] and [tex]\vec{v}[/tex]?

    If so then,

    [tex]\vec{p} = \vec{u}\times\vec{v}[/tex]


    [tex]\int f\vec{p} = -3\int t\vec{p} [/tex]

    Does this work?

    Attached Files:

  2. jcsd
  3. Jul 31, 2010 #2
    under what conditions should you think you cannot?
  4. Aug 1, 2010 #3


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    Yes, it is always true that [itex]\left(a\vec{u}\right)\times\vec{v}= a\left(\vec{u}\times\vec{v}\right)[/itex].

    (But [itex]\vec{u}\times\left(a\vec{v}\right)= -a\left(\vec{u}\times\vec{v}\right)[/itex].)
  5. Aug 2, 2010 #4

    [tex]\int f\vec{p}dt = -3\int t\vec{p}dt = \int \left(-3t^{3} - 3t\right)dt\hat{i} + \int \left(-6t^{2} + 3t\right)dt\hat{j} + \int \left(6t + 3t^{4} \right)dt\hat{k} [/tex]


    [tex]\int f\vec{p}dt = -3\int t\vec{p}dt = \left(\frac{-3}{4}t^{4} - \frac{3}{2}t^{2} \right)\hat{i} + \left( -2t^{3} + \frac{3}{2}t^{2} \right)\hat{j} + \left( 3t^{2} + \frac{3}{5}t^{5} \right)\hat{k} + C [/tex]

    Is this correct?
  6. Aug 2, 2010 #5


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    what about when a = 1...

    may be missing somethig, but I'd say [itex]\vec{u}\times\left(a\vec{v}\right)= a\left(\vec{u}\times\vec{v}\right)[/itex], if you change the order of u & v thats another story
    Last edited: Aug 2, 2010
  7. Aug 3, 2010 #6
    Bump, still looking for a quick check on my answer.
  8. Aug 3, 2010 #7


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    Yes, you are right. For some reason, I was thinking [itex]\vec{u}\times(a\vec{v})= -((a\vec{v})\times\vec{u}= -a(\vec{v}\times\vec{u})[/itex] but now we have to swap u and v back again!

    [itex](a\vec{u})\times\vec{v}= \vec{u}\times(a\vec{v})= a(\vec{u}\times\vec{v})[/itex]
  9. Aug 3, 2010 #8
    Bump, still looking to see if the route I took is the right one.
  10. Aug 4, 2010 #9


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    What part of "yes", in the second response, do you have trouble with?
  11. Aug 4, 2010 #10
    I was just curious whether or not I had done the integration correctly, that's all.
  12. Aug 4, 2010 #11
    Yep, that's right. Just make sure your C is a constant vector, not a real number constant.
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