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Flux of curl

  1. Apr 21, 2010 #1
    1. The problem statement, all variables and given/known data

    Question Attached

    2. Relevant equations



    3. The attempt at a solution

    So here I'm attempting b), I know [tex]\nabla \times \vec F\[/tex] is the curl, which in this case is defined by the matrix

    [tex] \left[ \begin {array}{ccc} x&y&z\\ \noalign{\medskip}{\frac {d}{{\it
    dx}}}&{\frac {d}{{\it dy}}}&{\frac {d}{{\it dz}}}\\ \noalign{\medskip}
    -z&-x&{y}^{2}\end {array} \right] [/tex]

    which gives me the vector [tex]\nabla \times \vec F\ = 2*yi+j-k[/tex]

    Since my given vector is a function of [tex]theta[/tex] and z, I apply the change of coordinates x=rcos(theta), y=rsin(theta), z=r which effectively changes the vector from

    r(theta,z)=(sqrt(z)*cos(theta) , sqrt(z)/2*sin(theta) , z)

    into

    r(theta,r)=(sqrt(r)*cos(theta) , sqrt(r)/2*sin(theta) , r)

    The equation of the integral being [tex]\int\int_S \nabla \times \vec F\cdot \hat n\, dS [/tex]

    so the [tex] \hat n\[/tex] is given by [tex]\hat n = \vec R_\theta \times \vec R_r[/tex] for downward pointing normal.

    Rtheta
    [tex]\left[ \begin {array}{c} -\sqrt {r}\sin \left( \theta \right)
    \\ \noalign{\medskip}1/2\,\sqrt {r}\cos \left( \theta \right)
    \\ \noalign{\medskip}0\end {array} \right] [/tex]

    Rr
    [tex] \left[ \begin {array}{c} 1/2\,{\frac {\cos \left( \theta \right) }{
    \sqrt {r}}}\\ \noalign{\medskip}1/4\,{\frac {\sin \left( \theta
    \right) }{\sqrt {r}}}\\ \noalign{\medskip}1\end {array} \right] [/tex]

    Thus taking the cross product and then yielding the normal

    n=
    [tex] \left[ \begin {array}{c} 1/2\,\sqrt {r}\cos \left( \theta \right)
    \\ \noalign{\medskip}\sqrt {r}\sin \left( \theta \right)
    \\ \noalign{\medskip}-1/4\end {array} \right] [/tex]


    Then I substitute the cylindrical coordinates x=rcos(theta), y=rsin(theta), z=r into my vector F to get.

    [tex] \left[ \begin {array}{c} 2\, \left( r \right) \sin \left( \theta
    \right) \\ \noalign{\medskip}1\\ \noalign{\medskip}-1\end {array}
    \right] [/tex]

    Thus finally, yielding the integral for the flux as.

    [tex]\int _{0}^{2\,\pi}\!\int _{0}^{h}\!{r}^{ 1.5}\sin \left( \theta
    \right) \cos \left( \theta \right) +{r}^{ 0.5}\sin \left( \theta
    \right) +1/4{dr}\,{d\theta}[/tex]

    Just want to know if everything I've done so far is correct(most of it were wild/educated guesses), and the domain for z is confusing me with the h>0, when I'm evaluating the integral, what would I put as h?

    Thank you.
     
    Last edited: Apr 21, 2010
  2. jcsd
  3. Apr 21, 2010 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    except that you forgot the attachment ;)
     
  4. Apr 21, 2010 #3
    Lol I'm so sorry, I was meant to click the add attachment button and went too fast and clicked the submit thread button too quick ><, editing it with all my working so far as we speak.
     
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