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Finding a surface form the intersection of two surfaces- Stokes' Thrm.

  1. May 31, 2013 #1
    1. The problem statement, all variables and given/known data

    Let [itex]\vec{F}=<xy,5z,4y>[/itex]

    Use Stokes' Theorem to evaluate [tex]\int_c\vec{F}\cdot d\vec{r}[/tex]

    where [itex]C[/itex] is the curve of intersection of the parabolic cylinder [itex]z=y^2-x[/itex] and the circular cylinder [itex]x^2+y^2=36[/itex]

    2. Relevant equations

    Stokes' Theorem, which says that [tex]\int_c\vec{F}\cdot d\vec{r}=\int\int_s ∇×\vec{F}\cdot d\vec{S}[/tex]

    3. The attempt at a solution

    Because we are in the chapter of Stokes' Theroem, I am supposed to integrate over a surface that is defined by the intersection of the two surfaces given in the problem. However, I am having a bear of a time coming up with what that surface would look like, and what an equation for it would be.

    I tried plugging the equations into a LiveMath module to see the intersection, but it would break as soon as I zoomed out far enough to see what I needed.

    When I have to do this with other problems (intersection of a parabloid and a cylinder), I can look at the graph, or just imagine it, and fairly quickly see the general shape, and then do some algebra to get the coefficients.

    If I could have some help coming up with the equation of that surface, I would be able to do some dot-and-cross products and be on my merry way.

    Thanks!
     
  2. jcsd
  3. May 31, 2013 #2

    LCKurtz

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    It is the bounding curve that is defined by the intersection of the surfaces. But the surface that is bounded by that curve is just your ##z = y^2 - x## parabolic surface. So you have that slanted parabolic surface whose domain in the xy plane is the the disk ##x^2+y^2\le 36##.

    You could set up the surface integral in terms of x and y and change the resulting integral to polar coordinates or parameterize the parabolic surface in terms of ##r,\theta## in the first place. Does that get you going?
     
    Last edited: May 31, 2013
  4. May 31, 2013 #3
    Yes, that was perfect. I used the surface [itex]z=g(x,y)=y^2-x[/itex], and got a normal vector from [itex]<-\frac{\partial g} {\partial x}, -\frac{\partial g}{\partial y}, 1> [/itex]. Did my curl and dot product calculations to get
    [tex] \int\int_d <1,-2y,1>\cdot<-1,0,-x>dA=\int\int_d -x-1dA[/tex]
    I converted to polar using
    [tex]x=rcos(\theta)[/tex]
    [tex]y=rsin(\theta)[/tex]

    and finally had an integral of
    [tex]\int_0^{2\pi}\int_0^6 (-1-rcos(\theta))rdrd\theta[/tex]

    Thanks so much for your help!
     
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