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Surface Integral involving Paraboloid

  1. May 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Evaluate the surface integral:

    ∫∫s y dS

    S is the part of the paraboloid y= x2 + z2 that lies inside the cylinder x2 + z2 =4.


    2. Relevant equations

    ∫∫sf(x,y,z)dS = ∫∫Df(r(u,v))*|ru x rv|dA


    3. The attempt at a solution

    I've drawn the region D in the xz-plane as a circle with radius 2. My limits of integration are r between 0 and 2, and t between 0 and 2∏.

    I've parametrized the surface as:
    r(r,t)= <rcos(t), r2, rsin(t)>

    I've gotten the vectors:
    rr=<cos(t), 2r, sin(t)>
    rt=<-rsin(t), 0, rcos(t)>

    Trying to find dS, I did:
    rr X rt = <2r2cos(t), -r, 2r2sin(t)>,
    and the magnitude of that is r√(4r2 +1).

    Overall, I end up with dS = r2*√(4r2+1)drdt.

    The integral becomes:

    ∫∫r4*√(4r2+1)drdt, with limits as mentioned above.

    However this is not the correct answer. I am having a difficult time with these surface integrals and I am not really sure why. I keep getting confused about whether or not I am parametrizing
    the surfaces correctly, so that may be where my error is. Any help would be much appreciated! Thanks
    -Luna
     
    Last edited: May 29, 2013
  2. jcsd
  3. May 30, 2013 #2

    haruspex

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    How did the r2 become r4? Not saying it's wrong, ... this is not an area I'm expert on.
     
  4. May 30, 2013 #3

    HallsofIvy

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    Yes, this is what I get.

    What? [itex]\int\int y dS= \int \int (r^2)dS= \int \int (r^2)(r\sqrt{4r^2+ 1})drd\theta[/itex]
    You should have [itex]r^3[/itex] outside the square root, not [itex]r^4[/itex].

     
  5. May 30, 2013 #4
    I believe it is r4 because:

    My paramatrization has y=r2, so f(r(r,t))=r2

    dS = |rr X rt|dA = r√(4r2 + 1) * rdrdt

    dA = r drdt (because of the relationship between polar and Cartesian coordinates) which is where the r4 comes from.

    Altogether, ∫∫sf(r(r,t))dS = ∫∫Dr2*r√(4r2 + 1) * rdrdt= ∫∫Dr4√(4r2+1) drdt

    This may be helpful: According to the book, the integral equals (∏/60)(391√(17) +1).
     
    Last edited: May 30, 2013
  6. May 30, 2013 #5
    So I tried solving the integral that HallsofIvy thought was correct, and it was! Thanks for the response.

    But that brings a new question to light: why does dA become drdt rather than r*drdt?

    I had the impression that every time we switched from Cartesian to polar, dA became r*drdt, but it seems that in this problem, that's not the case.

    Why?
     
  7. May 30, 2013 #6

    LCKurtz

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    The formula for surface area element for a surface parameterized as ##\vec R = \vec R(u,v)## is ##dS = |\vec R_u \times \vec R_v|dudv##. There is no extra ##u## or ##v##, and in your case, no extra ##r##.

    Suppose your surface had been a circle in the xy plane. You might parameterize it as ##\vec R(r,\theta) = \langle r\cos\theta,r\sin\theta,0\rangle##. If you calculate ##dS = |\vec R_r\times \vec R_\theta|drd\theta## you will find that that cross product is where the familiar ##r## in ##rdrd\theta## comes from. You don't add an extra ##r## when you parameterize it from the beginning.
     
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