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Multivariable Calculus - Surface integrals

  1. Jan 5, 2014 #1
    1. The problem statement, all variables and given/known data ∫∫S xz dS where S is the boundary region enclosed by the cylinder y2 + z2 = 9 and the planes x = 0 and x + y = 5.



    2. Relevant equation∫∫Sf(x,y,z)dS = ∫∫Df(r(u,v)) * |ru χ rv|dA

    3. The attempt at a solution
    I think I have broken this up into 3 surfaces. The bottom circle in the y-z plane. - S1

    The cylinder y2 + z2 = 9 that is intersected by the plane x + y = 5 - S2

    and the slanted ellipsoid resulting from that intersection - S3

    Calculations for S1

    The equation for r for the S1 is r = <0,y,z> - I think,
    ry(the partial derivative of r with respect to y) = <0,1,0>
    rx = <0,0,1>

    rx X(cross) ry = <1,0,0>. The magnitude is 1

    So we integrate the double integral of x * z dy dz = 0 & since x = 0

    Calculations for S2

    The equation for r for the S2 is r = <x,3cos(θ),3sin(θ)> - I think
    rθ = <0,-3sin(θ),3cos(θ)>
    rx = <1,0,0>

    rθ X(cross) rx = <0,3cos(θ),3sin(θ)>. The magnitude is 3

    The double integral becomes 3∫02∏05 - 3cos(θ) x(3sin(θ))dxdθ

    I have a question - do I need to apply a jacobian or something similar because I changed variables from z to θ?

    When I solved this double integral I got a value of 0.

    Calculations for S3

    I am not sure what r is supposed to be but I guessed r = <5-y,y,z>

    ry = <-1,1,0>
    rx = <0,0,1>

    ry χ rx = <1,1,0>. The magnitude is 2.5

    The double integral( without limits) becomes 2.5∫∫xz dydz ->

    2.5∫∫5z - yz dydz

    I am not sure where to go from here. If I convert to polar coordinates I only have one variable that is changing - θ

    Additionally I am not sure how jacobians figure into surface integrals.

    I am asking for help understanding this problem - validating that I am approaching it correctly. I also don't know how to parametrize and solve S3 (not that I am sure of my parametrizations for S1 or S2).


    Thanks
     
    Last edited: Jan 5, 2014
  2. jcsd
  3. Jan 6, 2014 #2

    joshmccraney

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    why are you not using polar coordinates for the entire problem?
     
    Last edited: Jan 6, 2014
  4. Jan 6, 2014 #3

    joshmccraney

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    nope. you are thinking of [itex]dA[/itex], in which case you would need the [itex]r[/itex] as a jacobian. however, your formula directs you to use [itex]dS[/itex], a parametized length, so don't worry about the jacobian.
     
  5. Jan 6, 2014 #4

    joshmccraney

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    well, it seems ideal to use polar coordinates, right??? thus if the "top" plane is [itex]x + y = 5[/itex], should we use coordinates, say [itex]x=r\cos\theta, y=r\sin\theta[/itex]? only remember, our coordinate system isn't around the z axis, its around the x-axis, so some slight modification should be used, which it seems you are aware of.

    so now, we should have [tex] \vec{r}=f(x,y,z)\hat{i}+r\cos\theta\hat{j} +r\sin\theta\hat{k}[/tex] and i think [itex]{f}=x=5-y=5-r\cos\theta[/itex] now perhaps integrate over this flat surface through [itex]r[/itex] and [itex]\theta[/itex] (after taking that messy cross product)

    you of course know when i use [itex]r[/itex] i speak of radius and [itex]\vec{r}[/itex] is in reference to your adopted r notation.
     
  6. Jan 6, 2014 #5

    joshmccraney

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    let me know if this doesn't make sense, but i think it's clear.
     
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