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Flux through oddly shaped cylinder

  1. Dec 9, 2013 #1
    1. The problem statement, all variables and given/known data
    gPaSNs4.jpg


    2. Relevant equations



    3. The attempt at a solution

    Here are my attempt.
    The top is when i tried using cylindrical coordinates. Instead of using dA=(cosθ i +sinθj)rdzdθ
    I used (cosθ i +sinθk)rdydθ since the cylinder is going along the y axis rather than z.
    The bottom is when i tried using dA=(-fx i - fy +k)dx dy

    The oddly shaped cylinder is what's throwing me off.

    1jeQBHT.jpg
     
  2. jcsd
  3. Dec 10, 2013 #2

    pasmith

    User Avatar
    Homework Helper

    I would suggest setting
    [tex]
    x = \frac 12 u \cos v \\
    z = u \sin v
    [/tex]
    for [itex]u \geq 0[/itex], [itex]v \in [0, 2\pi][/itex] so that [itex]4x^2 + z^2 = u^2[/itex].
     
  4. Dec 10, 2013 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You don't seem to have a firm grasp on "surface integrals". For one thing, there is no "r" here. If, as you appear to be doing, you want to use polar (it's certainly what I would do) then to get "[itex]4x^2+ z^2= 1[/itex]", you will need to use [itex]x= (1/2) cos(\theta)[/itex], [itex]z= sin(\theta)[/itex] (with [itex]\theta[/itex] from [itex]-\pi/2[/itex] to [itex]\pi/2[/tex] to keep z positive). That way, when x is squared, the "1/2" squared will cancel the "4". Of course, we can just take y= t, say, from -1 to 1 (or just keep "y" as the parameter).

    With [itex]x= (1/2)cos(\theta)[/itex], [itex]y= t[/itex], [itex]z= sin(\theta)[/itex] then we can write a "position vector" for any point on the surface as [itex]\vec{r}= (1/2)cos(\theta)\vec{i}+ t\vec{j}+ sin(\theta)\vec{k}[/itex] in terms of the two parameters t and [itex]\theta[/itex].

    The derivatives, with respect to t and [itex]\theta[/itex]
    [tex]\vec{r}_t= \vec{j}[/tex]
    [tex]\vec{r}_\theta= -(1/2)sin(\theta)\vec{i}+ cos(\theta)\vec{k}[/tex]
    are tangent to the surface.

    The cross product of those vectors, [itex]cos(\theta)\vec{i}- (1/2)sin(\theta)\vec{k}[/itex], is perpendicular to the surface and gives the "vector differential of surface area", [itex]d\vec{A}= (cos(\theta)\vec{i}- (1/2)sin(\theta)\vec{k})d\theta dt[/itex]

    The integrand vector is [itex]2xy^2z\vec{i}+ 2cos(z)\vec{j}+ 3\vec{k}[/itex] which, in terms of t and [itex]\theta[/itex] is [itex]cos(\theta)sin(\theta)t^2\vec{i}+ 2cos(sin(\theta))\vec{j} + 3\vec{k}[/itex] so that the integral becomes
    [tex]\int_{t= -1}^1\int_{\theta= -\pi/2}^{\pi/2} \left(cos^2(\theta)sin(\theta)t^2- \frac{3}{2} sin(\theta)\right)d\theta dt[/tex]
    [tex]= \left(\int_{-1}^1 t^2dt\right)\left(i\int_{-\pi/2}^{\pi/2} cos^2(\theta)sin(\theta)d\theta\right)- \frac{3}{2}\left(\int_{-1}^1 dt\right)\right(\int_{-\pi/2}{\pi/2} sin(\theta) d\theta\right)[/tex]
     
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