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How to find flux through certain sides of a surface with divergence theorem.

  1. Jan 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Given a vector field [itex]\textbf{F}[/itex] and a composite (with this I mean cuboids, cylinders, etc. and not spheres for example) surface [itex]S[/itex], how do I calculate the flux through only some of the sides of [itex]S[/itex]? I am interested in a general way to do this, but right now I am struggling with the following:

    Let [itex]\textbf{F}=\frac{xz}{x^2+y^2} \textbf{i}+\frac{yz}{x^2+y^2}\textbf{j}+ze^{x^2+y^2}\sqrt{x^2+y^2}\textbf{k},[/itex] defined on [itex]\mathbb{R}^3[/itex]

    and [itex]E\subset\mathbb{R}^3[/itex] be the solid body given by
    [itex]\sqrt{x^2+y^2}\leq z\leq2,\ x^2+y^2\geq 1.[/itex] (This is a sort of hollow stumped cone)

    a) A part of [itex]E[/itex]'s edge is the cylinder part [itex]S_0[/itex] given by
    [itex]x^2+y^2=1, 1\leq z\leq 2[/itex]
    Show that the flux of [itex]F[/itex] inward through [itex]S_0[/itex] is [itex]-3\pi.[/itex]

    2. Relevant equations
    The divergence theorem.


    3. The attempt at a solution
    [itex]div{\textbf{F}}=e^{x^2+y^2}\sqrt{x^2+y^2}-2z\frac{x^2-y^2}{x^2+y^2},[/itex]
    and with [itex]x=\cos{\theta}[/itex] and [itex]y=\sin{\theta}[/itex] we get
    [itex]div{\textbf{F}}=e-2z(1-\sin^2{\theta}).[/itex] Our triple integral then becomes

    [itex]\int_1^2\int_0^1\int_0^{2\pi}re-2rz(1-\sin^2{\theta})\,d\theta\,dr\,dz=2e\pi^2-\frac{3}{2}\pi[/itex].

    And that is how far I've gotten. Assuming all of this is correct, my problem now is that this is the flux through the entire cylinder, along with its top and bottom. How to I calculate the flux through these discs, so that I can subtract them from the total flux?

    Thanks in advance.

    SiggyYo
     
  2. jcsd
  3. Jan 7, 2012 #2

    LCKurtz

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    I didn't check all your work, but the limits on your last integral for r aren't correct. r doesn't go from 0 to 1; it goes from 1 to 2.

    The way you calculate flux through a surface S is to parameterize it:
    ##\vec R(u,v) = \langle x(u,v),y(u,v),z(u,v)\rangle##. Then you use the formula for a flux integral$$
    \iint_S \vec F\cdot d\vec S = \pm\iint_{(u,v)}\vec F(u,v)\cdot \vec R_u\times
    \vec R_v\, dudv$$ where the ##\pm## sign is chosen so that the direction of ##\pm\vec R_u\times\vec R_v## agrees with the orientation. In your problem, for the cylinder part you might use the parameterization$$
    \vec R(\theta,z)=\langle \cos\theta,\sin\theta,z\rangle$$to get started. Then you have two more surfaces to work similarly with their appropriate parameterizations.
     
    Last edited: Jan 7, 2012
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