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Net Flux across a surface

  1. May 22, 2015 #1
    1. The problem statement, all variables and given/known data

    Considering the vector Field F(x,y,z))(zx, zy, z2), and the domain whose boundary is provided by S=S1∪S2 with exterior orientation and
    S1={(x,y,z)∈ℝ3 : z=6-2(x2+y2), 0≤z≤6},
    S2={(x,y,z)∈ℝ3 : z=-6+2(x2+y2, -6≤z≤0}.
    Compute the total flux of F across S.
    2. Relevant equations
    Gauss' Theorem.
    3. The attempt at a solution
    What I have thought to do was to compute ∫Fds across the surface S. For this I applied Gauss' Theorem to obtain that ∫Fds=∫∫∫(∇⋅F) dV
    I would compute ∇⋅F= (∂/∂x, ∂/∂y, ∂/∂z) ⋅ (zx, zy, z2) = 4z
    And now integrate it using a cylindrical change of variable.
    With |J|=r
    For D={0≤θ≤2π, -6≤z≤6, 0 ≤r≤√(z-6)/2} The limits for r are taken from the paraboloid equation.

    For this I get a value of ∫Fds = ∫dθ∫dz∫√(z-6)/2r⋅4z⋅dr = 0 and the respective limits.
    But a collegue states it is 144π as he computed it to be two paraboloids including the lower cover and multiplying that by 2.
    Which result would be the correct one?
    Thank you for your time.
  2. jcsd
  3. May 22, 2015 #2


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    If I am reading your post correctly, the surfaces and vector field are given by:

    $$z = 6 - 2(x^2 + y^2), 0 \leq z \leq 6$$
    $$z = -6 + 2(x^2 + y^2), -6 \leq z \leq 0$$
    $$\vec F(x, y, z) = xz \hat i + yz \hat j + z^2 \hat k$$

    You cannot apply the divergence theorem to this problem, here is a graph to help understand why:

    Screen Shot 2015-05-22 at 11.55.07 AM.png

    The surface might look closed from the outside, but taking a peek inside the boundary:

    Screen Shot 2015-05-22 at 11.56.03 AM.png

    Notice the inside of the surface is completely hollow.

    You cannot apply the divergence theorem here because the interior of the closed surface is hollow.

    I think the surest way to do this problem is to do two surface integrals over each surface, i.e:

    $$\iint_S \vec F(x, y, z) \cdot d \vec S = \iint_{S_1} \vec F(x, y, z) \cdot d \vec S_1 + \iint_{S_2} \vec F(x, y, z) \cdot d \vec S_2$$

    Parametrize each surface ##S_1## and ##S_2## with ##\vec r_1(x, y)## and ##\vec r_2(x, y)##. You know what ##z_1(x, y)## and ##z_2(x, y)## are for each parametrization. Then apply a different theorem you are probably familiar with:

    $$\iint_S \vec F \cdot d \vec S = \iint_{S_1} \vec F \cdot d \vec S_1 + \iint_{S_2} \vec F \cdot d \vec S_2 = \iint_{D_1} \vec F \cdot \left(\vec r_{1_x} \times \vec r_{1_y} \right) \space dA_1 + \iint_{D_2} \vec F \cdot \left(\vec r_{2_x} \times \vec r_{2_y} \right) \space dA_2$$

    You will notice the limits for ##D_1## and ##D_2## are the same in polar co-ordinates.
  4. May 22, 2015 #3


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    The [itex]r[/itex] limit is incorrect; for [itex]z > 0[/itex] you need [itex]0 \leq r \leq \sqrt{(6-z)/2}[/itex] and for [itex]z < 0[/itex] you need [itex]0 \leq r \leq \sqrt{(6 + z)/2}[/itex].

    However it's easier to use [itex]0 \leq r \leq \sqrt{3}[/itex] and [itex]2r^2 - 6 \leq z \leq 6 - 2r^2[/itex]. Then [tex]
    \int \nabla \cdot \mathbf{F}\,dV = 8\pi \int_0^{\sqrt{3}} \int_{2r^2 - 6}^{6 - 2r^2} rz\,dz\,dr = 4\pi \int_0^{\sqrt{3}} r((6 - 2r^2)^2 - (2r^2 - 6)^2) \,dr = 0[/tex] since [itex](6 - 2r^2)^2 \equiv (2r^2 - 6)^2[/itex].

    The flux out of the top surface is [itex]72\pi[/itex]. The flux out of the bottom surface is [itex]-72\pi[/itex]. Your friend has made a sign error. (Note that the third component of the vector field is always non-negative, but the third component of the outward normal of [itex]S_2[/itex] has opposite sign from that of [itex]S_1[/itex].)

    @Zondrina: Your graphs are misleading. [itex]S_1 \cup S_2[/itex] is a closed surface; the intersection of the two surfaces is the circle of radius [itex]\sqrt{3}[/itex] centered at the origin and lying in the plane [itex]z = 0[/itex]. Sketching the curves [itex]z = 6 - 2r^2[/itex] and [itex]z = 2r^2 - 6[/itex] in the [itex]r-z[/itex] half-plane of cylindrical polar coordinates will clarify the situation.
    Last edited: May 22, 2015
  5. May 22, 2015 #4


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    That's a bit confusing. There was a prior thread in the math forum where a user wanted to perform a surface integral of a vector field for the givens: ##0 \leq z \leq H## and ##x^2 + y^2 = a^2##. It was reasoned that because the cylinder was not ##x^2 + y^2 \leq a^2##, the divergence theorem could not be applied (The interior of the surface is not solid).

    The divergence theorem wouldn't work in that situation. Rather you would have to do three surface integrals. Two of them would cancel anyway because of the orientation of the normal vector for the top and bottom of the cylinder.

    I thought in this situation with similar givens the divergence theorem wouldn't apply.

    You could still go the default route by parametrizing the surfaces in any case I'd assume.
  6. May 22, 2015 #5


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    The reason why the divergence theorem didn't work there is because the surface in question was the sidewall [itex]\{x^2 + y^2 = a^2, 0 \leq z \leq H\}[/itex]; the end caps [itex]\{ x^2 + y^2 \leq a^2, z = 0\}[/itex] and [itex]\{ x^2 + y^2 \leq a^2, z = H\}[/itex] were not included. Thus the surface wasn't closed: it had a hole at each end. The OP there used the divergence theorem to calculate the flux out of the volume [itex]\{x^2 + y^2 \leq a^2, 0 \leq z \leq H\}[/itex] but to determine the flux across the sidewall they should then have subtracted the contributions from the end caps, which as I recall is what they forgot to do. (If those contributions cancelled each other then the OP should have said that and justified the assertion. The error was not to address the point at all.)
  7. May 23, 2015 #6
    Thank you very much! :)
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