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Calculating flux using surface integrals.

  1. Jul 17, 2011 #1
    This isn't homework. I've been restudying vector calculus from the beginning to end on my free time and got stuck on this problem. I am not sure what I'm doing wrong, but it may be a calculation error since it has so much calculation involved.

    1. The problem statement, all variables and given/known data
    Evaluate the surface integral [itex]\int[/itex][itex]\int[/itex][itex]_{\Sigma}[/itex]f*d[itex]\sigma[/itex], where f(x, y, z) = x2i + xyj + zk and is the
    part of the plane 6x + 3y + 2z = 6 with x ≥ 0, y ≥ 0, and z ≥ 0, with the outward unit
    normal n pointing in the positive z direction.


    2. Relevant equations
    [itex]\int[/itex][itex]\int[/itex][itex]_{\Sigma}[/itex]f*d[itex]\sigma[/itex] = [itex]\int[/itex][itex]\int[/itex][itex]_{R}[/itex]f*n*d[itex]\sigma[/itex], where
    d[itex]\sigma[/itex] = |[itex]dr/du[/itex] X [itex]dr/dv[/itex]|dudv, and r is a vector parametricized for the surface.

    n = outward unit normal vector from the surface.

    3. The attempt at a solution
    First, I parametricized the surface as a vector in terms of u and v:
    x = u
    y = v
    z = 2 -3x - 1.5y
    [itex]\overline{r}[/itex](u,v) = <u, v, 2-3u - 1.5v>
    R is the region on the u, v plane under the line v = 2 -2u, u≥0, v≥0
    Took partial derivatives in terms of u and v:
    d[itex]\overline{r}[/itex]/du = <1, 0, -3>
    d[itex]\overline{r}[/itex]/dv = <0,1,-1.5>

    Calculated |[itex]dr/du[/itex] X [itex]dr/dv[/itex]|:
    i j k
    1 0 -3
    0 1 -1.5 =
    |3i + 1.5j + k| = (9 + 2.25 + 1)^1/2 = 3.5

    Calculated n:
    Since the plane is 6x + 3y + 2z = 6, perpendicular to the plane would be the vector v = <6, 3, 2>, so n would be v/(|v|), the vector divided by its magnitude.
    n = <6, 3, 2>/|7| = <6/7, 3/7, 2/7>

    Calculating f(u,v) * n:
    f(u,v) = <(u^2), (uv), (2 -3u - 1.5v)>* <6/7, 3/7, 2/7> =
    1/7 (6u^2 +3uv + 4 - 6u - 3v)

    Plugging into equation:
    [itex]\int[/itex][itex]\int[/itex][itex]_{R}[/itex]f*n*d[itex]\sigma[/itex] R=[(u,v): 0<u<1, 0<v<2-2u]

    [itex]\int[/itex][itex]\int[/itex][itex]_{R}[/itex] 1/7 (6u^2 +3uv + 4 - 6u - 3v) *(3.5) *dvdu =

    1/2 [itex]\int[/itex][itex]\int[/itex][itex]_{R}[/itex]6u^2 + 3uv -6u - 3v + 4 dv du =
    1/2 [itex]\int[/itex]6u^2v + 3uv^2 -3v^2/2 + 4v (evaluating v from 0 to 2-2u) dv

    Using wolfram alpha to evaluate (got tired of doing it by hand like 6 times XD) :
    1/2 [itex]\int[/itex]24u^3 + 30u^2 + 8u + 2 du from u = 0..1
    1/2 * (24u^4/4 + 30u^3/3 + 8u^2/2 + 2u) from 0 to 1
    1/2 (6 + 10+4+2) = 11

    The answer should be 15/4 though. I'm not sure what I did wrong, maybe I missed something. Can anyone see what I did wrong in this problem? Thanks much.
     
    Last edited: Jul 18, 2011
  2. jcsd
  3. Jul 17, 2011 #2

    vela

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    I don't know if it's just a typo, but you dropped the term -6u from the integrand on the step where you simplified 3.5/7 to 1/2 and reordered the terms.

    EDIT: I get 7/4 as well.
     
    Last edited: Jul 17, 2011
  4. Jul 17, 2011 #3
    Oops. I redid it and got 13/2 with the -6u term. It came out to 13/2 though. :(
     
  5. Jul 18, 2011 #4
    If I remember correctly, since the surface Σ is closed, the divergence theorem applies right?

    ∫∫Rf*n*dσ = ∫∫∫div(f) dV

    x is from 0 to one
    y is from 0 to the line -2x + 2
    z is from 0 to the plane 3 - 3x -3y/2

    f(x, y, z) = x2i + xyj + zk
    div(f) = d/dx i + d/dy j + d/dz k * x^2i + xyj + zk
    = 2x + x + 1 = 3x+1

    ∫∫∫div(f) dV = ∫∫∫3x+1 dzdydx
    ∫∫3xz+z | z = 0..3-3x-3y/2 dy dx
    ∫∫3x(3-3x-3y) + 3-3x-3y -[ 0] dydx
    ∫∫9x -9x^2 - 9xy +3 -3x - 3y dy dx
    ∫∫6x - 9x^2 -9xy +3 -3y dy dx
    ∫6xy - 9x^2y - 9xy^2/2 + 3y - 3y^2/2 | y = 0..-2x+2 dx
    ∫6x(-2x+2) - 9x^2(-2x+2) -9x(-2x+2)^2/2 +3(-2x+2) -3(-2x+2)^2/2 dx
    ∫-12x^2 + 12x +18x^3-18x^2-(9x/2) (4x^2 -8x + 4) -6x + 6 -3(4x^2 - 8x + 4) dx=
    ∫-12x^2 + 12x + 18x^3 -18x^2 -36x^3/2 + 72x^2/2 -36x/2 + 6 -12x^2 +24x - 12 dx=
    -12x^3/3 + 12x^2/2 + 18x^4/4 -18x^3/3 -36x^4/8 + 72x^3/6 - 36x^2/4 +6x -12x^3/3 +12x^2 -12x from x =0..1
    -12/3 + 12/2 + 18/4 -18/3 -36/8 +12 -9 +6 -4 + 12-12 =1


    :\ I'm frustrated with this problem.
     
  6. Jul 18, 2011 #5

    vela

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    You don't have a closed surface. You're calculating the flux only on one plane. The volume integral of the divergence would equal the total flux through the plane and the other three sides of the tetrahedron.
     
  7. Jul 18, 2011 #6
    Hmm.. But the problem states "...part of the plane 6x + 3y + 2z = 6 with x ≥ 0, y ≥ 0, and z ≥ 0, with the outward unit normal n pointing in the positive z direction." That doesn't mean that the flux is the net flux of the closed surface, but the flux of only that portion of the plane?
     
  8. Jul 18, 2011 #7

    vela

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    You solved for z incorrectly. It should be z = 3 - 3x - (3/2)y = 3 - 3u - (3/2)v.
    All this is fine.
    You propagated the error from above, but your method is okay. With the correct expression for z, you would get
    [tex]\frac{1}{2}\int_0^1\int_0^{2(1-u)} (6u^2+3uv+6-6u-3v)\,dv\,du = \frac{7}{4}[/tex]
     
  9. Jul 18, 2011 #8

    vela

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    Yes, I think so, otherwise it wouldn't specify which direction the normal points in.


    EDIT: It turns out you get the same answer, 7/4, because the vector field vanishes on the other three faces, so there's no flux through them.
     
  10. Jul 18, 2011 #9

    vela

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    You dropped the factor of 1/2 on the y term.

    I would pull the (3x+1) out front since it's a constant for the y and z integrations: [tex]\int_0^1 (3x+1)\int_0^{2(1-x)} \int_0^{3-3x-3/2 y}dz\,dy\,dx[/tex]It'll simplify the algebra a bit. After you integrate over z, you'll have[tex]\int_0^1 (3x+1)\int_0^{2(1-x)} (3-3x-3/2 y)\,dy\,dx = \int_0^1 3(3x+1)\int_0^{2(1-x)} [(1-x)-y/2]\,dy\,dx[/tex]
    If you keep things in terms of 1-x, I think the result of the y integral will be pretty simple. To do the x integration, you'll just have to bite the bullet and multiply it out, or try the substitution u=1-x.
    Either way, you should get 7/4 for the answer, so I wouldn't bother beating your head against the wall to try get 15/4. That answer is apparently wrong.
     
  11. Jul 19, 2011 #10
    You're right. I should just try the problems on my other calculus book. This book was a free online book which I think didn't have much revisement. Thanks much for your help. :]
     
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