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A lil help please?

  1. Jul 25, 2004 #1
    ok this probley seems simple but i just need to see how to do it, ok well how do u evaluate this...
    find the flux of
    the vector field...
    [tex]\vec{F}=<x,y,z>[/tex]

    throught this surface above the xy-plane..
    [tex]z = 4-x^2-y^2[/tex]

    how do u evaluate this with surface integrals method and the divergence therom

    [tex]dS = \sqrt{f_{x}^2+f_{y}^2+1}dA[/tex] for surface inergrals

    thanks very much to anyone who can show me this step by step, i keep getting the wrong answer and i don't know why...
     
    Last edited: Jul 25, 2004
  2. jcsd
  3. Jul 25, 2004 #2
    Hi phy_math,

    just to make sure I understand you, is this right:

    [tex]\begin{array}{l}
    \vec F = \vec x = x\hat x + y\hat y + z\hat z \\
    f(x,y,z) = z - (4 - x^2 - y^2) = 0\\
    \end{array}[/tex]

    Since then ...

    [tex]dS = dxdy\sqrt {(\frac{{\partial f}}{{\partial x}})^2 + (\frac{{\partial f}}{{\partial y}})^2 + 1}[/tex] makes sense

    ... where you can replace [itex]dA = dxdy[/itex].


    To get the total flux of [itex]\vec F[/itex] through the surface, you want the integral [itex]\phi = \int\limits_S {\vec F \bullet \hat ndS} [/itex], where [itex]\hat n = n_x \hat x + n_y \hat y + n_z \hat z[/itex] is the unit normal to the surface.

    So, to do this integral, you'll need to calculate the unit normal to the surface. You've already got [itex]\vec F[/itex] and can calculate [itex]dS[/itex] from the above formula.

    Have you seen the equation [itex]\nabla f \bullet \hat n = 0[/itex] before?

    Also, if you're calculating this flux over the whole of space remember that, by definition, you can write

    [tex]\int\limits_0^\infty {\int\limits_0^\infty {f(x,y)dxdy} } = \begin{array}{*{20}c}
    {\lim } \\
    {X,Y \to \infty } \\
    \end{array}\int\limits_0^Y {\int\limits_0^X {f(x,y)dxdy} }[/tex]

    That might help as well. See how you get on.
     
    Last edited: Jul 25, 2004
  4. Jul 25, 2004 #3
    Correct me if I am wrong, I just learned this last week.

    By the divergence theorem, [tex]\int_S \vec F \cdot \vec n \ dS = \int_V \nabla \cdot \vec F \ dV[/tex]

    Since [tex]\vec F = x \vec i + y \vec j + z \vec k[/tex], then [tex]\nabla \cdot \vec F = 3[/tex].

    A quick sketch of the surface shows a inverted parabolid. This volume should be easy to find via cylindrical coordinates.

    [tex]flux=\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{4-r^2} 3r \ dz \ dr \ d\theta[/tex]

    Side note: Does anyone know if there is a shortcut to writing the "[*tex]...[*/tex]" tags?
     
  5. Jul 25, 2004 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    By the way, a simpler way to find [itex]\int\limits_S {\vec F \bullet \hat ndS} [/itex]
    without calculating n and dS separately is this:

    The surface is given by z- (4- x2- y2)= 0 or z+ x2+ y2= 4, a constant. If we take F(x,y,z)= z+ x2+ y2 then the surface is a "level surface" of F and so grad F= <2x, 2y, 1> is perpendicular to the surface and n dS is <2x, 2y, 1>dxdy.

    Notice that we have "dropped down" to the xy-plane. If the z-component of grad F had not been 1, we would have had to make it 1.

    For example if the surface is x2+ y2+ z2= R2 (a sphere) we would take F(x,y,z)= x2+ y2+ z2 so that grad F= <2x, 2y, 2z>. Now we have to decide in what plane we want to do the integration (with a sphere it really doesn't matter!). If we choose to integrate in the xy-plane, then we divide the entire vector by 2z to "normalize" the z-component: <x/z, y/z, 1>dx dy. Of course, we would have to divide the problem into two parts, z positive or z negative, to integrate over the entire sphere.
     
  6. Jul 26, 2004 #5
    thanks all, its good to know ur all here to help! ya i did similar things, turns out i was just basicly using the wrong limits for the surface thanks everyone!
     
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