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How can I verify the Divergence Theorem for F=(2xz,y,−z^2)

  1. Mar 15, 2017 #1
    1. The problem statement, all variables and given/known data
    Verify the Divergence Theorem for F=(2xz,y,−z^2) and D is the wedge cut from the first octant by the plane z =y and the elliptical cylinder x^2+4y^2=16

    2. Relevant equations
    [tex]\int \int F\cdot n dS=\int \int \int divF dv[/tex]
    3. The attempt at a solution
    For the RHS
    r(u,v)=(4cosu,2sinu,v) where u=[0,2pi] and v=[0,1]
    [tex]\overrightarrow{r}u \times \overrightarrow{r}v =(2cosu,4sinu,0) [/tex]

    [tex]\left \| \overrightarrow{r}u \times \overrightarrow{r}v \right \|=\sqrt{4cos^{2}u+16sin^{2}v} [/tex]
    [tex]\int \int f(r(u,v))\cdot \left \| \overrightarrow{r}u \times \overrightarrow{r}v \right \| dudv [/tex]
    For the Div F . dv
    Div F=1
    What is the limit for the triple integral?
    How can I do the triple integration to verify the divergence theorem?

    thanks
     
    Last edited: Mar 15, 2017
  2. jcsd
  3. Mar 15, 2017 #2

    LCKurtz

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    That last integral should be$$
    \pm\int \int \vec F(r(u,v))\cdot \vec r_u \times \vec r_v~ dudv $$where the sign is chosen appropriate for the outer normal.
    ##u## doesn't go from ##0## to ##2\pi## for the first octant, and ##v## doesn't go from ##0## to ##1##. Also, I assume you are aware that the surface integral(s) must include all four surfaces.

    Have you drawn a picture of the wedge? Since the divergence is ##1## you are just doing a volume integral. You might pick some order of dz, dy, dx and set up a triple integral. Start with a picture.
     
    Last edited: Mar 15, 2017
  4. Mar 16, 2017 #3
    thanks for the reply
    Should u and v be [0,pi/2],[0,4] respectively ?
    thus, the integration is
    [tex]\int_{0}^{4}\int_{0}^{\frac{\pi}{2}} (4cosu,2sinu,v)\cdot (2cosu,4sinu,0) dudv[/tex]
    [tex]=\int_{0}^{4}\int_{0}^{\frac{\pi}{2}} (8cos^{2}u+8sin^{2}u)dudv[/tex]

    For the second part,
    If I use the shperical coordinate to do the integration,
    θ should be [0,pi/2] , ϕ,should be [0,pi/4] ?
    How can I find the limit of p ?
    z=y : psinϕ=psinθcosϕ
     
  5. Mar 16, 2017 #4

    LCKurtz

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    Yes for ##u##, no for ##v##. ##v## is just a rename of ##z## and it never gets as large as ##4##. And its range depends on what ##u## is, which determines ##x## and ##y##. You need a picture.

    No, per above. Also what I have highlighted in red doesn't look like ##\vec F(\vec r(u,v))##.

    Why in the world would you think of spherical coordinates? There is nothing "spherical" about this problem. You have apparently ignored my post where I suggested to draw a picture and set it up in rectangular coordinates. And what about the other three surfaces for part ##1##?
     
    Last edited: Mar 16, 2017
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