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Divergence Theorem checks

  1. Apr 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Folks, have I set these up correctly? THanks
    Use divergence theorem to calculate the surface integral \int \int F.dS for each of the following

    2. Relevant equations

    [tex] \int \int F.dS=\int \int \int div(F)dV[/tex]

    3. The attempt at a solution
    a) [tex]F(x,y,z)=xye^z i +xy^2z^3 j- ye^z k[/tex] and sigma is the surface of the box that is bounded by the coordinate planes and planes x=3, y=2 and z=1

    Attempt

    [tex]\int_0^3 \int_0^2 \int_0^1 2xyz^3 dzdydx[/tex] where [tex]div (F)=ye^z+2xyz^3-ye^z[/tex]


    b) [tex]F(x,y,z)=3xy^2 i+xe^zj+z^3k[/tex] and sigma is surface bounded by cylinder y^2+z^2=1 and x=-1 and x=2

    Attempt

    [tex]\int_0^{2\pi} \int_0^{1} \int_{-1}^{2} (3r^2) r dxdrd\theta[/tex] where [tex]div(F)=3y^2+3z^2[/tex]

    c)[tex] F(x,y,z)=(x^3+y^3)i+(y^3+z^3)j+(x^3+z^3)k[/tex] and sigma is sphere of r=2 and centre 0,0

    Attempt

    [tex]\int_0^{2\pi} \int_0^{\pi} \int_0^{2} 3p^4 sin(\phi) dp d\phi d\theta[/tex] where [tex]div(F)=3(x^2+y^2+z^2)[/tex]...?

    Thanks
     
  2. jcsd
  3. Apr 21, 2012 #2

    tiny-tim

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    hi bugatti79! :wink:

    yes, they all look fine :smile:

    (was anything worrying you about that? :confused:)
     
  4. Apr 22, 2012 #3
    Thanks. I just wanted to make sure before I proceeded. Here are 3 more trickier ones.

    d) ##F(x,y,z)=x^2 sin(y) i+x cos (y) j-xz sin(y) k## and sigma is the surface x^8+y^8+z^8=8
    Dont know how to tackle sigma. I thought I could manipulate it using the equation of a sphere.



    e) ##F(x,y,z)=x^4 i+x^3z^2 j +4y^2z k## and sigma is the surface of the solid that is bounded by the cylinder x^2+y^2=1 and the planes z=x+2 and z=0

    ##\displaystyle \int_0^{2\pi} \int_0^1 \int_0^{rcos(\theta)+2} (4x^3+4y^3)r dx dr d\theta=##

    ##\displaystyle\int_0^{2\pi} \int_0^1 \int_0^{rcos(\theta)+2} (4r^3cos^3(\theta)+4r^2sin^2(\theta))r dx dr d\theta##

    where ##x=rcos(\theta)##, ##y=rsin(\theta)## and ##div(F)=4x^3+4y^2##


    f)##F(x,y,z)=\frac{r}{||r||}## where ##r=x i+yj+zk## and sigma consists of the hemisphere ##z=\sqrt{1-x^2-y^2}## and the disc and ##x^2+y^2\le1## in the xy plane

    This looks like its going to be tedious...? ##\displaystyle div(F)=\frac{\partial}{\partial x}\frac{x}{(x^2+y^2+z^2)^{1/2}}+\frac{\partial}{\partial y}\frac{y}{(x^2+y^2+z^2)^{1/2}}+\frac{\partial}{\partial z}\frac{z}{(x^2+y^2+z^2)^{1/2}}##...........?

    Comments on above?
     
  5. Apr 22, 2012 #4

    tiny-tim

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    hi bugatti79! :smile:
    erm :redface:

    first, what's divF ? :biggrin:
    (i assume you meant 4x3 + 4y2zk and dz ?)

    yes, that looks ok :smile:
    just keep going, it's quite easy and it'll simplify out in the end :wink:
     
  6. May 11, 2012 #5
    ##div(F)=0##!! What does that mean physically and mathematically?


    Not sure I follow..but should it be

    ##\int_0^{2\pi} \int_0^{1} \int_0^{r \cos \theta +2} (4x^3+4y^3)dzdA## where ##da=r dr d\theta##..?

    Thanks
     
  7. May 11, 2012 #6

    tiny-tim

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    mathematically, it means the integral will be 0

    physically, it means F is a conserved flow
    my "zk" was a copy-and-paste error, it should have stopped at 4x3 + 4y2

    and yes the dx (in dA) should have been dz
     
  8. May 12, 2012 #7
    Thank you sir :-)
     
  9. May 12, 2012 #8

    HallsofIvy

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    You are aware that this is just [itex]\nabla F= 2xyz^3[/itex] aren't you?

    Your integral is just
    [tex]2\left(\int_0^3 x dx\right)\left(\int_0^2 y dy\right)\left(\int_0^1 z^3 dz\right)[/tex]

     
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