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Calculating the Flux of F

  1. May 24, 2004 #1
    Calculating the Flux of F (Vector Calculus)

    Let S be the surface of the solid which is enclosed by the cone


    [tex]\begin{align*} z = \sqrt{x^2+y^2} \end{align}[/tex]

    and the sphere [tex]x^2+y^2+(z-1)^2 = 1[/tex], and which lies above the
    cone and below the sphere.

    Let [tex]\begin{align*} \mathbf{F} = xz\mathbf{i}+yz\mathbf{j}-2\mathbf{k} \end{align}[/tex]

    Calculate the flux of [tex]\begin{align*} \mathbf{F} \end{align}[/tex] , outwards through S

    I am currently trying to work on the problem dont know were to approach it from exactly if any1 can help Great Appreciation :)
     
    Last edited: May 24, 2004
  2. jcsd
  3. May 24, 2004 #2
    I am thinking whether they want me to evaluate it as half an orange with a cone missing. any help is greatly appreciated
     
  4. May 24, 2004 #3

    arildno

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    The region is the solid "ice-cream" formed.
    (Letting x,y be fixed, the z-value will have its lowest value on the cone, and its highest value at a point on the spherical shell)

    You should in all probability use the divergence theorem (Gauss' theorem) to evaluate this integral.
     
  5. May 24, 2004 #4

    arildno

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    To help you on a bit, here's a derivation of the planes of intersection between the cone and the sphere:
    [tex]z=\sqrt{1-(z-1)^{2}}\rightarrow{z}^{2}=2z-z^{2}\rightarrow{z}=0,1[/tex]

    Hence, the portion of the sphere directly above the cone is a hemisphere!
     
  6. May 24, 2004 #5
    thanks Arildno
     
  7. May 24, 2004 #6
    umm Arildno do i have to do it in TWO parts? like asin Triple Integral over V1 UNION Triple Integral over V2???
    becuase of the fact that its a ICECREAM
     
  8. May 24, 2004 #7

    arildno

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    Not at all!
    Let z lie between the values:
    [tex]\sqrt{x^{2}+y^{2}}\leq{z}\leq{1}+\sqrt{1-x^{2}-y^{2}}, (x^{2}+y^{2}\leq{1})[/tex]
     
  9. May 24, 2004 #8
    ummmm Arildno, dunno mayb i'm not a quick learner..

    the DIV F i got to be 2z
    umm you have told me the bounds of integration for Z but for X and Y? like can you draw down the triple integral i need to evaluate, Appreciate it
     
  10. May 24, 2004 #9

    arildno

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    Well, I wrote the bounds down:
    [tex]x^{2}+y^{2}\leq{1}[/tex]
    Intgrating 2z between limits is easy (z^{2} evaluated on the given limits),
    while the disk in the x-y plane is most easily evaluated by polar coordinates.
    Hence, you get to evaluate the double integral:
    [tex]\int_{0}^{2\pi}\int_{0}^{1}((1+\sqrt{1-r^{2}})^{2}-r^{2})rdrd\theta[/tex]
     
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