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*screams in anger* ok, divergence theorem problem

  1. Oct 14, 2005 #1
    question says answer in whichever would be easier, the surface integral or the triple integral, then gives me(I'm in a mad hurry, excuse the lack of formatting...stuff)

    the triple integral of del F over the region x^2+y^2+z^2>=25
    F=((x^2+y^2+z^2)(xi+yj+zk)), so del F would be 3(x^2+y^2+z^2)?

    In this case, the region is a sphere, it'd be easier to do it with the triple integral over the volume, right? Well, regardless I tried it that way, converting to spherical coordinates(my book mixes up the traditional phi and theta placement, but whatever)

    triple integral of 3r^4*sin(theta)drd(theta)d(phi), and the limits of integration, going from the right integral to the left, 0-5, 0-pi, 0-2pi?

    And somewhere before there is where I messed up 'cuz I can do the integral I have there easily enough and I get like some huge square of 5 times pi, and the answer is 100pi

  2. jcsd
  3. Oct 14, 2005 #2
    is that actually a [itex]<=[/itex] ? (you say the region is a sphere...)
    I don't understand your notation. If you what you meant was [tex]\left( x^3 \hat{i} + y^3 \hat{j} + z^3 \hat{k} \right) [/tex], then your [tex]\nabla \mathbf{\vec{F}}[/tex] is correct. But what you wrote is [tex]\left( x^2 + y^2 + z^2 \right)\left( x \hat{i} + y \hat{j} + z \hat{k} \right)[/tex]
    [tex]=\left( x \left( x^2 + y^2 + z^2 \right) \hat{i} + y \left( x^2 + y^2 + z^2 \right) \hat{j} + z \left( x^2 + y^2 + z^2 \right)\hat{k} \right) [/tex], which has a different divergence.
    Last edited by a moderator: Oct 14, 2005
  4. Oct 14, 2005 #3
    Oh I love you, you're fast
    yes I mean <=
    but yah, the latter is what I wrote and meant, but I'm confused, isn't the (x^2+y^2+z^2) like...if you just had a constant A, times (xi+yj+zk)(or A<x,y,z>)it's be like <Ax,Ay,Az>, so can you not do that with the (x^2+y^2+z^2) out front?
    am I doing something stupid in the midst of night?
  5. Oct 14, 2005 #4
    F=((x^2+y^2+z^2)(xi+yj+zk)), so del F would be 3(x^2+y^2+z^2)?

    I'll throw in the intermediate step I did too,

    so F=<x^3+y^2+z^2,x^2+y^3+z^3,similarforz>

    and the divergence would be the same in that case as <x^3,y^3,z^3>
  6. Oct 14, 2005 #5
    Not quite:

    [tex]\left(x^2+y^2+z^2 \right)(x \mathbf{\hat{i}} \right))=\left(x^3+\mathbf{x}y^2+\mathbf{x}z^2 \right) \mathbf{\hat{i}}[/tex]
    and similarly...
  7. Oct 14, 2005 #6
    Well that was a dumb mistake. Remember kids, sleep is gooood
  8. Oct 14, 2005 #7
    So with that in mind

    how the devil is the answer what it is?
  9. Oct 14, 2005 #8
    because using the correct divergence and going through it I'm getting like 12500pi
  10. Dec 19, 2005 #9
    I hope you're awake after a long :zzz: .:wink: I was having trouble with the exact problem,from Boas' book I presume, and a search for "divergence theorem" brought me here. Anyway, I arrived at the answer given at the back of the book,viz. [tex]4\pi5^5[/tex], by solving the surface integral. You also get it when
    [tex]\nabla.\vec{F} = 3(x^2+y^2+z^2)[/tex],in the case of a volume integral, which is obviously wrong. So did you eventually solve it using the triple integral?

    Anyone's help will be appreciated.
    Last edited: Dec 19, 2005
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