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Surface Integral.

  1. Dec 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate ∫∫ z² dS where S is the part of the surface of the sphere x² + y² + z² = a² with z>= 0


    3. The attempt at a solution

    I get:

    ∫∫ (a² -x² - y²) . (1 + 4a²) a. da.dθ

    where dS = a. da.dθ

    I think I'm making a mistake somewhere, perhaps I'm getting confused with a² and r².



    Thanks.
     
    Last edited: Dec 26, 2009
  2. jcsd
  3. Dec 26, 2009 #2
    First let specify the angle:[tex]\phi=[/tex] angle between x-axis and the position vector [tex]\stackrel{\rightarrow}{r}[/tex] on the x-y plane.
    [tex]\theta=[/tex] angle from z-axis to [tex]\stackrel{\rightarrow}{r}[/tex] .
    [tex]|\stackrel{\rightarrow}{r}|=a[/tex]

    [tex]ds=r^{2}sin(\theta)d\theta d\phi[/tex]

    You have to also change x,y,z into spherical coordinates before you can integrate.

    [tex]x=acos(\phi)sin(\theta),y=asin(\phi)sin(\theta),z=acos(\theta)[/tex]

    Double check my work, it's been a long time!!! But this should give you a starting point before I have to hit the books to verify what I said, and I am drowning in my own books!!! :frown: Bottom line you have to match the coordinates first.
     
    Last edited: Dec 27, 2009
  4. Dec 27, 2009 #3
    I was wrong on [tex]\theta[/tex]

    [tex]\theta=[/tex] angle from z-axis to [tex]\stackrel{\rightarrow}{r}[/tex].

    I corrected in the original post already. Sorry!!
     
  5. Dec 27, 2009 #4
    First, you need to parameterize the surface, remembering that a complete parameterization has bounds for the parameters. One parametrization is:

    [tex]\vec{r}(x,y) = <x, y, \sqrt{a^{2}-x^{2}-y^{2}}>[/tex]

    [tex]-a \leq x \leq a, -\sqrt{a^{2}-x^{2}} \leq y \leq \sqrt{a^{2}-x^{2}}[/tex]

    Do you see how this describes the entire surface?

    Next, compute dS using the definition:

    [tex]dS = |\vec{r}_{x} \times \vec{r}_{y}|dA[/tex]

    Once you find dS, multiply by [tex]z^{2} = a^{2}-x^{2}-y^{2}[/tex] and integrate over the region described by the bounds given in the parameterization (a circle of radius a in the xy-plane). Polar coordinates are probably best here.
     
    Last edited: Dec 27, 2009
  6. Dec 27, 2009 #5

    LCKurtz

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    Yes, you are getting those confused. a is the radius of the sphere. r is the polar or cylindrical coordinate variable which will vary from 0 to a. You have to decide which variables you are going to use. If you want to use cylindrical coordinates you would use:

    [tex] x = r\cos\theta,\ y = r\sin\theta, z^2 = a^2 - r^2[/tex]

    Your surface becomes

    [tex]\vec R(r,\theta) = \langle r\cos\theta, r\sin\theta,\sqrt{a^2-r^2}\rangle[/tex]

    Now to get the correct dS on your surface use:

    [tex]dS = |\vec R_r \times \vec R_\theta|\, drd\theta[/tex]

    Try that substitution.
     
  7. Dec 27, 2009 #6
    Yep, you can do it in either one of the three coordinates, I choose spherical myself because it is most straight forward to me. the small patch of area [tex]ds=asin(\theta)d\phi X ad\theta[/tex].

    Important thing is to translate everything in the equation into one type of coordinate system, you cannot mix it. So it would be either in x-y-z, r[tex]-\phi-\theta[/tex] or [tex]r-\phi-z[/tex].
     
  8. Dec 27, 2009 #7

    LCKurtz

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    Although you have [itex]\phi[/itex] and [itex]\theta[/itex] reversed from the usual notation.
     
  9. Dec 27, 2009 #8

    I tried this but got no where.

    ∫∫ a² cos²θ . a² sinθ.dθ.dØ = a⁴∫∫cos²θ.sinθ.dθ.dØ

    since its a sphere so a=r

    When I integrate w.r.t to θ I get -(1/3). cos³θ

    from 2pi to 0 ..the terms cancel though..
     
    Last edited: Dec 27, 2009
  10. Dec 27, 2009 #9
    I did it in terms of cylindrical coordinates, my working out is the following:

    R_r = [cosθ, sin², -r(a² - r²)^-1/2 ]

    R_θ = [-rsinθ, rcosθ, 0]

    I got the cross product of both of them to be :

    r²(a² - r²)^-1/2 . (cosθ -sinθ) + r

    I then multiplied it by (a² - r²) to get the surface integral:

    ∫∫r².(a² - r²)^1/2.(cosθ -sinθ) + r dr.dθ

    I'm going on the right lines here?
     
  11. Dec 27, 2009 #10
    You use two times integrating from 0 to pi for Ø. A lot of time you integrate half of the surface and X2 for the complete surface.

    Where did you get all the symbols like ∫∫ a² cos²θ . a² sinθ.dθ.dØ = a⁴∫∫cos²θ.sinθ.dθ.dØ??!! I have to type with the special way to get the formula up!!!! Like for Ø, I have to type \phi inside the tex box!!!!
     
  12. Dec 27, 2009 #11
    I get zero when I integrate w.r.t to θ since the terms cancel, hence the whole integration will equate to zero, Im doing something wrong somewhere.

    lol, you can simply copy paste it from MS word or from charmap. I still havent managed to get myself around Latex, it just seems easier.
     
  13. Dec 27, 2009 #12
    It should be the same, instead of integrating θ from 0 to pi, make it 2X integrating from 0 to pi/2. This is a symetrical problem, we can do that.

    a⁴∫∫cos²θ.sinθ.dθ.dØ=2(pi)a⁴∫cos²θ.sinθ.dθ. Let u=cosθ

    The answer I have is [tex]\frac{4\pi a^{4}}{3}[/tex]

    Don't trust my answer, verify that, I am rusty on this!!!
     
  14. Dec 27, 2009 #13
    The correct answer is (2/3)(a⁴)Pi. I also got this by using the limits 0 to Pi/2 for θ as you said, and I used 0 to 2Pi for Ø. This I suppose will give me the surface integral for half the sphere. But how do I determine the limits in the first place. The question only told me to evaluate part of the surface, but didn't specify.

    Thanks!
     
  15. Dec 27, 2009 #14

    LCKurtz

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    I assume you meant sinθ for the second component of Rr. But when you take the cross product of Rr and Rθ you should get a vector. Then you need its magnitude to continue.
     
  16. Dec 27, 2009 #15
    I took your answer X2 to get the whole sphere. You never gave the limit, so I assume it is the complete sphere. If you have a specific limit, then is a different story.

    The question is not complete in your case!!!! I encounter this very problem on my Bessel expansion!!! They just assume you know the boundary condition!!! If what you have is the correct answer, then they must meant half the sphere!!!
     
  17. Dec 27, 2009 #16

    LCKurtz

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    The original post specifies [itex]z\ge 0[/itex].
     
  18. Dec 27, 2009 #17
    Yes, eyes getting too old!!! Then it is the top half of the sphere, no X2 needed.
     
  19. Dec 27, 2009 #18
    Ah! right!, thanks very much!
     
  20. Dec 27, 2009 #19
    See there is advantage of being old, I always have an excuse you young guys don't have......old and forgetful!!!:biggrin:
     
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