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Surface integral

  1. Nov 5, 2005 #1
    I need to evaluate the surface integral of f=x over a semi sphere.

    I know how to evaluate surface integral of a semi sphere but what are my steps in this case. As I found from books I should double integrate f = x with semi sphere limits.
    The problem is that I don't know how to start and what are these limits.
  2. jcsd
  3. Nov 5, 2005 #2


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    Start with the definition right?

    [tex]\int_S\int g(x,y,z)dS=\int_R\int g(x,y,f(x,y))\sqrt{(f_x)^2+(f_y)^2+1}dA[/tex]

    In your case g(x,y,z)=x

    f(x,y) is the surface of the sphere

    S is the region of the sphere to be integrated over

    R is the projection of that region onto the x-y plane.
  4. Nov 5, 2005 #3
    Are you talking about special case when surface is flat?
    If so, I don't think it would work here. Semi sphere is at z>0 with center at (0,0,0) and radius R=2.
  5. Nov 6, 2005 #4
    Just find a parametrization of the sphere, call it P if you want, substitute the correct values of P into F, and multiply that by the magnitude of [tex] (P_{x} \times P_{y}) [/tex], the rest you should know.
  6. Nov 6, 2005 #5


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    No, he's not talking about the surface being flat. That's the reason for having [tex]\sqrt{(f_x)^2+(f_y)^2+1}dA[/tex]. THEN project the surface down onto the xy-plane. On the surface of the sphere, centered at (0,0) with radius 2, [tex]z= \sqrt{4- x^2- y^2}[/tex].
    Even better, in polar coordinates, [tex]z= \sqrt{4-r^2}[/tex]. So
    [tex]\sqrt{(f_r)^2+(f_/theta)^2+1}dA= \frac{2}{\sqrt{4-r^2}rdrd\theta[/tex].
    The integral of [itex]x= r cos(\theta[/tex]over the top surface of the hemisphere is
    [tex]\int_{\theta=0}^{2\pi}\int_{r=0}^2 \frac{2r^2cos(\theta)drd\theta}{sqrt{4-r^2}[/tex]

    Notice that I said "over the top surface". If you are really integrating over the entire surface, don't forget to do the bottom, a circle in the xy-plane, also.
  7. Nov 6, 2005 #6


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    May I make some observations?

    f(x,y,z)=x is a density field symmetric about the x-z plane. That is, for any plane parallel to the y-z plane, the density is constant and is equal to the x-value of that plane. Imagine this plane moving towards the right and beginning to slice through that hemisphere sitting at the origin. When the plane is still in the negative region of x, say x=-1, all those density readings will be -1 right? Now imagine that plane passing through the origin and reaching the point x=1. All the values of f on that plane will be +1. Since the hemisphere is symmetrical about the y-z plane, what then does this say about the value of the surface integral?

    Edit: Actually symmetrical about the x-z plane but analysis still ok.

    So I'll go out on a limb and even before I calculate the integral, suggest that the value will be . . . well what do you think Sibiryk?
    Last edited: Nov 6, 2005
  8. Nov 6, 2005 #7
    I started with sphere parametrization:


    edit: and I tried latex for a first time and it doesn't work.
    Last edited: Nov 6, 2005
  9. Nov 6, 2005 #8
    Do you want to say that would be zero?
  10. Nov 6, 2005 #9


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    Yes, I believe it's zero. Can you set up the double integral using the formula I gave above?

    Edit: Hey Sibiryk, I can't analytically evaluate the integral (even when I switch to polar coordinates); the 1 messes things up.

    The function to be integrated is g(x,y,z)=x.

    The surface over which it is to be integrated is the top half of a sphere which has the equation:


    so the top half is just:



    And the projection of that hemisphere onto the x-y plane is the area inside of a circle of radius 2.
    Last edited: Nov 6, 2005
  11. Nov 6, 2005 #10
    The integral is too complex if I go with Cartesian coordinates. I parameterized sphere:
    x=r*cos(phi)*sin(theta); y=r*sin(phi)*sin(theta); y=r*cos(theta)
    From here I found that dA=(r^2)*sin(theta)
    Function g(x,y,z)=x I alco put in spherical coordinates as
    x=r*cos(phi)*sin(theta) (I'm not sure if I can do that) and plug it into surface double integral. Integration by d(phi) gave me zero because sin(2pi) and sin(0) equals zero.

    What do you think about my solution?
  12. Nov 6, 2005 #11


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    I'm not familiar with that method and I agree that the integral in cartesian coordinates (even polar coordinates) is too complex to evaluate.
  13. Nov 7, 2005 #12
    You parametrized your x and z wrong. Why are you using r? You have a sphere of radius 2, so [tex] x = 2 \sin\phi \cos\theta, z = 2 \cos\phi [/tex].
  14. Nov 7, 2005 #13
    I'll try that but what about y?
    Is my f=x parameterization correct?
  15. Nov 7, 2005 #14
    Your y is fine. Assuming you had set up the limits as 0<theta<2pi, 0<phi<pi/2, your answer should be right.
  16. Nov 7, 2005 #15
    Yes, I set limits as you say.My second problem in homework requires to do the same with volume integral. Same conditions. I also got zero.
    Would you think it will be the same for survace and for volume integrals?
  17. Nov 7, 2005 #16


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    Alright guys. Very clever (from my view anyway). Thanks. Now I know:

    [tex]\int_{0}^{2\pi}\int_0^{\pi/2} \rho^3 Sin^2(\phi)Cos(\theta)d\phi d\theta[/tex]

  18. Nov 7, 2005 #17
    What exactly is your second problem is asking?
  19. Nov 7, 2005 #18
    Evaluate volume integral f=x over sphere interior. Sphere at z>o, center at 0,0,0 and R=2. It looks to me pretty much the same as previous problem.
    It is only one extra integral from 0 to R= 2 for variable r , and dV instead of dA.
    Am I correct?
  20. Nov 7, 2005 #19
    That should be correct, one thing though, replace your radius two in your parametrization to rho or r, whichever. I don't see the book's point of having you do the same problem twice in a row.
  21. Nov 7, 2005 #20
    It's not a book. I think that professor set this up himself. It looks weird and also confuse me. Makes me think that I was doing some thing wrong. I got zero in both cases.
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