1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integral on the surface of a sphere - course notes

  1. Sep 8, 2013 #1
    Hello,
    In my electrodynamics course, there's a "maths" introduction and there's something i don't get !
    1. The problem statement, all variables and given/known data
    It says that :
    the integral on the surface of a sphere is
    ∫1/r da = 4πr'/3 with r=|r'-R|, R the vector from the element da to the center.
    r'=r'*z^


    3. The attempt at a solution
    For me, R=R*z^
    So, |r'-R|= (r'²-R²+2r'*R)1/2
    ∫1/r da = ∫1/((r'²-R²+2r'*R)1/2)) da
    And da= R²*cos dθ dphi z^
    And then i'm stuck !
    thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 8, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hello Dassinia! Welcome to PF! :smile:
    No, da is parallel to R, not to the constant z :wink:
     
  4. Sep 8, 2013 #3
    Hum.. I really don't get it .. !
    Thank you.
     
  5. Sep 9, 2013 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    (just got up :zzz:)

    da is parallel to the (outward) normal to any surface,

    so it's parallel to the vector from the centre of this sphere to the surface, ie to R :wink:
     
  6. Sep 9, 2013 #5
    Hi,
    So, |r'-R|= (r'²-R²+2r'*R*cosθ)1/2
    ∫1/r da = ∫1/((r'²-R²+2r'*R*cosθ)1/2)) da
    So da have a r^ , θ^ et phi^ component ?
     
  7. Sep 9, 2013 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Dassinia! :smile:
    Yes.

    That makes da horribly complicated in spherical coordinates (they really aren't designed for adding vectors in different directions :redface:).

    So you'll need to split it into x y and z components, and then integrate over r θ and φ …

    can you see any way of simplifying or ignoring some of the components? :wink:
     
  8. Sep 9, 2013 #7

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    I'd parametrize [itex]\vec{R}[/itex] in spherical coordinates, because these are the natural coordinates for the sphere. The surface normal vector was completely wrong in the original posting. It's
    [tex]\mathrm{d} \vec{a}= \hat{R} R^2 \sin \vartheta \mathrm{d} \vartheta \mathrm{d} \varphi[/tex]
    with
    [tex]\hat{R}=\hat{x} \cos \varphi \sin \vartheta + \hat{y} \sin \varphi \sin \vartheta + \hat{z} \cos \vartheta.[/tex]
    The integration ranges are [itex]\vartheta \in (0,\pi)[/itex] and [itex]\varphi \in (0,2 \pi)[/itex].

    Finally I'd also choose the coordinate system such that the polar axis (the [itex]z[/itex] axis in the usual definition of spherical coordinates) in direction of [itex]\vec{r}'[/itex] as given in the original posting. NB: You also have to rethink about the integrand again!
     
  9. Sep 9, 2013 #8
    Hello,
    So what I have to calculate is :
    ∫∫ R²*sinθ / (r'²+R²-2*r'*R*cosθ^(1/2) dθ dphi z^
    Theta from 0 to pi and Pi from 0 to 2*pi and I should find the good result ? ( It's obvious that there's only the z^ component that remains )
    Thanks !
     
    Last edited: Sep 9, 2013
  10. Sep 9, 2013 #9

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    correct :smile:

    (though it wouldn't hurt to give a brief reason)
    (where is the component factor? :confused:)

    try a reverse trig-substitution :wink:
     
  11. Sep 9, 2013 #10
    Oh yes, I forgot it !
    ∫∫ R²*sinθ*cosθ / (r'²+R²-2*r'*R*cosθ^(1/2) dθ dphi z^ !
    I replaced sinθ*cosθ by 1/2*sin(2θ) but it doesn't help..
     
  12. Sep 9, 2013 #11

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    try cosθ = x, sinθdθ = dx :smile:

    (and now i'm off to bed :zzz:)
     
  13. Sep 9, 2013 #12
    Thank you and good night !
     
  14. Sep 9, 2013 #13
    Can't get to the result ! :frown:
    I'm drowning under the calculations, I'm ending up with a really big result !
    I give up !
     
  15. Sep 10, 2013 #14

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    (just got up :zzz:)

    did you try cosθ = x ?

    show us how far you got :smile:
     
  16. Sep 10, 2013 #15
    Hello,
    x=cosθ
    dx=-sinθ dθ
    cos(0)=1
    cos(pi)=-1 x goes from 1 to -1

    ∫ -x dx / (r'²+R²-2*r'*R*x)1/2 (ignoring for the moment the factor 2π*R²)
    I've tryed integration by parts
    u= x
    u'=1
    v'= 1// (r'²+R²-2*r'*R*x)1/2
    v=-(r'²+R²-2*r'*R*x)1/2/(r'*R)

    u*v= (r'²+R²+2*r'*R)1/2/(r'*R)+(r'²+R²-2*r'*R)1/2/(r'*R)
    -∫u'*v = ∫(r'²+R²-2*r'*R*x)1/2/(r'*R) dx = -4(r'*R)(r'²+R²-2*r'*R*x)3/2/(3*r'*R)
     
  17. Sep 10, 2013 #16

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    oooh! :cry:

    try substituting r'²+R²-2*r'*R*x = y :smile:
     
  18. Sep 10, 2013 #17
    Huh ?
    So it's ∫-x/(y)1/2 DX ???
     
  19. Sep 10, 2013 #18

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    uhh? :confused:

    why didn't you substitute all of it?​
     
  20. Sep 10, 2013 #19
    I've tryed sooo many ways, it doesn't work.
    So, Thank you for your help tiny-tim !
     
  21. Sep 10, 2013 #20

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    substitue all of it,

    ie also the x on top and the dx …

    what do you get?​
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Integral on the surface of a sphere - course notes
  1. Surface of sphere (Replies: 2)

Loading...