• Support PF! Buy your school textbooks, materials and every day products Here!

Integral on the surface of a sphere - course notes

  • Thread starter Dassinia
  • Start date
  • #1
144
0
Hello,
In my electrodynamics course, there's a "maths" introduction and there's something i don't get !

Homework Statement


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^


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

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
249
Hello Dassinia! Welcome to PF! :smile:
… And da= R²*cos dθ dphi z^
No, da is parallel to R, not to the constant z :wink:
 
  • #3
144
0
Hum.. I really don't get it .. !
Thank you.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
249
(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:
 
  • #5
144
0
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 ?
 
  • #6
tiny-tim
Science Advisor
Homework Helper
25,832
249
Hi Dassinia! :smile:
So da have a r^ , θ^ et phi^ component ?
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:
 
  • #7
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
14,377
5,971
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!
 
  • #8
144
0
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:
  • #9
tiny-tim
Science Advisor
Homework Helper
25,832
249
( It's obvious that there's only the z^ component that remains )
correct :smile:

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

try a reverse trig-substitution :wink:
 
  • #10
144
0
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..
 
  • #11
tiny-tim
Science Advisor
Homework Helper
25,832
249
try cosθ = x, sinθdθ = dx :smile:

(and now i'm off to bed :zzz:)
 
  • #12
144
0
Thank you and good night !
 
  • #13
144
0
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 !
 
  • #14
tiny-tim
Science Advisor
Homework Helper
25,832
249
(just got up :zzz:)

did you try cosθ = x ?

show us how far you got :smile:
 
  • #15
144
0
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)
 
  • #16
tiny-tim
Science Advisor
Homework Helper
25,832
249
∫ -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 …
oooh! :cry:

try substituting r'²+R²-2*r'*R*x = y :smile:
 
  • #17
144
0
Huh ?
So it's ∫-x/(y)1/2 DX ???
 
  • #18
tiny-tim
Science Advisor
Homework Helper
25,832
249
uhh? :confused:

why didn't you substitute all of it?​
 
  • #19
144
0
I've tryed sooo many ways, it doesn't work.
So, Thank you for your help tiny-tim !
 
  • #20
tiny-tim
Science Advisor
Homework Helper
25,832
249
substitue all of it,

ie also the x on top and the dx …

what do you get?​
 
  • #21
144
0
You're telling me to calculate
y=r'²+R²-2*r'*R*x
dy=-2r'*R dx

∫-y/(r'²+R²-2*r'*R) *1/(y*2r'*R ) dy
∫-1/[(r'²+R²-2*r'*R)(2r'*R )] dy ?
 
  • #22
tiny-tim
Science Advisor
Homework Helper
25,832
249
that's almost completely wrong :redface:

what happened to the square-root?

what happened to the x ?​
 
  • #23
144
0
Oh my god !!!!
I'm sorry, i'm really heedless, I'm shocked by what I wrote ! :bugeye:
-(y-r'²-R²)/(2*r'*R)=x
-dy/(2r'*R )=dx

I replace x by -(y-r'²-R²)/(2*r'*R) and in the integral
∫(y-r'²-R²)/(2*r'*R)*1/y1/2*dy/(2r'*R )

and now ?
 
  • #24
tiny-tim
Science Advisor
Homework Helper
25,832
249
if you'd write it out more neatly, you'd see that it's (Ay + B)/√y,

which is just A√y + b/√y, which you can easily integrate! :smile:

(goodnight! :zzz:)
 

Related Threads on Integral on the surface of a sphere - course notes

Replies
7
Views
2K
Replies
1
Views
12K
Replies
2
Views
357
  • Last Post
Replies
1
Views
1K
Replies
4
Views
3K
  • Last Post
Replies
0
Views
2K
Replies
1
Views
2K
Top