# Integral on the surface of a sphere - course notes

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

## Answers and Replies

tiny-tim
Science Advisor
Homework Helper
Hello Dassinia! Welcome to PF!
… And da= R²*cos dθ dphi z^

No, da is parallel to R, not to the constant z

Hum.. I really don't get it .. !
Thank you.

tiny-tim
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

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 ?

tiny-tim
Science Advisor
Homework Helper
Hi Dassinia!
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 ).

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?

vanhees71
Science Advisor
Gold Member
I'd parametrize $\vec{R}$ 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
$$\mathrm{d} \vec{a}= \hat{R} R^2 \sin \vartheta \mathrm{d} \vartheta \mathrm{d} \varphi$$
with
$$\hat{R}=\hat{x} \cos \varphi \sin \vartheta + \hat{y} \sin \varphi \sin \vartheta + \hat{z} \cos \vartheta.$$
The integration ranges are $\vartheta \in (0,\pi)$ and $\varphi \in (0,2 \pi)$.

Finally I'd also choose the coordinate system such that the polar axis (the $z$ axis in the usual definition of spherical coordinates) in direction of $\vec{r}'$ as given in the original posting. NB: You also have to rethink about the integrand again!

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:
tiny-tim
Science Advisor
Homework Helper
( It's obvious that there's only the z^ component that remains )

correct

(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? )

try a reverse trig-substitution

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..

tiny-tim
Science Advisor
Homework Helper
try cosθ = x, sinθdθ = dx

(and now i'm off to bed :zzz:)

Thank you and good night !

Can't get to the result !
I'm drowning under the calculations, I'm ending up with a really big result !
I give up !

tiny-tim
Science Advisor
Homework Helper
(just got up :zzz:)

did you try cosθ = x ?

show us how far you got

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)

tiny-tim
Science Advisor
Homework Helper
∫ -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!

try substituting r'²+R²-2*r'*R*x = y

Huh ?
So it's ∫-x/(y)1/2 DX ???

tiny-tim
Science Advisor
Homework Helper
uhh?

why didn't you substitute all of it?​

I've tryed sooo many ways, it doesn't work.
So, Thank you for your help tiny-tim !

tiny-tim
Science Advisor
Homework Helper
substitue all of it,

ie also the x on top and the dx …

what do you get?​

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 ?

tiny-tim
Science Advisor
Homework Helper
that's almost completely wrong

what happened to the square-root?

what happened to the x ?​

Oh my god !!!!
I'm sorry, i'm really heedless, I'm shocked by what I wrote !
-(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 ?

tiny-tim
Science Advisor
Homework Helper
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!

(goodnight! :zzz:)