Triple Integral, Spherical Coordinates

[soofjan]

Homework Statement

Calculate:
integral B [ 1/sqrt(x^2+y^2+(z-a)^2) ] dx dy dz , when B is the sphere of radius R around (0,0,0), a>R.

Homework Equations

The Attempt at a Solution

I tried spherical coordinates for the integrand:
x=rsin(p)cos(t)
y=rsin(p)sin(t)
z=rcos(p)+a
The problem is when I try to find the boundaries of the integral, it gets extremely complicated.
If I try to choose the trivial spherical coordinates to make B easier, then the integrand gets very complicated. So I can't seem to find a way out.

Thanks!

 

[MathematicalPhysicist]

Well if you write it this way:

1/sqrt(x^2+y^2+z^2-2za+a^2) = 1/sqrt(r^2 - 2ra cos (p)+a^2)

and then you have an integral where r from 0 to R, t from 0 to 2pi and p from 0 to pi.

First integrate wrt t, you have an integral of the form:
sin(p)dp/sqrt(r^2-2ra cos(p) +a^2)= -d(cos(p))/sqrt(r^2+a^2-2ra cos(p))
which an integral of the form:
dx/sqrt(A+Bx)
which equals:
2 sqrt(A+Bx)/B

After that you get an integral for r which is of the form:
r^2/r [ sqrt((a+r)^2)-sqrt((r-a)^2)] dr
Now because r<R<a we have sqrt((r-a)^2) = a-r.

Well I think I gave you more than a hint.

 

[lanedance]

not sure how useful it is is but may give some insight
$$\int \int \int_V \frac{1}{\sqrt{(x^2+y^2+(z-a)^2}} dx dy dz$$

so how about letting
$$\vec{r} = \vec{x} + (0,0,a)^T$$

which centres the co-ords on the integrand symmetry, so now we have
$$\int_V dV \frac{1}{ |r|}$$

now consider a function in spherical coords
$$\vec{v} = (v(r),0,0)^T$$

calculating the divergence of v in spherical coords and equating with the integrand gives
$$\nabla \bullet \vec{v} =\frac{1}{r^2} \frac{\partial }{\partial r}(r^2 v(r)) = \frac{1}{r}$$

so choose v(r) = 1/2 and we can re-write the integral using divergence theorem
$$\int_V dV \nabla \bullet \vec{v} = \int_V dV \frac{1}{ r} = \oint_S \vec{v} \bullet \vec{dS}$$

though i think it would get pretty messy to go on..

 

[soofjan]

Well if you write it this way:

1/sqrt(x^2+y^2+z^2-2za+a^2) = 1/sqrt(r^2 - 2ra cos (p)+a^2)

and then you have an integral where r from 0 to R, t from 0 to 2pi and p from 0 to pi.

First integrate wrt t, you have an integral of the form:
sin(p)dp/sqrt(r^2-2ra cos(p) +a^2)= -d(cos(p))/sqrt(r^2+a^2-2ra cos(p))
which an integral of the form:
dx/sqrt(A+Bx)
which equals:
2 sqrt(A+Bx)/B

After that you get an integral for r which is of the form:
r^2/r [ sqrt((a+r)^2)-sqrt((r-a)^2)] dr
Now because r<R<a we have sqrt((r-a)^2) = a-r.

Well I think I gave you more than a hint.

Well, I tried integrating by dp before dr.. it seems to work.
I get:
sqrt{r^2+a^2-2ra cos(p)} / {a*r}
and when I put the limits 0..pi, I get that:
r^2 * {|a+r|-|a-r|} / {a*r}
since a>R>r, I can remove the absolute value and my final answer is:
{4*pi*R^3} / {3*a}

It never occoured to me changing the order of integration will solve this.. thanks a lot!