How to calculate a sink using spherical coordinates

[Addez123]

Homework Statement

$$A = grad(\frac{1} {\sqrt{(x-3)^2+(y+1)^2+z^2}} + xy^3)$$

Calculate the flow out of a sphere with radius 3, centered at (2, 1, 1)

Relevant Equations

Possibly gauss theorem

The issue is that the singularity is not in the center of the sphere.
So how would I calculate it?

I have a few questions:
1. Can I calculate the terms separately like so:
$$A = grad(a+b) = grad(a) + grad(b)$$

2. If I use a spherical coordinate system with the center being at the singularity I can calculate the gradient of first term as
$$grad(a) = -1/r^2 e_r$$
The second term, xy^3, can be calculated with normal coordinates:
$$grad(b) = (y^3, 3xy^2, 0)$$

This creates a few issues though.
I need to convert BOTH those vectors into spherical coordinates with the center being (2, 1, 1).
In grad(b) I recon all I have to do is replace x with $rsin\theta cos\phi - 2$?
Shouldn't I have to apply scale factors since I'm going from normal coordinates to sphericals?

in grad(a) I just have no idea how to do it.
e_r is not the same since I've moved the center of the spherical system.
So I have no idea how to solve it.

I COULD use spherical coordinates with center (2, 1, 1) and just insert those x, y, z values into A and brute force solve the equation but it really doesn't feel like what I'm suppose to do.

P.S. Sorry for posting so much, I have no class to go to or teacher to ask atm.

 

[Orodruin]

Hint: Divergence theorem.

 

[Addez123]

Hint: Divergence theorem.

Yes but how?
With spherical coordinates? If so what orgin (2,1,1) or (3, -1, 0)?

 

[Orodruin]

Yes but how?
With spherical coordinates? If so what orgin (2,1,1) or (3, -1, 0)?

The divergence theorem is coordinate independent.

 

[Addez123]

The divergence theorem is coordinate independent.

So I solve the div A by doing $1/r^2$ and $xy^2$ separatly.
$$div grad(1/r^2) = 2/r^4$$
$$div grad(xy^3) = 2xy$$

Now I have the results in two different coordinates, neither of which represents a spherical coordinate system with the orgin at (2,1,1). So calculating a div across a sphere with the center (2,1,1) will require some for of conversion that I dont know how to do..

 

[Orodruin]

It is 1/r, not 1/r^2.

 

[Addez123]

It is 1/r, not 1/r^2.

True!
$$div grad(1/r) = \frac 1{h_r h_\phi h_\theta} * (\frac d {dr} (\frac {h_\phi h_\theta}{h_r} \frac {dA}{dr}) + \frac d {d_\theta} ...) = 0$$
I excluded the $d/d_\theta$ and $d/d_\phi$ terms since they don't contribute to anything when A only has r components.

Since div = 0 the sink won't contribute to anything, nomatter what coordinate system i use?
In my textbook is says the contribution from the sink is $-4\pi$, so I must've done something wrong still.

 

[Orodruin]

Since div = 0 the sink won't contribute to anything, nomatter what coordinate system i use?

The expression for the divergence in spherical coordinates is only valid for r > 0. You need to treat the singularity of the coordinate system separately.