How to calculate a sink using spherical coordinates

In summary: 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.
  • #1
Addez123
199
21
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.
 
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  • #2
Hint: Divergence theorem.
 
  • #3
Orodruin said:
Hint: Divergence theorem.
Yes but how?
With spherical coordinates? If so what orgin (2,1,1) or (3, -1, 0)?
 
  • #4
Addez123 said:
Yes but how?
With spherical coordinates? If so what orgin (2,1,1) or (3, -1, 0)?
The divergence theorem is coordinate independent.
 
  • #5
Orodruin said:
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..
 
  • #6
It is 1/r, not 1/r^2.
 
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Likes Addez123
  • #7
Orodruin said:
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.
 
  • #8
Addez123 said:
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.
 

Related to How to calculate a sink using spherical coordinates

1. How do I convert Cartesian coordinates to spherical coordinates?

To convert from Cartesian coordinates (x, y, z) to spherical coordinates (r, θ, φ), use the following formulas:

r = √(x² + y² + z²)

θ = arccos(z / r)

φ = arctan(y / x)

2. What is the equation for calculating the volume of a spherical sink?

The equation for calculating the volume of a spherical sink is:

V = (π/6) * (3r - h) * (3r² + h²)

Where r is the radius of the sphere and h is the height of the sink from its base to the top of the sphere.

3. How do I find the surface area of a spherical sink?

To find the surface area of a spherical sink, use the formula:

A = 4πr² + 2πrh

Where r is the radius of the sphere and h is the height of the sink from its base to the top of the sphere.

4. Can I use spherical coordinates to calculate the depth of a sink at a specific point?

Yes, you can use spherical coordinates to calculate the depth of a sink at a specific point. The depth can be calculated using the formula:

d = r - h

Where r is the radius of the sphere and h is the height of the sink from its base to the top of the sphere.

5. How can I use spherical coordinates to determine the maximum capacity of a spherical sink?

To determine the maximum capacity of a spherical sink, you can use the formula:

C = (4/3) * π * r³

Where r is the radius of the sphere. This formula will give you the total volume of the sink, which is its maximum capacity.

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