# Homework Help: From spherical to cylindrical

1. Apr 26, 2012

### Niles

1. The problem statement, all variables and given/known data
Hi

I have an expression on the form
$$df(v, \theta, \phi) = v e^{-v^2/C}\cos(\theta)v^2\sin(\theta)\,dv\,d\theta\,d\phi$$
and I am trying to write it in cylindrical coordinates. Note that θ runs from 0..π, v is a velocity and C a real constant. So I wish to write it in terms of a radial and axial velocity, vr and vz.

First I thought of integrating out θ and $\phi$. $\phi$ is easily done, and it yields 2π. However, the integral over θ yields 0. Am I doing this wrong?

The result should yield something proportional to

$$v_re^{-v_r^2/C^2}v_ze^{-v_z^2/C^2}\,dv_r\,dv_z$$

Best,
Niles.

2. Apr 26, 2012

### marcusl

In spherical coordinates the integral over theta runs 0 to pi.

3. Apr 26, 2012

### Niles

Hi

Thanks for replying. The integral of cos(theta)sin(theta) over 0..pi gives 0.

4. Apr 26, 2012

### marcusl

Oh, right. I looked at the sin(theta) and didn't see the other term.

5. May 4, 2012

### Niles

OK, I am taking a different approach now. Now I use $v_r = v\cos(\theta)$ and $v_z = v\sin(\theta)$. This gives me
$$v_rv_zv\,dv\,d\theta$$
Then I use (from http://en.wikipedia.org/wiki/Spherical_coordinate_system) that $\arctan(v_r/v_z)$ and $v=\sqrt(v_r^2+v_z^2)$ to find dv/dvr and dθ/dvz. However, don't get the desired results.

Is my approach wrong?

Best,
Niles.

6. May 4, 2012

### sharks

Basically, what you're trying to do is express a complete sphere with radius = v, in terms of cylindrical coordinates, with density $v e^{-v^2/C}\cos(\theta)$. So, forget about the density for now, and focus on the conversion to cylindrical coordinates. Afterwards, plug in the density.

7. May 4, 2012

### Niles

Thanks for taking the time to help me. For the conversion I get
$$v e^{-v^2/C}\cos(\theta)v^2\sin(\theta) = v_rv_zv e^{-\sqrt{v_r^2+v_z^2}/C}$$
I have used the relations I linked to above. Can I get a hint to how I should take care of the density?

Best wishes,
Niles.

8. May 4, 2012

### Niles

Ahh, I think I get it now. So what you mean is that
$$v\,dvd\theta = dv_r\,dv_z$$

Then it adds up!

9. May 4, 2012

### HallsofIvy

In cylindrical coordinates r measures the distance from the origin to the foot of the projection from the point to the xy-plane. The line from the origin to the point in 3 dimensions, the line from the origin to the point at the foot of the projection and the projection form a straight line with hypotenuse of length $\rho$, angle $\pi/2- \phi$, "opposite side" of length z, and "near side" of length r. That is, the point that has spherical coordinates $\rho$, $\theta$, and $\phi$ has cylindrical coordinates $r= \rho sin(\phi)$, $\theta= \theta$ and $z= \rho cos(\phi)$.

Here, I have used the mathematics convention that $\theta$ is the "longitude" and $\phi$ is the "co-latitude". The physics convention reverses that.

10. May 4, 2012

### sharks

My understanding of the problem is in the attached graph below.
In this problem, $\phi$ is substituted by $\theta$, $\theta$ is replaced by $\phi$, and $\rho$ is replaced by v.

So, a description of the sphere would be:
For $\phi$ and $\theta$ fixed, $v$ varies from 0 to v.
For $\phi$ fixed, $\theta$ varies from 0 to ∏.
$\phi$ varies from 0 to 2∏.

The corresponding description in terms of cylindrical coordinates would be:
For $\phi$ and $r$ fixed, $z$ varies from $-\sqrt{v^2-v^2\sin^2\theta}$ to $\sqrt{v^2-v^2\sin^2\theta}$, where $\theta$ is $\arctan ({\frac {r}{z}})$.
For $\phi$ fixed, $r$ varies from 0 to v.
$\phi$ varies from 0 to 2∏.

For the integrand:
$$v e^{-v^2/C}\cos(\theta)$$Replace $\theta$ by $\arctan ({\frac {r}{z}})$ and $v$ by $\sqrt{r^2+z^2}$.

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• ###### spherical.PNG
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Last edited: May 4, 2012
11. May 4, 2012

### Niles

Thanks for the help, both of you.

Best,
Niles.