# Espheric to cartesian coordinates

1. Sep 8, 2010

### Telemachus

1. The problem statement, all variables and given/known data
Hi there. Well, I have the next exercise, which I've solved, but I don't know if the solution I got is the right one.

It says Given the next region on spheric coordinates find the expression for it in rectangular coordinates, and plot.

$$\phi\in{(0,\displaystyle\frac{\pi}{6}}),\rho\in{[1,2],\theta\in{(0,\displaystyle\frac{\pi}{2}})}$$

Well, In think it is a part of a cone, without a sphere of radius 1, with a sphere of radius 2 on the top.

2. Relevant equations
$$\begin{Bmatrix}{ x=\rho\cos\theta\sin\phi}\\y=\rho^2\sin\theta\sin\phi\\z=\rho\cos\phi \end{matrix}$$

3. The attempt at a solution
$$x^2+y^2=\rho^2\sin^2\phi(\cos^2\theta+\sin^2\theta)\Rightarrow{x^2+y^2=\rho\sin^2\phi}$$

$$z^2=\rho^2\cos^2\phi\Rightarrow{\rho^2=\displaystyle\frac{z^2}{\cos^\phi}}$$

$$x^2+y^2=z^2\displaystyle\frac{\sin^2\phi}{\cos^2\phi}\theta}\Rightarrow{x^2+y^2=z^2\tan^2\phi}\Rightarrow{x^2+y^2\leq{\displaystyle\frac{z^2}{3}}}$$

$$f(x,y)=\begin{Bmatrix}{ x^2+y^2\leq{\frac{z^2}{3}}\forall{}z\in{[\frac{\sqrt[ ]{3}}{2}},\sqrt[ ]{3}]}\\x^2+y^2+z^2\leq{4}\forall{z\in{(\sqrt[ ]{3},2]}}\\x^2+y^2+z^2\geq{1} \end{matrix},x>0,y>0$$

Is this right?

Last edited: Sep 8, 2010
2. Sep 8, 2010

### gabbagabbahey

You dropped a superscript, you should have $x^2+y^2 =\rho^2\sin^2\phi$.

What if $\cos\phi = 0$?

There's a much easier way to relate Cartesian coordinates to spherical coordinates, that is probably derived in your textbook/notes:

$$\rho = \sqrt{x^2+y^2+z^2}$$
$$\theta = \cos^{-1}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)$$
$$\phi = \tan^{-1}\left(\frac{y}{x}\right)$$

Your bounds on $r$, $\theta$ and $\phi$ each give you two inequalities (for a total of 6 inequalities) which you need to reduce/solve to find the region in closed.

For example, $\phi \in (0,\frac{\pi}{6})$ tells you that $0 < \tan^{-1}\left(\frac{y}{x}\right) < \frac{\pi}{6}$, or, taking the tangent of each side of this inequality (actually 2 inequalities combined into one expression),

$$0 < \frac{y}{x} < \frac{1}{\sqrt{3}}$$

You've correctly found that $x^2+y^2 < \frac{z^2}{3}$, but that only makes use of one of the 6 inequailties you have. You need to use them all to get the correct region.

Here I'm not sure what you're trying to say....there is no f(x,y) in this case, you are simply being asked to find a reduced set of inequalities for $x$, $y$ and $z$ that describe the region....Is there another part of the problem that involves some function f(x,y)?

3. Sep 8, 2010

### Telemachus

No, I've called f(x,y) to the region.

Im calling $$\phi,\theta,\rho$$ this way:

[PLAIN]http://www.math.montana.edu/frankw/ccp/multiworld/multipleIVP/spherical/convert3d.gif [Broken]

And $$\rho\geq{0}$$

Sorry, it was a typo, I'll correct it.

Then $$z=\rho$$

Thanks for your answer. I've found a set of inequalities (thats the last set of equations I've posted), I'll try to make a graph of "what I see".

Last edited by a moderator: May 4, 2017
4. Sep 8, 2010

### gabbagabbahey

That doesn't make any sense to me. In Cartesian coordinates, your region should be described by a set of inequalities for x,y and z (the same way it is desribed by a set of inequailties for $\rho$, $\theta$ and $\phi$ when working in spherical coordinates), not by some function f(x,y).

For example, if I wanted to describe the region the is bounded by the planes z=1, z=2, x=1, x=2 and y=0 and the parabaloid y=x^2, I would say the region is given by

$$1 \leq x \leq 2$$
$$0 \leq y \leq x^2$$
$$1 \leq z \leq 2$$

It would make no sense to say

$$f(x,y,z)= \left\{\begin{array}{lr}1 \leq x \leq 2 \\ 0 \leq y \leq x^2 \\1 \leq z \leq 2 \end{array}\right.$$

You can't set a function equal to an inequality or set of inequalities, and you can't describe a 3D region by a function.

Okay, then you have:

$$\rho = \sqrt{x^2+y^2+z^2}$$
$$\phi = \cos^{-1}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)$$
$$\theta = \tan^{-1}\left(\frac{y}{x}\right)$$

Last edited by a moderator: May 4, 2017
5. Sep 8, 2010

### Telemachus

Here it is:

$$x>0$$
$$y>0$$
$$x^2+y^2\leq{\frac{z^2}{3}}\forall{}z\in{[\frac{\sqrt[ ]{3}}{2}},\sqrt[ ]{3}]}$$
$$x^2+y^2+z^2\leq{4}\forall{z\in{(\sqrt[ ]{3},2]}}$$
$$x^2+y^2+z^2\geq{1}$$

Sorry, it took some time to make the draw. Anyway, its limited by the planes zy and zx, and by the sphere x²+y²+z²=1.