# (Multivariable Calculus) Convert $\rho = \sin \phi$ to cylindrical and rectangular

Question: Consider the surface given in spherical coordinates by $\rho = \sin(\phi)$. Convert to rectangular coordinates and cylindrical coordinates. Identify the surface.

By graphing the function, I've found that it is a horn torus (circles in cross section are tangent to each other). Using some conversion formulas, I got this:

$$r = \sin^2(\phi)$$ $$\theta = \theta$$ $$z = \frac{\sin(2\phi)}{2}$$

And then for rectangular (using those cylindrical values):

$$x = \sin^2(\phi)\cos(\theta)$$ $$y = \sin^2(\phi)\sin(\theta)$$ $$z = \frac{\sin(2\phi)}{2}$$

I would say this is wrong as I'm probably supposed to get it in terms of $(r, \theta, z)$, and then also into $(x, y, z)$. Unless this is correct?

Thanks for the reading.

• Looks good to me. – Chris Custer Sep 16 '18 at 2:02
• I thought so too, but after looking at how a cardioid is written in rectangular coordinates and seeing that it is implicit, I decided to find implicit solutions to this problem. I'm going to leave the answers I got above on my work though. – BiaxialObject48 Sep 16 '18 at 2:18

## 2 Answers

You shouldn't have a formula with spherical coordinate variables when you made the change to rectangular and cylindrical coordinates!

To convert to rectangular, use $\rho=\sqrt{x^2+y^2+z^2}$ and $\phi=\sin^{-1}\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}$. So you get $$\sqrt{x^2+y^2+z^2}=\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}$$ Or $$x^2+y^2+z^2 = \sqrt{x^2+y^2}.$$

Now use $r=\sqrt{x^2+y^2}$ to have $$r^2+z^2=r.$$

• I marked you as the answer because you answered earlier, but both you and Villa gave the same solution, which was correct according to my graphing utility. – BiaxialObject48 Sep 16 '18 at 1:47

Well, there are different conventions for the meaning of $\phi$ in spherical coordinates, but I will use:

$$x=\rho \cos{\theta}\sin{\phi}$$ $$y=\rho \sin{\theta}\sin{\phi}$$ $$z=\rho \cos{\phi}$$

It is easy to see that $\sin{\phi}=\frac 1\rho \sqrt{\rho^2-z^2}$

And because $\rho = \sqrt{x^2+y^2+z^2}$, in cartesian $\sin{\phi}=\rho$ becomes:

$$\sqrt{x^2+y^2}=x^2+y^2+z^2$$

And in cylindrical:

$$r=r^2+z^2$$

Note that $\theta$ does not appear because of the symmetry with respect to the $z$ axis.