[Math] (Multivariable Calculus) Convert $\rho = \sin \phi$ to cylindrical and rectangular

cylindrical coordinatesmultivariable-calculusproof-verificationspherical coordinates

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.

Best Answer

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

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