[Math] What does a plane in spherical coordinates look like as seen in cartesian coordinates

geometryplane-geometryspheresspherical coordinates

In working with spherical coordinates $\rho$, $\phi$,$\theta$, where $\phi$ is the spherical angle, one could plot a simple flat plane in the space of the spherical coordinates.

For example:

In the basis ($\rho$, $\phi$,$\theta$), I have three points. They are
(1,0,0), (0,$\pi/2$,0), (0,0,$\pi/4$). Clearly these 3 points define a plane/triangular region in this basis.

My question is, however, what does this region look like in cartesian space? Is it simply a spherical patch of area?

Is the area of this region given by one half the cross product of the vectors:

Area = (1/2) | (<1,0,0> – <0,$\pi/2$,0>) x (<0,0,3> – <0,$\pi/2$,0>) |?

I have been thinking about this for quite some time now and have not yet come up with an answer. Help would be appreciated.

Best Answer

It won't just be a spherical patch: since $r$ is going from $0$ to $1$, the shape will sweep from the origin out to a point along a curved surface.

It ends up looking like a seashell:

parametric surface

I drew this using GeoGebra with the following command:

Surface(((1-u)(1-v)sin(2u*(1-v))cos(3v),
         (1-u)(1-v)sin(2u*(1-v))sin(3v),
         (1-u)(1-v)cos(2u*(1-v))),
        u,0,1,
        v,0,1)
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