sphere of radius 1 as parametric function

Question

I know that the equation or a sphere of radius 1 in spherical coordinates is:

$$\rho = 1$$

how to mathematically convert this above equation into one that looks like this:

$$ f(\rho, \theta, \phi) = (\rho \sin \phi \cos \theta,\ \ \rho \sin \phi sin \theta, \ \ \rho \cos \phi)$$

Answer 1

Assuming your question is regaurding the derivation of the spherical coordinates:

Let $\phi$ represent the angle between the z-axis and your point. Similarly let $\theta$ represent the angle between the x axis and your point.

Notice that you get:

$$cos(\phi) = \frac{z}{\rho}$$

$$z = \rho \space cos(\phi)$$

Now we still have our equations:

$$x=rcos(\theta)$$ $$y=rsin(\theta)$$

So we need to find how $r$ is related to $\rho$. To do this we notice that

$$sin(\phi)=\frac{r}{\rho}$$ Thus $r =\rho \space sin(\phi)$

Plugging in this result you get the equations:

$$x = \rho \space sin(\phi) cos(\theta)$$ $$y = \rho \space sin(\phi) sin(\theta)$$ $$z = \rho \space cos(\phi)$$

Answer 2

$$ f(ρ,θ,ϕ) = (ρ \sin ϕ \cos θ,\ \ \ ρ \sin ϕ \sin θ,\ \ \ ρ \cos ϕ) $$

The preceding function is how you can reach any point in a 3-dimensional coordinate system, using a spherical coordinate "address" for each point in the 3-d space. The function $f(ρ,θ,ϕ)$ works by returning the (x,y,z) coordinate given a spherical coordinate $(ρ,θ,ϕ)$.

Next we have the equation for a sphere of radius 1 when specified using spherical coordinate variables of $\rho$, $\theta$, and $\phi$:

$$ρ=1$$

$$\theta = unconstrained$$

$$\phi = unconstrained$$

Thus, given the above information, the parametric function for a sphere of radius 1 is:

$$ f(θ,ϕ) = (\sin ϕ \cos θ,\ \ \ \sin ϕ \sin θ,\ \ \ \cos ϕ) $$

Allowing us to get to any point on a sphere by choosing coordinates of $f(\theta, \phi)$ and returning the (x,y,z) coordinate of this point:

Another example, just to be extra clear, the parametric function for a sphere of radius 2 is:

$$ f(θ,ϕ)=(2 \sin ϕ \cos θ,\ \ \ 2 \sin ϕ \sin θ,\ \ \ 2 \cos ϕ) $$