Consider the surface $S$ in $\mathbb{R^3}$ given by the equation $$x^2 + y^2 + z^2 = 1$$
Let $φ ∈ [0, 2π)$ and $θ ∈ [0, π]$
Give a surface element $f : [0, 2π) × [0, π] → \mathbb{R^3}$ for $S$ in terms of the angles $φ, θ$
Could someone shove me in the right direction if possible? Need some confidence on the direction to head.
The surface $S$ is a sphere, since the equation $x^2+y^2+z^2=1$ describes a surface in $\Bbb R^3$ where all points have radial distance of $1$. One can use spherical coordinates to easily describe the surface $S$. The spherical to Cartesian coordinate conversion is as follows: $$\begin{cases}x=\rho\sin(\theta)\cos(\phi)\\y=\rho\sin(\theta)\sin(\phi)\\z=\rho\cos(\theta)\end{cases}$$ where $\theta\in[0,\pi]$, $\phi\in[0,2\pi)$, and $\rho^2=x^2+y^2+z^2$. So in the case of this sphere $S$, the radial distance is constant, so $\rho=1$. So the surface can be described by only the two parameters $\theta$ and $\phi$, which would have parametric equations as follows: $$\begin{cases}x(\phi,\theta)=\sin(\theta)\cos(\phi)\\y(\phi,\theta)=\sin(\theta)\sin(\phi)\\z(\phi,\theta)=\cos(\theta)\end{cases}$$