# Area element of a spherical surface

I am trying to find out the area element of a sphere given by the equation:

$$r^2= x^2 +y^2+z^2$$

The sphere is centered around the origin of the Cartesian basis vectors $(e_x,e_y,e_z)$. The spherical-polar basis vectors are $(e_r,e_\theta,e_\phi)$ which is related to the cartesian basis vectors as follows:

$$e_r=(\sin \theta \cos \phi )e_x+(\sin \theta \sin \phi )e_y+(\cos \theta )e_z \tag 1$$ $$e_\theta =(\cos \theta \cos \phi )e_x+(\cos \theta \sin \phi )e_y−(\sin \theta )e_z \tag 2$$ $$e_\phi =(−\sin \phi )e_x+(\cos \phi )e_y \tag 3$$

Reference image:

this would give a rough idea of the setup but the basis vectors are set as in this image: .

The vector surface element can be given as the cross product of $d\overrightarrow r_1$ and $d\overrightarrow r_2$ i.e. : $$dS = d\overrightarrow r_1×d\overrightarrow r_2$$

and by taking its magnitude I would get its area.

What I think is that the vectors $d\overrightarrow r_1$ and $d\overrightarrow r_2$ can be given by linear combination of the basis vectors $e_r$ $e_\phi$ and $e_\theta$ and therefore: $$d\overrightarrow r= a_r e_r + a_\phi e_\phi + a_\theta e_\theta$$

so, $d\overrightarrow r_1$ and $d\overrightarrow r_2$ can be given as: $$d\overrightarrow r_1= 0 e_r + d\phi e_\phi + 0 e_\theta \tag 4$$ and $$d\overrightarrow r_2= 0 e_r + 0 e_\phi + d\theta e_\theta \tag 5$$

I can then substitute equation $(1),(2)$ & $(3)$ in $(4)$ & $(5)$ to get $d\overrightarrow r_1$ and $d\overrightarrow r_2$ in terms of the Cartesian basis vectors $(e_x,e_y,e_z)$.

My question is:

1. Am I using the correct values of the coefficients $a_r, a_\phi$ and $a_\theta$?

2. Is this approach valid and how/where do I make use of the equation of sphere $r^2= x^2 +y^2+z^2$ in $d\overrightarrow r_1$ and $d\overrightarrow r_2$ so that I can further find their cross product?

I would appreciate any help.