# Confused with a spherical coordinate system surface element

I can not understand how a particular surface element is derived in spherical coordinates.

The equation expressing the surface element vector is given as

$$r_s = (\sqrt{R^2-z_s^2} \cos \phi,\sqrt{R^2-z_s^2}\sin \phi, z_s)$$

And then the differential surface element is given as

$$dr_s = R d\phi dz_s$$

The center of the sphere is the origin and its radius is $$R$$. The z axis coordinate of the surface element is $$z_s$$.

I can not find a reference on the web that shows this particular surface element. I figured out that it is the formula for an 'infinitesimal' spherical zone of height $$dz_s$$ if $$d\phi$$ can be integrated out by symmetry to yield $$2\pi$$, (that is a spherical zone area = $$2\pi Rh$$ where here $$h=dz_s$$).

So in an integral where there is symmetry around the z axis, this differential surface area can be used to integrate a function over a sphere by summing up infinitesimal spherical zones over the z axis.

My question is: How is this surface element derived?

In the first equation you have the Cartesian coordinates. Let's allow $$R$$ to vary, and call it $$r$$. Then you can write the volume element as $$dV=dx\ dy\ dz=|J|dr\ dz\ d\phi$$ Here J is the Jacobian of the transformation from $$(x,y,z)$$ to $$(r,z,\phi)$$. The surface is perpendicular to $$r$$ for the sphere, so the surface element will be then just $$ds=\frac{dV}{dr}=|J|dz\ d\phi$$ The Jacobian is $$J=\begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} & \frac{\partial z}{\partial r} \\ \frac{\partial x}{\partial z} & \frac{\partial y}{\partial z} & \frac{\partial z}{\partial z} \\ \frac{\partial x}{\partial \phi} & \frac{\partial y}{\partial \phi} & \frac{\partial z}{\partial \phi} \end{vmatrix}=\begin{vmatrix} \frac{r}{\sqrt{r^2-z^2}}\cos\phi & \frac{r}{\sqrt{r^2-z^2}}\sin\phi & 0\\ \frac{-z}{\sqrt{r^2-z^2}}\cos\phi & \frac{-z}{\sqrt{r^2-z^2}}\sin\phi & 1\\ -\sqrt{r^2-z^2}\sin\phi& \sqrt{r^2-z^2}\cos\phi &0 \end{vmatrix}=-r$$
Now plugging in $$r=R$$, you get $$ds=R\ dz\ d\phi$$
• I think so. But maybe someone else might disagree. For spherical surfaces I write the volume as the surface area times the thickness. And for the volume element I use the Jacobian. Alternatively, you can find a set of vectors perpendicular to $\vec r$, but you will end up in the same place, meaning you need to calculate the derivatives of the components. Apr 25 '19 at 20:44