General condition for the surface element be the same as the volume element, up to a dt

Question

The surface element in spherical coordinates is $r \sin \theta \mathrm{d}\theta \mathrm{d}\varphi$, and the volume element is $r \sin \theta \mathrm{d}\theta \mathrm{d}\varphi \mathrm{d}r$. We see that the last element is equal to the former multiplied by $dr$. I ask if there is some general condition on a change of variable that allows to assert that the volume element is just the surface element multiplied by the differential of the remaining variable, say $dt$.

Answer 1

Let $r(x_1, x_2, ..., x_n)$ parametrize the volume, and $S(x_n)$ be the surface described by $r$ at fixed $x_n$.
Generally, the volume element associated to $r$ will be equal to $dx_n$ times the surface element corresponding to $S(x_n)$ through $r$, if $\partial r\over \partial x_n$ is orthogonal to $S(x_n)$ at every point of $S(x_n)$, and of unit length.
This can be seen easily for the case of $\mathbb R^3$, because the surface element is then $\bigg|\bigg|{\partial r\over \partial x_1}\times {\partial r\over \partial x_2}\bigg|\bigg|\mathrm{d}x_1\mathrm{d}x_2$ and the volume element is $\left|\det\left({\partial r\over \partial x_1},{\partial r\over \partial x_2},{\partial r\over \partial x_3}\right)\right|\mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3$. Now, by the triple product formula, we have $$\left|\det\left({\partial r\over \partial x_1},{\partial r\over \partial x_2},{\partial r\over \partial x_3}\right)\right| = \left|\left({\partial r\over \partial x_1}\times {\partial r\over \partial x_2}\right)\cdot {\partial r\over \partial r_3}\right| = \bigg|\bigg|{\partial r\over \partial x_1}\times {\partial r\over \partial x_2}\bigg|\bigg| \bigg|\bigg|{\partial r \over \partial x_3}\bigg|\bigg|= \bigg|\bigg|{\partial r\over \partial x_1}\times {\partial r\over \partial x_2}\bigg|\bigg|$$ (if one assumes ${\partial r\over \partial x_3}$ orthogonal to $S(r)$ and of unit length).

More generally, if $\partial r\over \partial x_n$ is orthogonal to $S(r)$ but not necessarly of unit length, then the element of volume will be equal to the element of surface multiplied by a scalar function of $r$, namely, by $\bigg|\bigg|{\partial r\over \partial x_n}\bigg|\bigg|$.

Answer 2

No general condition, but we look to how volume is computed in parts of a sphere.. in two ways.

Direct differentials

Volume of differential brick is found in the same way for a flat faced cuboid. Instead we multiply lengths of a differential curvilinear orthogonal sides to obtain volume of curvilinear shell/solid element

($ \sin \phi $ is component in radial direction for cone generator)

$$ = y\times r d \phi \times r d \theta = r \sin \phi\; dr \; \times r d \phi \times r d \theta = r^2 \sin \phi \;dr \;d \phi\; d \theta $$

Consideration of associated Jacobian of transformation in changing coordinate variables

$$ (dx,dy,dz) \to J(du,dv,dw) \; $$

where

$$ J= \left(\begin{matrix}\frac{\partial x}{\partial u}&\frac{\partial y}{\partial u}&\frac{\partial z}{\partial u}\\ \frac{\partial x}{\partial v}&\frac{\partial y}{\partial v}&\frac{\partial z}{\partial v}\\ \frac{\partial x}{\partial w}&\frac{\partial y}{\partial w}&\frac{\partial z}{\partial w}\end{matrix}\right) $$

The volume determinant evaluates to

$$ r^2 \sin \phi\; dr\;d\phi\;d \theta$$

Area element is simpler, with no thickness $dr$ to multiply.

Volume Element