Volume element

A simple example of a volume element can be explored by considering a two-dimensional surface embedded in n-dimensional Euclidean space. Such a volume element is sometimes called an area element. Consider a subset ${\displaystyle U\subset \mathbb {R} ^{2}}$ and a mapping function

${\displaystyle \varphi :U\to \mathbb {R} ^{n}}$

thus defining a surface embedded in ${\displaystyle \mathbb {R} ^{n}}$. In two dimensions, volume is just area, and a volume element gives a way to determine the area of parts of the surface. Thus a volume element is an expression of the form

${\displaystyle f(u_{1},u_{2})\,\mathrm {d} u_{1}\,\mathrm {d} u_{2}}$

that allows one to compute the area of a set B lying on the surface by computing the integral

${\displaystyle \operatorname {Area} (B)=\int _{B}f(u_{1},u_{2})\,\mathrm {d} u_{1}\,\mathrm {d} u_{2}.}$

Here we will find the volume element on the surface that defines area in the usual sense. The Jacobian matrix of the mapping is

${\displaystyle \lambda _{ij}={\frac {\partial \varphi _{i}}{\partial u_{j}}}}$

with index i running from 1 to n, and j running from 1 to 2. The Euclidean metric in the n-dimensional space induces a metric ${\displaystyle g=\lambda ^{T}\lambda }$ on the set U, with matrix elements

${\displaystyle g_{ij}=\sum _{k=1}^{n}\lambda _{ki}\lambda _{kj}=\sum _{k=1}^{n}{\frac {\partial \varphi _{k}}{\partial u_{i}}}{\frac {\partial \varphi _{k}}{\partial u_{j}}}.}$

The determinant of the metric is given by

${\displaystyle \det g=\left|{\frac {\partial \varphi }{\partial u_{1}}}\wedge {\frac {\partial \varphi }{\partial u_{2}}}\right|^{2}=\det(\lambda ^{T}\lambda )}$

For a regular surface, this determinant is non-vanishing; equivalently, the Jacobian matrix has rank 2.

Now consider a change of coordinates on U, given by a diffeomorphism

${\displaystyle f\colon U\to U,}$

so that the coordinates ${\displaystyle (u_{1},u_{2})}$ are given in terms of ${\displaystyle (v_{1},v_{2})}$ by ${\displaystyle (u_{1},u_{2})=f(v_{1},v_{2})}$. The Jacobian matrix of this transformation is given by

${\displaystyle F_{ij}={\frac {\partial f_{i}}{\partial v_{j}}}.}$

In the new coordinates, we have

${\displaystyle {\frac {\partial \varphi _{i}}{\partial v_{j}}}=\sum _{k=1}^{2}{\frac {\partial \varphi _{i}}{\partial u_{k}}}{\frac {\partial f_{k}}{\partial v_{j}}}}$

and so the metric transforms as

${\displaystyle {\tilde {g}}=F^{T}gF}$

where ${\displaystyle {\tilde {g}}}$ is the pullback metric in the v coordinate system. The determinant is

${\displaystyle \det {\tilde {g}}=\det g\left(\det F\right)^{2}.}$

Given the above construction, it should now be straightforward to understand how the volume element is invariant under an orientation-preserving change of coordinates.

In two dimensions, the volume is just the area. The area of a subset ${\displaystyle B\subset U}$ is given by the integral

${\displaystyle {\begin{aligned}{\mbox{Area}}(B)&=\iint _{B}{\sqrt {\det g}}\;\mathrm {d} u_{1}\;\mathrm {d} u_{2}\\&=\iint _{B}{\sqrt {\det g}}\left|\det F\right|\;\mathrm {d} v_{1}\;\mathrm {d} v_{2}\\&=\iint _{B}{\sqrt {\det {\tilde {g}}}}\;\mathrm {d} v_{1}\;\mathrm {d} v_{2}.\end{aligned}}}$

Thus, in either coordinate system, the volume element takes the same expression: the expression of the volume element is invariant under a change of coordinates.

Note that there was nothing particular to two dimensions in the above presentation; the above trivially generalizes to arbitrary dimensions.

For example, consider the sphere with radius r centered at the origin in R³. This can be parametrized using spherical coordinates with the map

${\displaystyle \phi (u_{1},u_{2})=(r\cos u_{1}\sin u_{2},r\sin u_{1}\sin u_{2},r\cos u_{2}).}$

Then

${\displaystyle g={\begin{pmatrix}r^{2}\sin ^{2}u_{2}&0\\0&r^{2}\end{pmatrix}},}$

and the area element is

${\displaystyle \omega ={\sqrt {\det g}}\;\mathrm {d} u_{1}\mathrm {d} u_{2}=r^{2}\sin u_{2}\,\mathrm {d} u_{1}\mathrm {d} u_{2}.}$