Curvilinear Coordinates Volume Element

Mar 2015
1,720
126
New Jersey
Given r=(x(u,v,w), y(u,v,w), z(u,v,w))

How do you get from

\(\displaystyle dx=x_{u}du+x_{v}dv+x_{w}dw\)
\(\displaystyle dy=y_{u}du+y_{v}dv+y_{w}dw\)
\(\displaystyle dz= z_{u}du+z_{v}dv+z_{w}dw\)

to

\(\displaystyle dxdydz=\biggr\vert\frac{\partial (x,y,z)}{\partial (u,v,w)}\biggr\vert dudvdw\)

Look up Jacobian
 
Mar 2015
1,720
126
New Jersey
\(\displaystyle \vec{r}=x\vec{i}+y\vec{j}+z\vec{k}\)

\(\displaystyle d\vec{r}_{u}\equiv\frac{\partial \vec{r}}{\partial u}du=x_{u}du\vec{i}+y_{u}du\vec{j}+z_{u}du\vec{k}\)

\(\displaystyle d\vec{r}_{v}\equiv\frac{\partial \vec{r}}{\partial v}dv=x_{v}dv\vec{i}+y_{v}dv\vec{j}+z_{v}dv\vec{k}\)

\(\displaystyle d\vec{r}_{w}\equiv\frac{\partial \vec{r}}{\partial w}dw=x_{w}dw\vec{i}+y_{w}dw\vec{j}+z_{w}dw\vec{k}\)

\(\displaystyle d\vec{r}_{u}, d\vec{r}_{v}, d\vec{r}_{w}\) are distance vectors in x,y,z coordinate system and as such you can calculate the element of volume they form from

\(\displaystyle dV=|d\vec{r}_{u}\cdot d\vec{r}_{v}\times d\vec{r}_{w}| = Jdudvdw\)

This is not dxdydz which is obvious if at any point you draw dx, dy, dz and \(\displaystyle d\vec{r}_{u}, d\vec{r}_{v}, d\vec{r}_{w}\)

However, \(\displaystyle \int_{V}dxdydz=\int_{V}Jdudvdw\) which gives rise to the common fiction that dxdydz=Jdudvdw.

dxdydz is given by:
\(\displaystyle dx=x_{u}du+x_{v}dv+x_{w}dw\)
\(\displaystyle dy=y_{u}du+y_{v}dv+y_{w}dw\)
\(\displaystyle dz=z_{u}du+z_{v}dv+z_{w}dw\)
which in principle would give the same volume if you could do the integral.