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September 24th, 2018, 11:03 AM   #1
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Curvilinear Coordinates Volume Element

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
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September 27th, 2018, 11:30 AM   #2
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$\displaystyle \vec{r}=x\vec{i}+y\vec{j}+z\vec{k}$

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

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

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

$\displaystyle d\vec{r}_{u}, d\vec{r}_{v}, d\vec{r}_{w}$ 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.
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