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September 24th, 2018, 11:03 AM  #1 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  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 
September 27th, 2018, 11:30 AM  #2 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
$\displaystyle \vec{r}=x\vec{i}+y\vec{j}+z\vec{k}$ $\displaystyle d\vec{r}_{u}\equiv\frac{\partial \vec{r}}{\partial u}du=x_{u}du\vec{i}+y_{u}du\vec{j}+z_{u}du\vec{k}$ $\displaystyle d\vec{r}_{v}\equiv\frac{\partial \vec{r}}{\partial v}dv=x_{v}dv\vec{i}+y_{v}dv\vec{j}+z_{v}dv\vec{k}$ $\displaystyle d\vec{r}_{w}\equiv\frac{\partial \vec{r}}{\partial w}dw=x_{w}dw\vec{i}+y_{w}dw\vec{j}+z_{w}dw\vec{k}$ $\displaystyle d\vec{r}_{u}, d\vec{r}_{v}, d\vec{r}_{w}$ 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. 

Tags 
coordinates, curvilinear, element, volume 
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