October 21st, 2018, 11:38 PM  #1 
Senior Member Joined: Apr 2017 From: New York Posts: 119 Thanks: 6  rdrdθ
Hi to all, How can I prove dA=rdrdθ in cylindrical coordinates, using the theorem attached? Note: I know how to prove in polar coordinates. And I know that conversion from Cartesian to cylindrical is X=ρcosθ Y=ρsinφ Z=z Second problem is again proving dA=rdrdθ In elliptic coordinate system in R^2 defined X=coshμcosυ Y=sinhμsinυ Determine dA in elliptic coordinates. I have worked on part 1 and 2, but I couldn't complete. Last edited by skipjack; October 22nd, 2018 at 03:59 AM. 
October 21st, 2018, 11:40 PM  #2 
Senior Member Joined: Apr 2017 From: New York Posts: 119 Thanks: 6 
Better picture

October 22nd, 2018, 12:18 AM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 535 Thanks: 306 Math Focus: Dynamical systems, analytic function theory, numerics 
If you know how to do it in polar I don't see how you could be confused here. You compute the determinant of the Jacobian of the coordinate transform, just like for polar coordinates (or any other coordinate transform). In your case, the transform is exactly the one you have listed. Just compute its Jacobian and take the determinant. Done. 
October 22nd, 2018, 06:09 AM  #4 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,644 Thanks: 119 
You have given spherical coordinates. In cylindrical coordinates: x=rcos$\displaystyle \theta$ y=rsin$\displaystyle \theta$ z=z This has a threedimensional volume element and you need a 3d Jacobian. In polar coordinates, in your Jacobian's notation, r=u and v=$\displaystyle \theta$. In cylindrical coordinates, dV=dzdA and you can consider dA a transformation in polar coordinates. In elliptic coordinates, $\displaystyle \mu$=u and u=v, and then calculate the Jacobian. Presumably you can calculate the partial derivatives. The interesting question is why does area or volume transformation lead to Jacobian? Or really, more generally, transformation of differential products. 
October 22nd, 2018, 06:36 AM  #5  
Senior Member Joined: Sep 2016 From: USA Posts: 535 Thanks: 306 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
A fundamental fact from linear algebra is that if $A$ is diagonalizable, then it has $n$many eigenvectors, $\{v_i\}$, which of course form a basis for $\mathbb{R}^n$ since eigenvectors are linearly independent. Each $v_i$ has an associated eigenvalue, $\lambda_i$, which represents a scaling (and possibly reflection) in the direction of $v_i$. It follows that upon mapping a volume element of $\mathbb{R}^n$ through $A$, the entire volume is scaled by the $\Pi \lambda_i = $ det$(A)$. Of course, the assumption that $A$ is diagonalizable is merely to aid intuition. The result that the determinant of a linear operator is a quantitative measure of volume dilation holds for all finite dimensional linear operators.  
October 22nd, 2018, 07:36 AM  #6 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,644 Thanks: 119 
Unintelligible jargon posing as a derivation. Please illustrate in polar coordinates. Your last sentence is not a given, it's what's to be shown. Last edited by zylo; October 22nd, 2018 at 07:38 AM. 
October 22nd, 2018, 10:34 AM  #7 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,644 Thanks: 119 
$\displaystyle \vec{r}=x(u,v,w)\vec{i}+y(u,v,w)\vec{j}+z(u,v,w)\v ec{k}$ $\displaystyle d\vec{r}=\vec{r}_{u}du+\vec{r}_{v}dv+\vec{r}_{w}dw$ $\displaystyle d\vec{r}=(x_{u}\vec{i}+y_{u}\vec{j}+z_{u}\vec{k})d u+(x_{v}\vec{i}+y_{v}\vec{j}+z_{v}\vec{k})dv+(x_{w }\vec{i}+y_{w}\vec{j}+z_{w}\vec{k})dw$ The components of $\displaystyle d\vec{r}$ form a parallelipiped whose volume is $\displaystyle d\vec{r}_{u}\cdot d\vec{r}_{v}\times d\vec{r}_{w}=Jdudvdw$ $\displaystyle J=\begin{bmatrix} x_{u} &y_{u} & z_{u}\\ x_{v} &y_{v} & z_{v}\\ x_{w} &y_{w} & z_{w} \end{bmatrix}$ Note $\displaystyle dxdydz\neq Jdudvdw$ but $\displaystyle \int_{V}dxdydz=\int_{V}Jdudvdw$ As for higher dimensions, I suppose you could define dxdudvdw=Jdrdsdtdu, but it would only be meaningful if $\displaystyle \int _{V}dxdydzdw=\int _{V}Jdrdsdtdw$ But how do you show that? How do you even define a meaningful volume in n dimensions with meaningful limits of integration, other than a "cube"? Last edited by skipjack; October 23rd, 2018 at 12:51 AM. 
October 22nd, 2018, 12:59 PM  #8 
Senior Member Joined: Apr 2017 From: New York Posts: 119 Thanks: 6 
Ok I’m quiet new on this stuff so I want to keep it simple. I have substituted u=r and v=Θ At the given equation in the first original picture. So Dudu turned into drdθ Partial derivative of (x(r,θ)... Is with respect to dr or dθ? Or both consecutively. And what is next then? All I need to do is to turn dA= rdrdθ but still couldn’t see the patterns 
October 22nd, 2018, 01:13 PM  #9 
Senior Member Joined: Apr 2017 From: New York Posts: 119 Thanks: 6 
Ok I got it. Almost done. I will post. Thanks

October 22nd, 2018, 01:42 PM  #10 
Senior Member Joined: Apr 2017 From: New York Posts: 119 Thanks: 6 
I did in polar system Cylindrical system And in elliptic system I stuck with partial derivative of the components. Is everything. Alright on polar and cylindrical? And setting on the elliptic ? 