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October 21st, 2018, 10:38 PM   #1
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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.
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Last edited by skipjack; October 22nd, 2018 at 02:59 AM.

October 21st, 2018, 10:40 PM   #2
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Better picture
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 October 21st, 2018, 11:18 PM #3 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 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. Thanks from topsquark and SenatorArmstrong
 October 22nd, 2018, 05:09 AM #4 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 You have given spherical coordinates. In cylindrical coordinates: x=rcos$\displaystyle \theta$ y=rsin$\displaystyle \theta$ z=z This has a three-dimensional 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. Thanks from topsquark and SenatorArmstrong
October 22nd, 2018, 05:36 AM   #5
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Quote:
 Originally Posted by zylo The interesting question is why does area or volume transformation lead to Jacobian? Or really, more generally, transformation of differential products.
The Jacobian is nothing more than the derivative. Specifically, for a $C^1$ function $f: \mathbb{R}^n \to \mathbb{R}^n$, and $x \in \mathbb{R}^n$, there is a linear operator, $A$, which is the best linear approximation of $f$ in a neighborhood of $x$. In canonical coordinates, $A$ is a matrix and the Jacobian is an explicit formula for computing this matrix. It is commonly expressed as $J(x)$ or $Df(x)$.

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, 06:36 AM #6 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 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 06:38 AM.
 October 22nd, 2018, 09:34 AM #7 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 $\displaystyle \vec{r}=x(u,v,w)\vec{i}+y(u,v,w)\vec{j}+z(u,v,w)\v ec{k}$ $\displaystyle d\vec{r}=\vec{r}_{u}du+\vec{r}_{v}dv+\vec{r}_{w}dw$ $\displaystyle d\vec{r}=(x_{u}\vec{i}+y_{u}\vec{j}+z_{u}\vec{k})d u+(x_{v}\vec{i}+y_{v}\vec{j}+z_{v}\vec{k})dv+(x_{w }\vec{i}+y_{w}\vec{j}+z_{w}\vec{k})dw$ 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 22nd, 2018 at 11:51 PM.
October 22nd, 2018, 11:59 AM   #8
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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
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 October 22nd, 2018, 12:13 PM #9 Senior Member   Joined: Apr 2017 From: New York Posts: 155 Thanks: 6 Ok I got it. Almost done. I will post. Thanks
October 22nd, 2018, 12:42 PM   #10
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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 ?
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