Converting vector field from cartesian to cylindrical and finding the curl

1k Views Asked by At

The question is this:

(a) Find the curl of the vector field $ = y \hat{x} + \hat{} + \hat{}$ in Cartesian coordinates. (b) Rewrite in cylindrical coordinates. (c) Find ∇× explicitly in cylindrical coordinates.

I've worked out (a) to be $x \hat{x} - y\hat{y} + (2x-2y)\hat{z}$ but I feel like I keep messing up on converting the vector field to cylindrical coordinates.

I know $ \rho = \sqrt{x^2 + y^2}$, $\theta = arctan(y/x)$, and $z = z$ and have done $\rho = \sqrt{y^4+x^4}$, $\theta = arctan(x^2/y^2)$ and $z = xy = r^2cos\theta sin\theta$

I feel like i'm missing something to use this though because of the partial derivatives in the $\nabla$xV formula.

Any guidance would be appreciated.

1

There are 1 best solutions below

1
On

$\vec{V}= (y, x, xy)$

Please note the $\hat{z}$ component of the curl is zero and not $(2x-2y)$.

So, $ \ \nabla \times \vec{V} = (x,-y,0)$.

In cylindrical coordinates,

$\left \{\begin{array} {l} x = \rho \cos\varphi, y = \rho \sin\varphi, z = z \\ \hat{x} = \cos\varphi \hat{\rho} - \sin\varphi \hat{\varphi} \\ \hat{y} = \sin\varphi \hat{\rho} + \cos\varphi \hat{\varphi} \\ \hat{z} = \hat{z} \end{array} \right.$

Refer wiki

So the vector field can be re-written in cylindrical coordinates as

$\vec{V} = \rho \sin \varphi (\cos\varphi \hat{\rho} - \sin\varphi \hat{\varphi}) + \rho \cos \varphi (\sin\varphi \hat{\rho} + \cos\varphi \hat{\varphi}) + \rho^2 \sin \varphi \cos\varphi \hat{z}$

Rearrange this in $\hat{\rho}, \hat{\varphi}, \hat{z} \ $ components and that is your vector field in cylindrical coordinates.

Again refer to the same link that gives you formula to find curl of the vector field in cylindrical coordinates as the question asks you to explicitly find curl in cylindrical coordinates which means you cannot convert the curl found in cartesian coordinates to cylindrical using the above conversion I showed.