0
$\begingroup$

I know the Surface area element for polar co-ordinates is:

dA = r dr d(theta/phi)

whereas... what is the area vector? I know the area vector is perpendicular to the to the surface plane but between x, y, r and theta/phi

which vector matches that description? I assumed the position vector/radial vector as being the "direction" for the area vector.. but i have nowhere to confirm this.

$\endgroup$
5
  • $\begingroup$ so what is dA ⃗ $\endgroup$ Apr 13 '17 at 21:51
  • $\begingroup$ $dxdy=rdrd\theta$ in $\mathbb{R}^2$ so what do you mean by surface area? Perpendicular to the plane in $\mathbb{R}^2$ would either "into" or "out of" the screen/paper. $\endgroup$ Apr 13 '17 at 21:58
  • $\begingroup$ my reason being some refer to them as being area elements while others say surface area elements. $\endgroup$ Apr 13 '17 at 22:04
  • $\begingroup$ I mean are you actually working in $\mathbb{R}^2$ or are you secretly using cylindrical coordinates or something? $\endgroup$ Apr 13 '17 at 22:07
  • $\begingroup$ I am not restricted to R2 only that polar co-ordinates are 2-dimensional. So i am assuming an area vector wouldn't exist in R2 $\endgroup$ Apr 13 '17 at 22:10
1
$\begingroup$

Polar coordinates consist of three "directions": $r,\varphi, z$ as opposed to spheric coordinates, which are two angles and one radius, $r,\varphi,\theta$. So your Area vector is as you said perpendicular to the area, so you have $$d \vec{A}=d A \cdot \vec{e}_z=r\cdot dr \thinspace d\varphi\cdot \vec{e}_z$$

$\endgroup$
1
  • 1
    $\begingroup$ i didn't think of adding the 3rd dimension z for some reason as i wanted to stay in a 2-dimension real field but it is definitely the only logical answer. $\endgroup$ Apr 13 '17 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.