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.

  • $\begingroup$ 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

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 $$d \vec{A}=d A \cdot \vec{e}_z=r\cdot dr \thinspace d\varphi\cdot \vec{e}_z$$

  • 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

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