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"This can only be defined for flat surfaces"
Really? So what about that?:
http://farside.ph.utexas.edu/teaching/316/lectures/node4.html
Fitzpatrick says that it's possible to define vector area for curved surfaces by summing up infinitesimal differential vector areas dS (vector sum), which is equivalent for integrating vector area contributions over the whole surface.
It can be easily proved:
Imagine two rectangular surfaces, 1 by sqrt(2) each, connected with one border and at an angle of 90 degrees to each other (45 and 135 degrees to the floor). The area of each half of the roof is 1*sqrt(2)=sqrt(2), so the whole roof area = 2*sqrt(2). But the magnitudes of area vectors of each half = sqrt(2), which (when added vectorily) gives a wector with magnitude 2, perpendicular to the floor. Two triangle areas on both sides of the roof have the area 2*1/2=1, but their normals are pointing in opposite directions, so they cancel each other. The roof, when projected on the floor, has the area 1+1=2, which is exactly the vector area's magnitude! Not the roof surface's magnitude. Fitzpatrick also shows that the vector area S is dependent only on the surface area of the projections of the rim viewed from each axis, but not dependent on the shape of the surface. Two surfaces attached to the same rim would have the same vector area. Returning to the example with roof, we can see that even if the roof's area is changing (e.g. when we have steeper roof), its projection along the floor's normal is not changing, so as the vector area.
It's worth to mention that the rim could be any closed path. It's not necessary for it to lay on a flat surface.

All these are very important notions in the light of Maxwell's equations application of vector areas.

Who's right then? -- SasQ777

Vector area is well defined for curved surfaces. See, e.g., the Spiegel reference I've put into the article. This article could really do with some good graphics to illustrate these points. Alma Teao Wilson (talk) 06:51, 7 August 2009 (UTC)Reply