Undergrad How does the change in area compare to the differential area element?

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SUMMARY

The discussion clarifies the distinction between the differential area element, denoted as ##dA##, and the change in area resulting from displacements in the x and y directions. The differential area element is expressed as ##dx \, dy##, while the change in area is calculated as ##d(xy) = ydx + xdy##. The participants emphasize that ##dA## represents an infinitesimal surface area used in integration, whereas the change in area accounts for perturbations over time. This distinction is crucial in applications such as the Young-Laplace equation, where changes in pressure relate to changes in surface area.

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  • Understanding of differential calculus
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  • Knowledge of the Young-Laplace equation
  • Basic concepts of surface area and perturbations
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member 428835
Hi PF!

Suppose we have a differential area element ##dA##. This can be expressed as ##dx \, dy##. However, a change in area ##dA## seems different. Take positions ##x## and ##y## and displace them by ##dx## and ##dy## respectively. Then the change in area ##dA = (x+dx)(y+dy)-xy = xdy+ydx## (ignoring higher order terms). How is the change of area and the differential element different (clearly they must be, right?). Or is it as I've said: one is the CHANGE in surface area and the other is a DIFFERENTIAL area element?

Thanks!
 
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Hi,

Make a drawing to verify that what you think you mean with ##dA## is actually what you write. An area element ##dA## is not the same as ##d(xy)##.
 
BvU said:
Hi,

Make a drawing to verify that what you think you mean with ##dA## is actually what you write. An area element ##dA## is not the same as ##d(xy)##.
Good idea, and I did, and what I'm saying is that ##dA## is an infinitesimal surface area, like something we'd have when integrating a surface. Examples would be ##dx\,dy##, ##r \,dr\,d\theta##, and ##r^2\sin\phi\,d\theta\, d\phi## for Cartesian surface, cylindrical surface normal to ##\hat{z}## and spherical surface normal to ##\hat{r}##.

The change in surface area would be if we had a surface that was perturbed in time, so assume it was displaced some amount ##dx## in ##\hat{x}## and ##dy## in ##\hat{y}##. Then the change in surface area would be ##d(xy) = ydx+xdy## (or how I explained it earlier).

Does this look correct?
 
##d(xy)## would apply to a rectangular area that was stretched from ## (0,0), \ (x,y)\ ## to ## (0,0), \ (x+dx,y+dy)\ ##

If it were displaced from ## (0,0), \ (x,y)\ ## to ## (dx,dy), \ (x+dx,y+dy)\ ## the change in area would be ##0## :smile:
 
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Likes member 428835
Consider the following diagram. A = xy is the area of the solid rectangle. A + dA = (x + dx) (y + dy) = xy + x dy + y dx + dx dy is the area of the larger rectangle. So from that we have dA = x dy + y dx + dx dy as you say. But if you look at the extra area beyond the solid rectangle, it can be broken up into three rectangles with areas x dy on the top left, dx dy on the top right and y dx on the right, as given by the equation for dA.

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Thank you both! This was very helpful. I was in fact talking about a stretched area (not translated), and so I am neglecting that triangle, at least until we take the limit as ##dx## and ##dy \to 0##.

In case you're wondering, this questions manifested in the Young-Laplace equation, where a change in pressure times volume is proportional to the change in surface area.
 
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