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

[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!

 

[BvU]

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 $d(xy)$.

 

[member 428835]

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?

 

[BvU]

$d(xy)$ 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 $0$

 

[pixel]

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.

 

[member 428835]

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.