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$\vec F$ = force vector,
$\vec A$ = area vector,
$P$ = pressure

Mathematically $\vec F = P \vec A$. By product rule we get,

$$ {\rm d}\vec F = P {\rm d}\vec A + \vec A {\rm d}P $$

Why do we often compute Force over a surface as $\vec F = \int P {\rm d}\vec A$ whilst ignoring the term $\int \vec A {\rm d}P$ ?

Qmechanic
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Chaser
  • 29

2 Answers2

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If the area under consideration is a more complicated surface than a flat plane, and, if the pressure varies with position spatially on the surface, then the force per unit area at a given location on the surface is $p\mathbf{n}$, where $\mathbf{n}$ is the unit normal to the surface, and the differential element of area over which the force is applied is dA, so the vectorial force on the surface is $\mathbf{F}=\int{p\mathbf{n}dA}$.

Chet Miller
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The definition $\vec F = P \vec A$ is only valid if P is a constant, and in such case your second term is zero. The definition that makes physical sense when P varies with position is $d\vec F = P d\vec A$.