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In some formulas in physics having a ratio, for example $ pressure={F \over\ A}$, the denominator is chosen to be a small quantity ($\Delta A$) and is written like, $$P= {\Delta F\over \Delta A}.$$ Then it is said that 'we take element $ \Delta A$ small enough that the pressure is independent of the size of the element' or 'pressure at a point is the limiting value of this ratio as the area element becomes small'.

What I don't understand is:

The meaning of above two statements in quotation marks and why it is necessary to take a small difference of $A$? Why we can't just take a small value of $A$?

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  • $\begingroup$ It is in fact a small value of A. The point is that a difference between areas is an area. And you can make it as small as you like. Your delta is an area element . It is not like the actual difference between Vfinal V star, which perhaps confuses you. But the math is the same. When studying limits it will be clearer. $\endgroup$ – Alchimista Jan 31 '18 at 14:24
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    $\begingroup$ It is in fact a small value of A. The point is that a difference between areas is an area. And you can make it as small as you like. Your delta is an area element . It is not like the actual difference between Vfinal V start, which perhaps confuses you. But the math is the same. When studying limits it will be clearer. $\endgroup$ – Alchimista Jan 31 '18 at 14:24
  • $\begingroup$ @Alchimista I think that should be an answer. Why don't you post it as one? $\endgroup$ – sammy gerbil Jan 31 '18 at 15:51
  • $\begingroup$ @samny gerbil because I cannot quickly elegantly formulate it without formatting a lot and I don't know really how to. Plus it is clear in my mind but difficult to write down in a more mathematical and physical way than how actually the comment looks. $\endgroup$ – Alchimista Jan 31 '18 at 15:55
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You're right: this is a different interpretation of $\Delta$ (or $d$) than usual, because there's no natural sense in which $\mathbf{A}$ is changing. Similarly, people also get confused by the moment of inertia, $$I = \int r^2 \, dm$$ because there's nothing about the mass $m$ that's changing. Now, you can always cook up a picture where everything is changing (as I do for $I$ here) but it might be clunky. For instance, to interpret $$\mathbf{F} = \int P \, d\mathbf{A}$$ I could say that I'm totaling up the force by gradually looking at more and more of the total area, so the 'total area considered' really does go from 'none' to 'all' in steps of $d\mathbf{A}$. Similarly I could imagine $P = dF/dA$ as telling me how the force on some surface would change if the surface were to expand.

Neither of these interpretations are very compelling, but we use the $d$ notation anyway because it's useful; it suggests things we can do. For instance, I can also calculate the force by splitting the surface into tiny $dz$ pieces. In our notation it looks very natural; it's just splitting fractions, $$\mathbf{F} = \int P \frac{d\mathbf{A}}{dz} \, dz.$$ Now $z$ is a perfectly good coordinate we can regard as changing; if we put an upper bound of $z_0$ then $\mathbf{F}(z_0)$ just gives the force on everything below $z = z_0$, and so on. It's by virtue that we can convert the $d\mathbf{A}$ to an actual differential (in this case, by parametrizing the area by $z$) that we use the differential notation for it too.

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The difference is between localized pressure($P_L$) and average pressure($P_A$). Localized pressure is defined at a given point with area $\Delta A \to 0$.

$$P_L(x,y)=\lim_{\Delta A\to 0}\frac{\Delta F}{\Delta A}=\frac{\mathrm{d}F}{\mathrm{d}A}$$

$$P_A=\frac{\iint P_L(x,y)\mathrm{d}x\mathrm{d}y}{\iint \mathrm{d}x\mathrm{d}y}$$ Since $$P_L(x,y)=\frac{\mathrm{d}F}{\mathrm{d}A}\\ \implies \text{Total Force}=\iint\mathrm{d}F=\iint P_L(x,y)\mathrm{d}A=\iint P_L(x,y)\mathrm{d}x\mathrm{d}y$$ Thus $$P_A=\frac{\text{Total Force}}{\text{Total Area}}$$

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