# Why is the pressure on top of a slab of water greater than the pressure on the bottom?

In the image below, my book seems to be saying that if you have a cup of water in complete equilibrium, then if you have a small cube of water inside the cup given by $$dm=ρAdy$$ there is a downward force of $$-(p+dp)A$$ acting on the superior surface, a force $$pA$$ acting on the inferior surface, and $$-dw$$ which is the weight.

What I don't get is why the pressure on top is greater than the pressure on the bottom? Isn't this supposed to be the other way around? If anything, they're deriving Pascal's Law here.

Source: University Physics with Modern Physics - Sears & Zemansky

• You are saying the pressure on top must be greater since $p + dp > p$? May 19, 2014 at 23:39
• Yes. I think that makes sense, or it doesn't?
– DLV
May 19, 2014 at 23:41

Nothing says $\mathrm{d}p$ has to be positive.

The reason the pressure on the top is $p + \mathrm{d}p$, even though $\mathrm{d}p$ is not necessarily positive, is that the distance on the top is $y + \mathrm{d}y$. In other words, when you're dealing with differentials like this, you have to be consistent in which location you subtract from which other location. So if you decide that $\mathrm{d}y$ will be defined by

$$y_\text{top} - y_\text{bottom} = \mathrm{d}y\tag{1}$$

then it also has to be the case that

$$p_\text{top} - p_\text{bottom} = \mathrm{d}p\tag{2}$$

Or if it were temperature, you'd have to do

$$T_\text{top} - T_\text{bottom} = \mathrm{d}T$$

and so on. Always top minus bottom. Hopefully you can see the reason for this: derivatives are defined as

$$\frac{\mathrm{d}p}{\mathrm{d}y} = \lim_{\text{height}\to 0}\frac{p_\text{top} - p_\text{bottom}}{y_\text{top} - y_\text{bottom}}$$

and in particular, the ordering (top minus bottom) is the same in both the numerator and denominator.

You can switch it around in both places,

$$\frac{\mathrm{d}p}{\mathrm{d}y} = \lim_{\text{height}\to 0}\frac{p_\text{bottom} - p_\text{top}}{y_\text{bottom} - y_\text{top}}$$

which is equivalent to changing both equations (1) and (2) to

\begin{align} y_\text{bottom} - y_\text{top} &= \mathrm{d}y\\ p_\text{bottom} - p_\text{top} &= \mathrm{d}p \end{align}

but you have to change them both together so the signs cancel out when you calculate $\frac{\mathrm{d}p}{\mathrm{d}y}$.

In this particular case, pressure goes down as height goes up, so when $\mathrm{d}y$ is positive (assuming $y$ increases upward), $\mathrm{d}p$ will be negative. If it really bothers you, you could change to a coordinate system in which $y$ increases going downward, so that $p$ would increase with increasing $y$, but it's generally not worth the effort to figure out which direction will give you a positive derivative.

• Wait... If ptop-pbottom = dp actually gave a positive value, then the whole derivative dp/dy would be giving the wrong impression. Right? I say this because at the end of the derivation they say dp/dy=-ρg
– DLV
May 20, 2014 at 0:00
• In this case (assuming you still have $y$ increasing upwards), then $p_\text{top} - p_\text{bottom}$ giving a positive value would be wrong. It's not positive, it's negative. I'm not sure I see where you're going with this... May 20, 2014 at 0:05
• Oh I see, if it were the case that dp>0 then the equivalency wouldn't hold (dp/dy = -ρg). Right? I don't know, it's just that it seems to weird to me that the right relationship is springing out of here... I guess it really has nothing mysterious, but I just can't come to grips with it. By the way where can I find information about how to get the math format?
– DLV
May 20, 2014 at 0:09