Pressure at a certain depth when density varies

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pressurised
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Homework Statement


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Homework Equations


dp/dz=-ρg
p=ρgh

The Attempt at a Solution


I've found the density at depth 3m using ρ=ρ0+kh, which gave me 1060kgm-3. I then put this value into ρgh to get 31195.8Pa which seems to be ≈+1000 off the answer.

What is the correct mathematical way of solving this as I am not quite sure how to form the equation for variation of pressure using that.
 
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Think of summing the pressure contributions from thin layers stacked one on top of the next from the surface down to the desired depth. Does that remind you of anything?
 
gneill said:
Think of summing the pressure contributions from thin layers stacked one on top of the next from the surface down to the desired depth. Does that remind you of anything?

I can't quite think of the name of it but I understand what you mean:sorry:.

I would have thought ∫-(ρ0+kh)g dz would have been okay to use?
 
pressurised said:
I would have thought ∫-(ρ0+kh)g dz would have been okay to use?
That's the idea. You'll need to relate the h to your integration variable z. You should pay attention to the "direction" that your integration path takes as it affects the sign of the dz differential element and hence whether that leading minus sign is warranted.
 
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gneill said:
That's the idea. You'll need to relate the h to your integration variable z. You should pay attention to the "direction" that your integration path takes as it affects the sign of the dz differential element and hence whether that leading minus sign is warranted.

Thank you!