Proving P1>Po: A Mathematical Analysis

AI Thread Summary
The discussion centers on proving that P1 is greater than Po using mathematical equations related to fluid pressure. The user derives the equation P1 - Po = (rho)(g)(del(y)), indicating that since rho(g) is a positive constant, P1 must be greater than Po. The user expresses uncertainty about the next steps in the proof but concludes that the derived equation supports the claim that P1 > Po. A diagram is referenced to aid in understanding the problem. The analysis effectively demonstrates the relationship between the pressures based on the given parameters.
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Homework Statement


Prove that P1>Po

Homework Equations


dp=-rho(g)dy
integral(Po,P1)dp=-integral(Y2,Y1) rho(g)dy

The Attempt at a Solution


dp=-rho(g)dy
intgrl(Po,P1) dp=-integral(Y2,Y1) rho(g)dy
Po-P1=-(rho(g))(Y2-Y1)
Po-P1=(rho(g))(Y1-Y2)
Y2-Y1-->Y1-Y2<0 defined as -del(y)
Po-P1=-(rho)(g)(del(y))
P1-Po=+(rho)(g)(del(y))

not sure where to proceed or if this proves that P1>Po. Please help.

Oh and the attached file has a diagram on it.
 

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Since rho(g) is a positive constant, the equation P1-Po=+(rho)(g)(del(y)) implies that P1 is bigger than Po. Thus, P1>Po.
 
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