# Pressure at the bottom of a swimming pool [closed]

Why is atmospheric pressure it not taken into account when finding the force exerted by incompressible fluids onto a surface within the fluid? If you don't understand my question, then consider the example below:

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• it looks like there's air all around the swimming pool, so that atmospheric pressure acts on the other side (the dry side) of the floor of the swimming pool Nov 24 at 21:45

I find this text a bit confusing. The pressure field: $$P(z)=P_0+\rho gz$$ assuming that $$\rho$$ is constant. If $$z=0$$ is the surface of the water, then $$P_0$$ is the atmospheric pressure (the pressure field is continuous).
The text you're quoting says "due to the water". The author probably sees the above formula as a sum: $$P(z)=\underbrace{P_0}_{\text{atm}}+\underbrace{\rho gz}_{\text{water}}$$ In other words, $$\rho gz$$ is the additional pressure due to the water.
Assuming that the bottom of the pool is exposed to atmospheric pressure under it (for example because there's a tunnel letting air in from the surface), there yes, as you said, $$P_0$$ is present on both sides of the bottom and vanishes from the equation.