Contact pressure of piston rod onto a surface?

[randombill]

I have two images below.

This is a single piston with pressure inside, the variables are as listed.

P_in = 2700 Pa.
P_out = 0 Pa.
R1 = Radius 1 = 10 cm.
R2 = Radius 2 = 5 mm.
P_contact = ?

My solution to P_contact is as follows.

Area A1 of piston at R1 = .0314m^2 and area A2 at R2 = 7.85x10^-5 m^2. The force exerted by P_in on the face of the piston with A1 is 84.78 N. Therefore P_contact = 84.78/ A2 = 1080000 Pa. I'm just wondering if this is correct?

My second picture has a dual piston setup and my question is whether the P_contact variable on either side is halved?

 

[nasu]

Assuming that the black area is a fluid, P_in will be the same as P_contact. The forces on the two sides will be different though.

 

[randombill]

Assuming that the black area is a fluid, P_in will be the same as P_contact. The forces on the two sides will be different though.

Oh no, forgot to mention the black area would be a solid like steel or something. The Pressure for P_in would be due to a gas like air and the grey area would be a solid too like steel or wood.