is greater than 90º, but that is not relevant to the, projection of the
upper part of the dome and the lower part is the same.
The projected area of the dome is a circle with radius R.
We have
F
z
=
−
p dA
z
A
Dome
∫
=
−
p(z)rdrd
θ
0
R
∫
0
2
π
∫
.
A cylindrical coordinate system is being used to do the integral over the circular projected area.
Using
the expression above for the pressure, the z-force is (absent the contribution of atmospheric pressure)
F
z
=
−ρ
g
L
−
z
(
)
rdr
0
R
∫
0
2
π
∫
d
θ
.
This integral yields the volume of seawater above the dome, which is the volume of a cylinder minus
the volume of a half-sphere, so
F
z
=
−ρ
g
π
R
2
L
−
2
π
3
R
3
⎡
⎣
⎢
⎤
⎦
⎥
.
The atmospheric pressure is present in the seawater, so its effect should be added to yield
F
z
=
−
P
atm
π
R
2
− ρ
g
π
R
2
L
−
2
π
3
R
3
⎡
⎣
⎢
⎤
⎦
⎥
.
The atmospheric contribution could have been included in the expression for p(z) and this result would
have been obtained naturally.
Note:
Since the question asks for the force due to hydrostatic pressure, and atmospheric pressure is in
the hydrostatic pressure, I think it is best to include it in this result.
If the question asked for “the net

force,” then the pressure inside the dome would have to be included.
If the pressure inside the dome