Force and position on inclined door

The figure is only partly legible. Assumed reading: a circular door of diameter 3 m (radius $R=1.5$ m, length 3 m along the slope), inclined at $\theta=60^\circ$ to the horizontal, with its upper edge 1.5 m below the water surface. The air above the water has gauge pressure $p_0=800$ kPa. $g=9.81$.

   air 800 kPa
  ---------------- water surface
   1.5 m
    \
     \  door 3 m along slope
      \   60 deg

Depth of centroid:

$$\bar h=1.5+R\sin60^\circ=1.5+1.5(0.866)=2.799\ \text{m}$$

Pressure at the centroid:

$$p_c=p_0+\gamma\bar h=800+9.81(2.799)=827.46\ \text{kPa}$$

Area: $A=\pi R^{2}=\pi(1.5)^{2}=7.069\ \text{m}^2$.

Force:

$$F=p_cA=827.46\times7.069=5849\ \text{kN}$$

Position. The air pressure is equivalent to an extra head $p_0/\gamma=81.55$ m of water above the surface. Distance of the centroid from this imaginary free surface along the slope:

$$\bar y=\frac{81.55+2.799}{\sin60^\circ}=97.40\ \text{m}$$ $$\bar y_{cp}-\bar y=\frac{I_G}{A\bar y}=\frac{\pi R^{4}/4}{\pi R^{2}\,\bar y}=\frac{R^{2}}{4\bar y}=\frac{2.25}{4(97.40)}=0.0058\ \text{m}$$

Answer: $F\approx5.85\times10^{3}$ kN (5849 kN), acting normal to the door at about 5.8 mm below the centre of the door, measured along the slope. The large air pressure makes the centre of pressure almost coincide with the centroid.