Suppose a solid, right-circular, cylindrical body has a long axis of length $L$ and end-faces each having a diameter $D$ was placed fully submerged into a fluid having a hydrostatic pressure of $p$, as shown in the figures below.
If one wanted to know the stress distribution within the cylinder's body one might bisect the cylinder in two ways as shown in the figure. One way would be to bisect along the cylinder's long axis, denoted as Surface 1, where the surface $A_1$ is equal to $LD$. Another way would be to bisect the cylinder diametrically, denoted as Surface 2 in the figure, where its surface area $A_2$ is equal to $\frac{\pi D^2}{4}$.
At first thought, if the confining fluid pressure acting on the cylinder's outer surfaces is hydrostatic, then I would think the resulting stress induced within the cylinder's body would be isostatic. However, would this be true for any length $L$ to end-face diameter $D$ relationship? How might one show the resulting axial (along the cylinder's long axis) and lateral (or radial-direction) internal stress using a free body diagram?
My attempt at this is as follows. For Surface 1 I can imagine the resulting internal forces having to counter act the normal forces $F_n$ acting on the plane caused by the external pressure. I might show this using the figure below.
Assuming some infinitesimal area for the external pressure to act on, I would think I could find the force acting on Surface 1 by first integrating the pressure force acting on the quarter-circle and then multiplying by 2 and then by the length of the cylinder $L$, i.e.,
$$F_n=\int_0^{\frac{\pi}{2}}F_R \cos\theta ~\mathrm d\theta$$
evaluating the integral,
$$F_n = F_R\left[\sin\left(\frac{\pi}{2}\right)-\sin(0)\right]=F_R$$
multiplying by 2 and by $L$,
$$F_n = 2F_RL$$
Since pressure is force over area, $F_R$ is equated to pressure in the following manner
$$F_R=pA=p\left(\frac{\pi D}{2}\right)L$$
where $\left(\frac{\pi D}{2}\right)$ is the arc-length of the half-circle. Substituting, we solve for the stress acting on surface 1,
$$\sigma_1=\frac{F_{n_1}}{A_1}=\frac{2p(\frac{\pi D}{2})L^2}{DL}=p\pi L$$
The stress for surface 2 due to the pressure acting on the end face would be
$$\sigma_2=p$$
Therefore, the cylinder's length to diameter stress relationship might be said to be equal to
$$\frac{p\pi L}{p}=\pi L$$