A cylinder has a bad direction. Around the girth it carries pd/2t, along the axis only half of that, so half the wall's strength along the axis is never called on. A sphere has no bad direction. It curves the same way whichever way you cut it, so the wall carries one stress, the same in every direction: $\sigma = \dfrac{pr}{2t}$, read aloud sigma equals p r over two t.

Name the letters. $\sigma$ is the membrane stress in the wall, in MPa. $p$ is the internal gauge pressure in MPa. $r$ is the internal radius in mm, and $t$ is the wall thickness in mm. Solve it for whichever one the question withholds. The millimetres cancel, so the stress arrives in MPa with no conversion owed.

Watch the letter, because this chapter has now used both. The cylinder lessons were written on the DIAMETER, pd/2t. This one is written on the RADIUS, and on the diameter it reads pd/4t. That is numerically the cylinder's longitudinal stress, and it is no coincidence: cut a sphere in half and the pressure on the flat face is resisted by one full ring of wall, which is exactly the cut that gave the cylinder its axial stress. Feed a diameter into the radius form and you get pr/t, a cylinder's hoop stress, and you have doubled your answer.

The practical upshot is worth money. For the same pressure, the same size and the same allowable stress, a sphere needs half the wall a cylinder does. That is why large gas storage vessels are spheres, and why a cylindrical receiver is closed with domed heads and never with flat plates. A hemispherical head may be rolled from plate half as thick as the shell it closes. As with the cylinder, this is thin-wall theory: it assumes the stress is uniform through the wall, and it holds while $t$ is no more than about a tenth of $r$.