Correct solutions were received from Joshua T. and Andrew I. Congratulations to both of you. I have used a mixture of the solutions to form the basis of what follows.

The first thing to observe is that by rotating the triangle about the hypotenuse I obtain two cones having the same base and opposite vertices.

Triangle ready to rotate

To find the volume of a cone you need its height and the radius of its base. So you need to calculate the radius and the height of each cone.

v = third pi r squared times h

The radius of the base of the cones is the altitude of the right-angled triangle (r), and the corresponding heights are AX and CX.

Working directly on the general case - as it seems simpler. Let the sides of the triangle be a, c and b (b- the hypotenuse).

Maths workings, leading to V equals a third pi r squared times b.

Finding the area of a right-angled triangle in two ways:

r equals a times c over b.

Substituting for r in the equation for the volume you get:

V equals pi times (a times c) squared over b.
From here you can substitute for the particular case given at the start of the problem.