In a spin

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.

Image

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.

$$V=\frac{1}{3}\pi r^{2}h$$

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).

$$\begin{array}{rcl}V& =& \frac{1}{3}\pi r^{2}h\\ & =& \frac{1}{3}\pi r^2{h}_1+\frac{1}{3}\pi r^2{h}_2\\ & =& \frac{1}{3}\pi r^2(h_1+h_2)\\ & =& \frac{1}{3}\pi r^{2}b\end{array}$$

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

$$\frac{ac}{2}=\frac{rb}{2}$$

Therefore $$r=\frac{ac}{b}$$

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

$$\begin{array}{rcl}V& =& \frac{1}{3}\pi r^{2}b\\ & =& \frac{1}{3}\pi {\left(\frac{ac}{b}\right)}^{2}b\\ & =& \frac{1}{3}\pi \frac{{(ac)}^2}{b}\end{array}$$

From here you can substitute for the particular case given at the start of the problem.