A cone

Let the original right triangle be solvable by the Pythagorean Theorem: $(3x)^2 + (4x)^2 = 25^2 \Rightarrow 25x^2 = 25^2 \Rightarrow x = 5.$ , which gives us the $15-20-25$ triplet. Let $r$ be the length of the altitude drawn from the right angle to the hypothenuse (which also serves as the common radius of the double-cone of revolution) such that it divides the hypothenuse into lengths $y$ and $25-y$ . This radius is equal to:

$\sqrt{15^2-y^2} = r = \sqrt{20^2 - (25-y)^2} \Rightarrow y = 9 \Rightarrow r = 12.$

Finally, the double-cone volume is computed per:

$V = \frac{\pi}{3} \cdot r^2 (y + (25-y)) = \frac{\pi}{3}r^2(25) = \frac{\pi}{3}(12)^2 (25) = \boxed{1200\pi}.$