Calculative or not?

If the base and height of the right triangle be b and h respectively then, The volume when the triangle is rotated about its base will be = 1/3πh^2b. And the volume when the triangle is rotated about its height will be = 1/3πb^2h. Hence the ratio of the volumes will be = h/b. In this case, the ratio of volumes will be = 18/24 = 3/4. So it is not calculative.

After revolving, it forms two regular cones. Solving for their volumes

$V_1 = \frac{1}{3}\pi(18)^2(24) = \frac{1}{3}\pi(7766)$

$V_2 = \frac{1}{3}\pi(24)^2(18) = \frac{1}{3}\pi(10368)$

Solving for the ratio of their volumes

$\frac{V_1}{V_2} = \frac{7776}{10368} = \frac{3}{4}$

$Given \space a \space right \space \triangle \space with \space base \space= \space 24 \space and \space height \space= \space 18$

$After \space revolving \space the \space \triangle \space along \space either \space of \space its \space arms \space , \space we \space will \space obtain \space a \space \boxed{CONE}$

$Case \space 1 \space:$

$Radius \space of \space cone \space = \space height \space of \space \triangle \space= \space 18$

$Height \space of \space cone \space = \space base \space of \space \triangle \space= \space 24$

$\therefore \space Volume \space (V_{1}) \space=\space \frac{1}{3}\space\pi (18)^{2} \times 24$

$Case \space 2 \space:$

$Radius \space of \space cone \space = \space base \space of \space \triangle \space= \space 24$

$Height \space of \space cone \space = \space height \space of \space \triangle \space= \space 18$

$\therefore \space Volume \space (V_{2}) \space=\space \frac{1}{3}\space\pi (24)^{2} \times 18$

$\Rightarrow \space Volume \space Ratio \space = \space \frac{V_{1}}{V_{2}} \space=\space \frac{ \frac{1}{3}\space\pi (18)^{2} \times 24}{\frac{1}{3}\space\pi (24)^{2} \times 18} \space = \space \frac{3}{4}$

$\therefore \space \boxed{Volume \space Ratio \space = \space 3 \space : \space 4}$