Spinning Card

If side lengths be $x$ and $y$ , which $x \leq y$ ,

and the hypotenuse be $r=\sqrt{x^2 +y^2}$

Then using Calculus or finding the volume of a few cones, we can found that the volume can be expressed by

$V=\frac{\pi x^2 (8y^4 -r^4)}{12y^2 r}$

For this question, x=12, y=16, r=20.

So volume is $\frac{4269}{5} \pi$

$4269+5=\boxed{4274}$ .

---

Now I will reveal a solution using geometry.

In the graph, apparently the total volume is twice the volume that $V_a$ and $V_b$ make.

Area of triangle:

$\frac{1}{2}DB \times AC =\frac{1}{2} AB \times BC$

$DB=\frac{12 \times 16}{20}=9.6$

$EC=10$ , $\frac{EF}{EC}=\frac{DB}{DC}=\frac{AB}{BC}$ , $EF=7.5$

Half of the volume can be expressed by

$V_a+V_b=\frac{1}{3} \pi DB^2 AC -V_c=\frac{1}{3} \pi 9.6^2 \times 20 - \frac{1}{3}\pi EF^2 EC=\frac{1}{3} \pi 9.6^2 \times 20 - \frac{1}{3} \pi 7.5^2 \times 10$

$\frac{1}{2}V=426.9\pi$

$V=853.8 \pi=\frac{4269}{5}\pi$

$4269+5=\boxed{4274}$