Geometry problem - volume calculation

Label the cross-section as follows:

Then $OB = OD = OE = 5$ and $OA = 13$ , and by the Pythagorean Theorem on $\triangle OBA$ , $AB = AD = 12$ .

Since $\triangle OCB \sim \triangle OBA$ by AA similarity, $\frac{OC}{BC} = \frac{OB}{AB}$ , or $\frac{OC}{BC} = \frac{5}{12}$ , and by the Pythagorean Theorem on $\triangle OCB$ , $OC^2 + BC^2 = OB^2$ or $OC^2 + BC^2 = 25$ . These equations solve to $OC = \frac{25}{13}$ and $BC = \frac{60}{13}$ . By segment addition, $AC = AO - OC = 13 - \frac{25}{13} = \frac{144}{13}$ , and $EC = OE - OC = 5 - \frac{25}{13} = \frac{40}{13}$ .

The desired volume is the difference of the volume of the cone with a radius of $BC = \frac{60}{13}$ and a height of $AC = \frac{144}{13}$ , and the volume of the spherical cap ( $V_{\text{cap}} = \frac{1}{3}\pi h^2 (3R - h)$ ) where $h = EC = \frac{40}{13}$ and $R = OB = 5$ .

Therefore, $V = V_{\text{cone}} - V_{\text{cap}} = \frac{1}{3}\pi (\frac{60}{13})^2 \frac{144}{13} - \frac{1}{3}\pi (\frac{40}{13})^2 (3 \cdot 5 - \frac{40}{13}) = \frac{1600}{39}\pi$ , which means $a = 1600$ , $b = 39$ , and $a + b = \boxed{1639}$ .

In the cross section, we have the following situation,

The semi-vertical angle of the cone is $\theta = \tan^{-1}(\frac{5}{12})$ , therefore angle $\phi = \frac{\pi}{2} - \theta =\tan^{-1}(\frac{12}{5})$ . The radius of the base of the cone is $r = 5 \sin \phi = 12 \sin \theta = \dfrac{60}{13}$ .

The surface area of lateral surface of the cone is

$S_1 = \pi r s = \dfrac{60}{13} (12) \pi = \dfrac{720}{13} \pi$

and the surface area of the spherical cap is given by,

$S_2 = \displaystyle 2 \pi (5)^2 \int_{0}^{\phi} \sin t dt = 50 \pi (1 - \cos \phi) = 50 \pi (1 - \frac{5}{13} ) = \frac{400}{13} \pi$

Hence, the volume contained between them is given by,

$V = \frac{1}{3} ( 5 S_1 - 5 S_2 ) = \dfrac{5}{3} \pi (\dfrac{320}{13}) = \dfrac{1600}{39} \pi$

Therefore, the answer is $1600 + 39 = \boxed{1639}$ .