$\dfrac{12}{L} = \dfrac{1}{3}$
$L = 36 ~ \text{in.}$
$V_1 = \text{Frustum of a right circular cone}$
$V_1 = \frac{1}{3}\pi h(R^2 + r^2 + Rr)$
$V_1 = \frac{1}{3}\pi (12)[ \, 6^2 + 2^2 + 6(2) \, ]$
$V_1 = 208\pi ~ \text{in.}^3$
$V_2 = \text{Right circular cylinder}$
$V_2 = pi r^2 h$
$V_2 = \pi (2^2)(6)$
$V_2 = 24\pi ~ \text{in.}^3$
$V_3 = \text{Frustum of a right circular cone}$
$V_3 = \frac{1}{3}\pi L(r^2 + R^2 + rR)$
$V_3 = \frac{1}{3}\pi(36) [ \, 2^2 + 6^2 + 2(6) \, ]$
$V_3 = 624\pi ~ \text{in.}^3$
Total volume
$V = V_1 + V_2 + V_3$
$V = 208\pi + 24\pi + 624\pi$
$V = 856\pi ~ \text{in.}^3$
$V = 2689.2 ~ \text{in.}^3$ answer
