water in the pail

The volume of a frustum of a right circular cone is given by $v=\dfrac{h}{3}\left(A_1+A_2+\sqrt{A_1 \times A_2}\right)$ where $A_1$ and $A_2$ are the base areas and $h$ is the height. The base areas are

$A_1=\dfrac{\pi}{4}(14^2) \approx 153.938~in^3$

$A_2=\dfrac{\pi}{4}(10^2) \approx 78.54~in^3$

Since it is $\dfrac{3}{4}$ full of water, we have

$v=\dfrac{3}{4}\left(\dfrac{18}{3}\right)(153.938+78.54+\sqrt{153.938 \times 78.54})\approx \color{#D61F06}\boxed{1541~in^3}$