Force Exerted By a liquid integration problem

[B3NR4Y]

Homework Statement

A rectangular swimming pool is 20 feet wide and 40 feet long. The depth of the water varie uniformly from 3 feet at one end to 9 feet at the other end. Find the total force exerted at the bottom of the pool.

Homework Equations

The force F exerted by a liquid of constant density "p", where the functions f and g are continuous on [c,d], is
\begin{equation}
F=\int_c^d \, p(k-y)[f(y)-g(y)]dy
\end{equation}
The equation of the line making up the bottom of the pool is
\begin{equation}
\frac{-3}{10}x+6=y
\end{equation}

The Attempt at a Solution

The depth of any rectangle below the surface would be (9-y), I reasoned.
I tried to do the integration
\begin{equation}
\begin{split}
F&=62.5\int_0^6 (9-y)(\frac{10(y-6)}{-3})dy\\
&=26250\\
\end{split}
\end{equation}
The book says this answer is wrong, as I suspected. I can't seem to find a way to think of it. Ways I've tried:
Finding the volume of water above the wedge caused by the incline + the wedge volume * 62.5

 

[awkward]

You don't really need calculus for this problem. The total force is just the volume of the pool times the density of water. Notice that a cross section of the pool is a trapezoid.