# Homework Help: Force Exerted By a liquid integration problem

1. Oct 28, 2012

### B3NR4Y

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

The force F exerted by a liquid of constant density "p", where the functions f and g are continuous on [c,d], is

F=\int_c^d \, p(k-y)[f(y)-g(y)]dy

The equation of the line making up the bottom of the pool is

\frac{-3}{10}x+6=y

3. 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{split}
F&=62.5\int_0^6 (9-y)(\frac{10(y-6)}{-3})dy\\
&=26250\\
\end{split}

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

Last edited: Oct 28, 2012
2. Oct 29, 2012

### 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.