We are tasked with setting up the integral that represents the work required to pump all the water out of a cylindrical tank.

Problem Breakdown:

1. Tank Dimensions:

   • Height $h = 10$ ft.
   • Radius $r = 3$ ft.
   • Filled with water up to 4 ft (depth of water is $4$ ft).

2. Water Properties:

   • Weight density of water $\rho = 62.4 \, \text{lb/ft}^3$.

3. Reference:

   • $x = 0$ is the bottom of the tank.
   • The water must be pumped to a height of $11$ ft above the bottom.

Work Formula:

The work $W$ to pump a small slice of water at height $x$ to the top of the tank is given by: $\text{Work for slice} = (\text{Weight of slice}) \cdot (\text{Distance to be pumped})$ We'll integrate over the height of the water to find the total work.

Step-by-Step Setup:

1. Volume of a Thin Slice: A thin slice of water at height $x$ has:

   • Thickness $dx$.
   • Cross-sectional area of the tank $A = \pi r^2 = \pi (3^2) = 9\pi \, \text{ft}^2$.

   Volume of the slice: $\text{Volume} = A \cdot dx = 9\pi \, dx.$

2. Weight of the Slice: The weight of the slice is the volume multiplied by the weight density: [ \text{Weight} = \rho \cdot \text{Volume} = 62.4 \cdot 9\pi