To calculate the work $W$ required to pump the water out of the tank, we need to apply the following physics and calculus principles:

1. Work and Force Relationship:
   Work is defined as the integral of force applied over a distance. In this case, the force is due to the weight of water, and the distance depends on the height the water needs to be pumped.

2. Volume and Weight of Water:
   The weight of a small slice of water depends on its volume and density. Water has a density of $1000 \, \text{kg/m}^3$, and the weight per unit volume (due to gravity) is $\rho \cdot g = 1000 \cdot 9.8 \, \text{N/m}^3 = 9800 \, \text{N/m}^3$.

3. Tank Geometry:
   The tank appears to have a non-standard shape, so we need to carefully calculate the volume of each horizontal slice of water. However, the general approach for calculating the work remains the same.

Now, let’s set up the integral for the work calculation.

Assumptions:

1. The dimensions of the tank:

   • $a = 4 \, \text{m}$ (possibly the width of the base).
   • $b = 4 \, \text{m}$ (possibly the length of the base).
   • $c = 18 \, \text{m}$ (the height of the tank).
   • $d = 2 \, \text{m}$ (possibly the spout's height above the tank).

2. Let $x$ represent the depth from the spout at the top, meaning the water at depth $x$ needs to be lifted by a distance of $c + d - x$.

3. The cross-sectional area $A$ of the tank remains constant at $a \times b = 4 \times 4 = 16 \, \text{m}^2$.

Step 1: Elemental Work for a Slice at Depth $x$

The volume of a thin horizontal slice of water at depth $x$ is: $\Delta V = A \, dx = 16 \, dx$ The weight of this slice is: $\Delta F = \rho g \Delta V = 9800 \cdot 16 \, dx = 156800 \, dx$

Step 2: Work to Pump the Slice

The work to pump this slice out of the spout involves lifting it a distance of $(c + d - x) = (18 + 2 - x) = (20 - x)$ meters. Thus, the elemental work is: $dW = 156800 \cdot (20 - x) \, dx$

Step 3: Total Work

Now, integrate this from the bottom of the tank ($x = 0$) to the top of the tank ($x = 18$): $W = \int_0^{18} 156800 \cdot (20 - x) \, dx$ We can now evaluate this integral.

$W = 156800 \int_0^{18} (20 - x) \, dx$ $W = 156800 \left[ 20x - \frac{x^2}{2} \right]_0^{18}$ $W = 156800 \left[ (20 \cdot 18) - \frac{18^2}{2} \right]$ $W = 156800 \left[ 360 - \frac{324}{2} \right]$ $W = 156800 \left[ 360 - 162 \right] = 156800 \times 198 = 31046400 \, \text{Joules}$

Thus, the total work required to pump the water out of the tank is: $W = 31,046,400 \, \text{Joules}$

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Would you like further clarification or details on any steps?

Here are five related questions for further practice:

1. How would the work change if the tank height was increased to 20 meters?
2. What if the cross-sectional area of the tank was circular instead of rectangular?
3. How much work is required to pump half of the water out of the tank?
4. How does changing the density of the liquid in the tank affect the work required?
5. How would the work change if the tank was not full initially?

Tip: When solving similar problems, visualizing the geometry of the tank and the lifting distances is crucial for setting up the correct integral.