The work required to pump the water to the ground level can be calculated using the concept of integration, where we find the work done to lift each slice of water and sum it over the height of the tank.

Given:

• The height of the cylindrical water tank is 4 meters.
• The radius of the tank is 2 meters.
• The top of the tank is 1 meter below the ground level.
• The water weighs 9,800 newtons per cubic meter.

We will divide the tank into thin slices of water at height $y$, each of thickness $\Delta y$, and compute the work done to lift each slice to the ground level. The total work is then the integral of the work for each slice.

Step 1: Coordinates and Slices

• The water level ranges from $y = 1$ (the top of the tank) to $y = 5$ (the bottom of the tank).
• The volume of a thin slice of water at height $y$ with thickness $\Delta y$ is the area of the cross-section of the cylinder times $\Delta y$. The cross-sectional area is the area of a circle with radius 2 meters:
  $A = \pi r^2 = \pi (2^2) = 4\pi \, \text{m}^2.$
• The volume of each slice is therefore: $dV = 4\pi \, \Delta y.$

Step 2: Weight of the Slice

The weight of each slice is the volume times the weight density of water, which is 9,800 newtons per cubic meter: $dW_{\text{weight}} = 9,800 \cdot 4\pi \, \Delta y = 39,200\pi \, \Delta y \, \text{newtons}.$

Step 3: Distance to Ground Level

The distance that each slice of water at height $y$ must be lifted to reach the ground level (which is at $y = 1$) is $1 - y$.

Step 4: Work Done on Each Slice

The work done to lift each slice is the weight of the slice times the distance it is lifted: $dW = 39,200\pi \cdot (1 - y) \, \Delta y.$

Step 5: Total Work

To find the total work, we integrate this expression from $y = 1$ (the top of the tank) to $y = 5$ (the bottom of the tank): $W = \int_1^5 39,200\pi \cdot (1 - y) \, dy.$

Step 6: Solving the Integral

We solve the integral: $W = 39,200\pi \int_1^5 (1 - y) \, dy.$ The integral of $1 - y$ is: $\int (1 - y) \, dy = y - \frac{y^2}{2}.$ Evaluating this from 1 to 5:

= \left( 5 - 12.5 \right) - \left( 1 - 0.5 \right) = (-7.5) - (0.5) = -8.$$ ### Step 7: Total Work Now multiply by $$39,200\pi$$: $$W = 39,200\pi \cdot (-8) = -313,600\pi \, \text{newton-meters}.$$ Approximating $$\pi \approx 3.1416$$: $$W \approx -313,600 \cdot 3.1416 = -985,960 \, \text{newton-meters}.$$ Since work is typically positive, we take the absolute value: $$W \approx 985,960 \, \text{newton-meters}.$$ Rounding to the nearest integer: $$W \approx 985,960 \, \text{newton-meters}.$$ Thus, the total work required to pump the water to the ground level is approximately **985,960 newton-meters**. Would you like more details or have any questions? Here are 5 related questions: 1. How would the work change if the radius of the tank were doubled? 2. What if the tank were only half-filled with water, how would that affect the work required? 3. Can you explain how to generalize this method for different tank shapes? 4. How does the weight density of water affect the amount of work? 5. What would happen if the tank were buried 2 meters below the ground instead of 1? **Tip**: When dealing with work integrals in physics problems, always pay attention to the limits of integration, as they determine the range of movement in your system.