A cylindrical water tank 4 meters high with a radius of 2 meters is buried so that the top of the tank is 1 meter below ground level (see figure). How much work (in newton-meters) is done in pumping a full tank of water up to ground level? (The water weighs 9,800 newtons per cubic meter. Round your answer to the nearest integer.)

The x y-coordinate plane is given. There is a cylinder and a circular cross section of the cylinder.
The base of the cylinder is a circle centered at the origin that passes through the points (2, 0) and (−2, 0).
The height of the cylinder is 4.
The circular cross section of the cylinder has thickness Δy.
Ground level is at y = 5.
The distance between ground level and the cross section is labeled 5 − y.

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:
\[
\left[ y - \frac{y^2}{2} \right]_1^5 = \left( 5 - \frac{5^2}{2} \right) - \left( 1 - \frac{1^2}{2} \right) = \left( 5 - \frac{25}{2} \right) - \left( 1 - \frac{1}{2} \right)
= \left( 5 - 12.5 \right) - \left( 1 - 0.5 \right)
= (-7.5) - (0.5) = -8.
\]

### 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.

A cylindrical water tank 4 meters high with a radius of 2 meters is buried so that the top of the tank is 1 meter below ground level. How much work (in newton-meters) is done in pumping a full tank of water up to ground level? (The water weighs 9,800 newtons per cubic meter. Round your answer to the nearest integer.)

Learn how to calculate the work required to pump water from a full cylindrical tank (4 meters high, 2-meter radius) buried 1 meter below ground. This problem applies integration, calculus, and the work-energy theorem to determine the total work of 985,960 newton-meters, given the water weight of 9,800 newtons per cubic meter.

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