Use this pi value in rounding to solve for this.
A cylindrical tank has a radius of 8m . The tank is 7m high, and the depth of the water is 4m . How much work, in Joules, is required to pump all water over the top of the tank? (If necessary, round to two decimal places.) F(x) = _____ J

To solve this problem, we need to calculate the work required to pump the water out of the cylindrical tank. We'll use the concept of work in physics, which is defined as:

\[
W = \int F(x) \, dx
\]

Where \( F(x) \) is the force at a given height \( x \), and the limits of integration will depend on the depth of the water.

### Given:
- Radius \( r = 8 \, \text{m} \)
- Height of the tank \( h = 7 \, \text{m} \)
- Depth of water \( d = 4 \, \text{m} \)
- Gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \)
- Density of water \( \rho = 1000 \, \text{kg/m}^3 \)
  
### Step-by-Step Solution:

1. **Volume of a cylindrical slice of water at height \( x \)**:
   The volume of a thin slice of water with thickness \( dx \) at height \( x \) is given by:
   \[
   dV = \pi r^2 dx
   \]
   Since the radius \( r \) is constant at 8 meters:
   \[
   dV = \pi (8^2) dx = 64\pi \, dx
   \]

2. **Mass of the water slice**:
   The mass of the water in this slice is the volume multiplied by the density of water \( \rho \):
   \[
   dm = \rho dV = 1000 \times 64\pi \, dx = 64000\pi \, dx
   \]

3. **Force required to lift the water slice**:
   The force required to lift this slice of water is the weight of the water slice, which is mass times gravitational acceleration:
   \[
   F(x) = dm \cdot g = 64000\pi \cdot 9.8 \, dx = 627200\pi \, dx
   \]

4. **Distance to pump the water**:
   To calculate the work done to pump the water over the top of the tank, the slice at height \( x \) needs to be lifted a distance of \( (7 - x) \) meters, since the tank is 7 meters high.

5. **Work to pump the water slice**:
   The work to pump a single slice of water is:
   \[
   dW = F(x) \cdot \text{distance} = 627200\pi (7 - x) \, dx
   \]

6. **Total work**:
   The total work is the integral of \( dW \) from \( x = 0 \) (bottom of the tank) to \( x = 4 \) (the depth of the water):
   \[
   W = \int_0^4 627200\pi (7 - x) \, dx
   \]

   Let’s compute this integral.

\[
W = 627200\pi \int_0^4 (7 - x) \, dx
\]
\[
W = 627200\pi \left[ 7x - \frac{x^2}{2} \right]_0^4
\]
\[
W = 627200\pi \left[ (7(4) - \frac{4^2}{2}) - (0 - 0) \right]
\]
\[
W = 627200\pi \left[ 28 - 8 \right] = 627200\pi \times 20
\]
\[
W = 12544000\pi
\]

7. **Substitute the value of \( \pi \)**:
   Using \( \pi \approx 3.1416 \):
   \[
   W = 12544000 \times 3.1416 = 39422502.4 \, \text{J}
   \]

### Final Answer:
\[
F(x) = 39422502.4 \, \text{J}
\]

Would you like more details on any part of the solution?

Here are 5 related questions for practice:
1. What would the work be if the depth of the water was 5 meters instead of 4?
2. How would the work change if the radius of the tank was reduced to 6 meters?
3. What is the formula for the volume of a cylindrical tank?
4. How much work is required to pump only half the water out of the tank?
5. What is the potential energy of the water at the top of the tank?

**Tip**: When solving work problems in physics, always carefully consider the distance the object moves and the force required over that distance.

A cylindrical tank has a radius of 8m. The tank is 7m high, and the depth of the water is 4m. How much work, in Joules, is required to pump all water over the top of the tank? (If necessary, round to two decimal places.) F(x) = _____ J

This detailed solution explains how to calculate the work required to pump water from a cylindrical tank with a radius of 8 meters and a water depth of 4 meters. Using calculus, we integrate the force over the height of the tank and apply physics formulas to determine the total work in Joules. Suitable for physics or calculus students at the undergraduate level.

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