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To solve this problem, we need to determine the work done to pump all the seawater from a conical tank to a height of 2 feet above the tank's rim, using the given specific weight of seawater.

### Steps:

1. **Setup the Geometry and Variables**:
   - The tank is a cone with height \( H = 4 \) ft and radius \( R = 3 \) ft.
   - The water fills the tank to a height of \( h = 1 \) ft.
   - Specific weight of seawater \( \gamma = 64 \frac{\text{lb}}{\text{ft}^3} \).

2. **Find the Relation between Water Level and Radius**:
   Since the cone's shape is consistent, the radius at any water level \( y \) can be described as \( r(y) = \frac{3}{4}y \) based on similar triangles (scaling the full tank radius to height).

3. **Volume of Water Element**:
   The differential volume \( dV \) of water at height \( y \) with thickness \( dy \) in the cone is:
   \[
   dV = \pi r(y)^2 \, dy = \pi \left(\frac{3}{4}y\right)^2 \, dy = \frac{9\pi}{16} y^2 \, dy
   \]

4. **Work to Lift Differential Volume**:
   The work \( dW \) to lift this water element to 2 ft above the rim is:
   \[
   dW = \gamma \times dV \times (6 - y) = 64 \times \frac{9\pi}{16} y^2 \, dy \times (6 - y)
   \]
   Here, \( (6 - y) \) is the lifting height because the water needs to be raised from height \( y \) to \( 6 \) ft above the ground (4 ft to the rim and additional 2 ft).

5. **Total Work**:
   Integrate \( dW \) from the bottom of the tank \( y = 0 \) to the top of the water level \( y = 1 \):
   \[
   W = \int_{0}^{1} 64 \cdot \frac{9\pi}{16} y^2 (6 - y) \, dy
   \]
   \[
   W = \frac{576\pi}{16} \int_{0}^{1} y^2 (6 - y) \, dy
   \]

   Let's compute this integral:
   \[
   \int_{0}^{1} y^2 (6 - y) \, dy = \int_{0}^{1} (6y^2 - y^3) \, dy
   \]
   Integrating term by term:
   \[
   \int_{0}^{1} 6y^2 \, dy = 6 \left[\frac{y^3}{3}\right]_0^1 = 2
   \]
   \[
   \int_{0}^{1} y^3 \, dy = \left[\frac{y^4}{4}\right]_0^1 = \frac{1}{4}
   \]
   \[
   \text{So,} \int_{0}^{1} (6y^2 - y^3) \, dy = 2 - \frac{1}{4} = \frac{7}{4}
   \]

   Plugging back:
   \[
   W = \frac{576\pi}{16} \times \frac{7}{4} = \frac{1008\pi}{16} = 63\pi \text{ ft-lb}
   \]

### Conclusion:
The work needed to pump all the water to a level 2 feet above the rim of the tank is \( 63\pi \) foot-pounds.

If you need further explanation or have any more questions, feel free to ask! Here are some topics you might consider:

1. Why did we use the relationship \( r = \frac{3}{4}y \)?
2. How is specific weight used in the calculation?
3. Could we change the lifting height for a different scenario?
4. What would happen if the tank's dimensions were different?
5. How do changes in the specific weight affect the final work?

**Tip**: Remember that the work required is directly proportional to the weight of the water and the vertical distance it needs to be moved. This understanding can help in evaluating scenarios with different fluids or lifting heights.

A conical tank contains seawater to a height of 1 ft. The tank measures 4 ft high and 3 ft in radius. Find the work needed to pump all the water to a level 2 ft above the rim of the tank. The specific weight of seawater is 64 lb/ft³. Give the exact answer (reduced fraction) in function of π.

This problem explores the work needed to pump water from a conical tank filled to a height of 1 ft. Using the specific weight of seawater (64 lb/ft³), integral calculus, and principles of physics, this solution provides an exact answer in terms of π. Suitable for calculus-based physics and mathematics students.

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