A cylindrical tank has a height of 6 m and a radius of 1 m. How much work is required to pump all of the water out of the tank if the tank is filled halfway? Round your answer to the nearest tenth of a Joule.

The problem asks to calculate the work required to pump all of the water out of a cylindrical tank that is filled halfway. 

The height of the tank is 6 meters, and the radius is 1 meter. The tank is half-filled with water, so the water height is \( 3 \) meters. We need to compute the work required to pump the water to the top of the tank.

### Key Formulas:
1. **Work Formula**:
   \[
   W = \int_a^b \text{Force} \cdot \text{Distance} \, dy
   \]
   Here:
   - Force = weight of the water in a thin slice of the tank at height \( y \),
   - Distance is how far the slice must be pumped (from height \( y \) to the top of the tank).

2. **Volume of a thin slice**:
   A thin slice of water at height \( y \) has a thickness \( dy \), and its volume is:
   \[
   dV = \pi r^2 dy
   \]
   where \( r = 1 \) m is the radius of the tank.

3. **Mass and Force**:
   - The density of water is \( \rho = 1000 \, \text{kg/m}^3 \),
   - Weight of the water slice is:
     \[
     dF = \text{mass} \cdot g = \rho dV \cdot g = 1000 \cdot \pi \cdot 1^2 \cdot dy \cdot 9.8
     = 9800\pi \, dy \, \text{Newtons}.
     \]

4. **Work for each slice**:
   The distance the slice at height \( y \) needs to be pumped is \( 6 - y \) (since the top of the tank is at 6 meters), so the work for a small slice is:
   \[
   dW = 9800 \pi (6 - y) dy.
   \]

### Total Work:
The total work to pump out all the water from the bottom to \( y = 3 \) meters (since the tank is half-filled) is the integral:
\[
W = \int_0^3 9800 \pi (6 - y) \, dy
\]

Now, let's calculate this integral.

\[
W = 9800\pi \int_0^3 (6 - y) \, dy
\]

This integral simplifies to:
\[
W = 9800\pi \left[ 6y - \frac{y^2}{2} \right]_0^3
\]
Substituting the limits of integration:
\[
W = 9800\pi \left[ 6(3) - \frac{3^2}{2} - 0 \right]
= 9800\pi \left[ 18 - 4.5 \right]
= 9800\pi \times 13.5
\]

Now, compute the numerical value of this expression:
\[
W = 9800 \times 13.5 \times \pi = 132,300\pi \, \text{Joules}
\]

Using \( \pi \approx 3.1416 \), we find:
\[
W \approx 132,300 \times 3.1416 = 415,767.8 \, \text{Joules}
\]

### Final Answer:
The work required to pump the water out of the tank is approximately \( \boxed{415,767.8} \) Joules, or rounded to the nearest tenth, \( 415,767.8 \, \text{Joules} \).

Would you like a detailed breakdown of any part of this solution?

### Related Questions:
1. How would the work change if the tank were full?
2. What would the work be if the radius were doubled but the height stayed the same?
3. How would the work differ if the tank were a cone instead of a cylinder?
4. What effect does the gravitational constant \( g \) have on the work calculation?
5. How does the height from which the water is pumped influence the total work?

### Tip:
When solving work problems involving integrals, carefully consider how the distance changes with height, as it is a critical part of the calculation.

Learn how to calculate the work required to pump water from a half-filled cylindrical tank with a height of 6 meters and a radius of 1 meter. This solution involves key concepts from calculus and physics, including definite integrals and the work-energy theorem. Suitable for advanced high school students in Grades 11-12.

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