To solve this problem, we need to calculate the work required to pump half of the water out of the aquarium. The aquarium is a rectangular tank, so we can model the work as the integral of the force needed to pump the water over the distance it needs to be lifted.

Step-by-step breakdown:

1. Volume of the aquarium: The aquarium has dimensions of 3 m (length), 1 m (width), and 1 m (depth). The total volume $V$ of the tank is: $V = 3 \times 1 \times 1 = 3 \text{ m}^3$ Since we're pumping half of the water out, the volume of water to pump is: $V_{\text{half}} = \frac{3}{2} = 1.5 \text{ m}^3$

2. Weight of water: The density of water is given as $1000 \, \text{kg/m}^3$, and gravity is $9.8 \, \text{m/s}^2$. Therefore, the weight density of water is: $\text{Weight density} = 1000 \times 9.8 = 9800 \, \text{N/m}^3$

3. Riemann sum setup: Let $x$ be the height below the top of the tank, so that $x = 0$ corresponds to the top of the tank and $x = 1$ corresponds to the bottom. Each infinitesimal slice of water at height $x$, of thickness $\Delta x$, needs to be lifted a distance $x$ to the top of the tank.

   • The area of each horizontal slice is $3 \times 1 = 3 \, \text{m}^2$.
   • The volume of a thin slice of water at height $x$ is approximately $3 \Delta x \, \text{m}^3$.
   • The weight of that slice is: $\text{Weight of slice} = 3 \times 9800 \times \Delta x = 29400 \Delta x \, \text{N}$
   • The distance this slice needs to be lifted is $x$.

4. Work required to pump the slice: The work to pump this slice of water is: $\Delta W \approx \text{Weight of slice} \times \text{Distance lifted} = 29400 \Delta x \times x = 29400 x \Delta x \, \text{J}$

5. Riemann sum for the total work: To approximate the total work needed to pump half the water, we sum the work for slices from $x = 0.5$ to $x = 1$, since we're pumping out the top half of the water. This gives us the Riemann sum: $W \approx \sum_{i=1}^{n} 29400 x_i^* \Delta x$ where $x_i^*$ is the height of the $i$-th slice, and $\Delta x$ is the thickness of each slice.

6. Integral for exact work: In the limit as $\Delta x \to 0$, the Riemann sum becomes an integral. The total work $W$ is given by: $W = \int_{0.5}^{1} 29400 x \, dx$

7. Solve the integral: The integral is straightforward: $W = 29400 \int_{0.5}^{1} x \, dx = 29400 \left[ \frac{x^2}{2} \right]_{0.5}^{1}$ $W = 29400 \left( \frac{1^2}{2} - \frac{0.5^2}{2} \right) = 29400 \left( \frac{1}{2} - \frac{0.25}{2} \right) = 29400 \times \frac{0.75}{2}$ $W = 29400 \times 0.375 = 11025 \, \text{J}$

Thus, the work required to pump half of the water out of the aquarium is 11,025 joules.

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Would you like further details or have any other questions?

5 Relative Questions:

1. How would the result change if the aquarium were taller but had the same base area?
2. How would you approach the problem if the water had a different density?
3. What would be the work required to pump all the water out of the tank?
4. Can you set up a similar Riemann sum for a tank of different shape, such as a cylinder?
5. What is the physical significance of integrating the height over which water is pumped?

Tip:

Remember that when solving work problems involving lifting, the force and distance must be integrated, and the variable representing the changing distance should be carefully chosen based on the setup.