To tackle this problem, we need to set up integrals for the work done to pump water out of the tank in various scenarios. The work done ($W$) is calculated as the product of force ($F$) and distance ($d$), and for a continuous distribution of mass, the integral approach is used:

Key Elements

• Density of water ($\rho$): $1,000 \, \text{kg/m}^3$
• Acceleration due to gravity ($g$): $9.8 \, \text{m/s}^2$
• Radius of hemisphere: $3 \, \text{m}$
• Volume of a thin horizontal slice of water ($dV$): $\pi \cdot r^2 \, dy$, where $r$ is the radius of the slice, and $y$ is the vertical position of the slice.

The work required to pump out water is determined by integrating the gravitational force over the vertical distance the water needs to travel.

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Part (a): Full tank, water pumped out through a 1-meter spout

1. Tank Description: Full tank implies water extends from $y = -3$ (bottom of hemisphere) to $y = 0$ (top of hemisphere).
2. Distance Water is Pumped: A slice at height $y$ is pumped to $1 - y$ meters.
3. Radius of Slice ($r$): From the equation of a hemisphere, $r = \sqrt{9 - y^2}$.
4. Volume of Slice: $dV = \pi (\sqrt{9 - y^2})^2 \, dy = \pi (9 - y^2) \, dy$.
5. Mass of Slice: $\rho \, dV = 1,000 \pi (9 - y^2) \, dy$.
6. Force on Slice: $F = g \, \rho \, dV = 9,800 \pi (9 - y^2) \, dy$.
7. Work for Slice: $W = F \cdot d = 9,800 \pi (9 - y^2)(1 - y) \, dy$.

Integral for Work: $W_a = \int_{-3}^{0} 9,800 \pi (9 - y^2)(1 - y) \, dy$

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Part (b): Half-full tank, water pumped out through a 1-meter spout

1. Tank Description: Water extends from $y = -3$ to $y = -1.5$.
2. Work for Slice: Same as in part (a), with the limits of integration adjusted.

Integral for Work: $W_b = \int_{-3}^{-1.5} 9,800 \pi (9 - y^2)(1 - y) \, dy$

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Part (c): Full tank, water pumped out over the top

1. Tank Description: Full tank, water pumped to $0 - y$ meters.
2. Distance Water is Pumped: $d = -y$.
3. Work for Slice: Similar to part (a), but with $d = -y$.

Integral for Work: $W_c = \int_{-3}^{0} 9,800 \pi (9 - y^2)(-y) \, dy$

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Part (d): Full tank, pump out half the water over the top

1. Tank Description: Full tank, but only half the water is pumped out. Use the volume of a hemisphere to determine the corresponding height.
2. Volume of Hemisphere: $V = \frac{2}{3} \pi r^3 = 18 \pi$.
3. Half Volume: $9 \pi$. Solve for height $y_1$ corresponding to $9 \pi = \int_{-3}^{y_1} \pi (9 - y^2) \, dy$.

Once $y_1$ is determined, the limits of integration are $y = -3$ to $y = y_1$, and the distance is $d = -y$.

Integral for Work: $W_d = \int_{-3}^{y_1} 9,800 \pi (9 - y^2)(-y) \, dy$

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Key Steps for Solving

1. Simplify the integrals: Multiply out the terms and simplify expressions.
2. Evaluate definite integrals: Use calculus techniques or numerical integration if necessary.

Would you like detailed computation for one or more parts? Here are some additional questions to deepen understanding:

1. How do you derive the radius of the slice in the hemisphere?
2. What are the physical implications of the spout's height on work done?
3. How does the density of water influence the result?
4. Could we extend the problem to account for pumping rate or efficiency?
5. What would change if the tank were a full sphere instead of a hemisphere?

Tip: Always sketch the tank and identify the slice geometry before setting up your integral!