To calculate the work needed to pump all the water out of a semi-spherical tank, we must integrate the work done in pumping infinitesimally small amounts of water from different heights to the top of the tank.

Steps:

1. Define Variables:

   • Radius of the tank $R = 7 \, \text{m}$
   • Density of water $\rho = 1000 \, \text{kg/m}^3$
   • Gravitational acceleration $g = 9.8 \, \text{m/s}^2$

2. Volume of a Thin Horizontal Slice (Disk): Consider a horizontal slice of water at a height $y$ from the bottom of the tank (where $y = 0$ is the bottom of the tank and $y = 7$ is the top). The radius of the slice at height $y$ is given by: $r(y) = \sqrt{R^2 - y^2}$ The volume of this thin slice of thickness $dy$ is: $dV = \pi r(y)^2 \, dy = \pi (R^2 - y^2) \, dy$

3. Mass of the Slice: The mass of the water in this thin slice is: $dm = \rho \, dV = \rho \, \pi (R^2 - y^2) \, dy$

4. Work to Pump the Slice: To pump this slice of water to the top of the tank (at height $7 \, \text{m}$), the height it needs to be lifted is $(7 - y)$. The work required to pump this slice is: $dW = g \, dm \, (7 - y) = g \, \rho \, \pi (R^2 - y^2) (7 - y) \, dy$

5. Total Work: To find the total work, integrate from $y = 0$ to $y = 7$: $W = \int_0^7 g \, \rho \, \pi (R^2 - y^2) (7 - y) \, dy$ Substituting $g = 9.8$, $\rho = 1000$, and $R = 7$: $W = \int_0^7 9.8 \times 1000 \times \pi (49 - y^2)(7 - y) \, dy$ Simplifying constants: $W = 9800 \pi \int_0^7 (49 - y^2)(7 - y) \, dy$

6. Expansion of the Integrand: Expand $(49 - y^2)(7 - y)$: $(49 - y^2)(7 - y) = 343 - 49y - 7y^2 + y^3$ Thus, the integral becomes: $W = 9800 \pi \int_0^7 (343 - 49y - 7y^2 + y^3) \, dy$

7. Integration: Now, integrate each term: $\int_0^7 343 \, dy = 343 \times 7 = 2401$ $\int_0^7 49y \, dy = \frac{49 \times 7^2}{2} = \frac{49 \times 49}{2} = 1200.5$ $\int_0^7 7y^2 \, dy = \frac{7 \times 7^3}{3} = \frac{7 \times 343}{3} \approx 800.33$ $\int_0^7 y^3 \, dy = \frac{7^4}{4} = \frac{2401}{4} = 600.25$

8. Combine the Results: $W = 9800 \pi \left( 2401 - 1200.5 - 800.33 + 600.25 \right)$ $W = 9800 \pi (1000.42)$ $W \approx 9800 \times 3.1416 \times 1000.42 \approx 30782190.15 \, \text{Joules}$

Final Answer:

The work needed to pump all the water out is approximately: $W \approx 30,782,190.15 \, \text{J} \quad (\text{rounded to two decimal places})$

Would you like further details or clarifications?

Here are 5 related questions to expand your understanding:

1. How does changing the radius of the tank affect the work required to pump the water?
2. What would the work be if the tank were fully spherical instead of semi-spherical?
3. How does changing the height the water is pumped to (e.g., higher or lower than the top) affect the result?
4. How does the density of the fluid (e.g., oil or mercury) impact the total work required?
5. What is the relationship between the shape of the tank and the work required to pump the water?

Tip: When solving integration problems involving physics, always focus on setting up the correct infinitesimal element (e.g., slice or shell) and express all quantities in terms of known variables.