To calculate the work required to pump all of the oil to the surface, we will use the formula for work in the context of fluid mechanics:

$W = \int_{a}^{b} F(y) \, dy$

Where:

• $W$ is the total work required to pump the oil.
• $F(y)$ is the force exerted at a given height $y$.
• $a$ and $b$ are the limits of integration, in this case, the range of oil depth.

Step-by-Step Calculation

1. Volume of the tank at a given height:
   The tank is a cylinder, and the volume of a thin slice of oil at height $y$ (where $y$ is the distance from the bottom of the tank) is given by the area of the cross-section times the thickness of the slice. The cross-sectional area is constant and equals the area of a circle:

   $A = \pi r^2 = \pi \times (8)^2 = 64\pi \text{ square feet}.$

2. Force due to the weight of the oil:
   The weight of the oil in a thin slice at height $y$ (with thickness $dy$) is:

   $dF = \text{(weight density)} \times \text{(volume of the slice)} = 50 \times 64\pi \, dy = 3200\pi \, dy \text{ pounds}.$

3. Distance to pump the oil:
   The distance that a slice of oil at height $y$ must be pumped to the surface is $10 + 24 - y = 34 - y$ feet, because the top of the tank is 10 feet below ground level, and the height of the oil column is $y$.

4. Work for a slice:
   The work to pump this slice to the surface is the force $dF$ times the distance to the surface:

   $dW = (3200\pi \, dy) \times (34 - y).$

5. Total work:
   Now, to find the total work, we integrate from $y = 0$ to $y = 23$ (since the oil level is 23 feet deep):

   $W = \int_0^{23} 3200\pi (34 - y) \, dy.$

Step-by-Step Integration

Let’s carry out the integration:

$W = 3200\pi \int_0^{23} (34 - y) \, dy.$

First, we expand the integrand:

$W = 3200\pi \left[ \int_0^{23} 34 \, dy - \int_0^{23} y \, dy \right].$

Now we solve each part:

• $\int_0^{23} 34 \, dy = 34y \Big|_0^{23} = 34 \times 23 = 782$,
• $\int_0^{23} y \, dy = \frac{y^2}{2} \Big|_0^{23} = \frac{23^2}{2} = \frac{529}{2} = 264.5$.

So, the total work is:

$W = 3200\pi \left( 782 - 264.5 \right) = 3200\pi \times 517.5.$

Finally, we calculate:

$W \approx 3200 \times 3.1416 \times 517.5 \approx 5,225,964.2 \text{ foot-pounds}.$

Final Answer:

$W \approx 5,225,964 \text{ foot-pounds}.$

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Would you like more details or have further questions on this? Here are some related questions to explore:

1. How would the work change if the oil level were deeper?
2. What is the role of the shape of the tank in determining the work required?
3. How does the weight density of the oil affect the result?
4. How can you calculate the total volume of oil in the tank?
5. What other factors (like friction or temperature) could affect the work calculation?

Tip: When dealing with problems involving liquids or fluids, always remember that the work formula takes into account both the force (due to weight) and the distance (height to pump).