A fuel oil tank is an upright cylinder, buried so that its circular top is 8 feet beneath ground level. The tank has a radius of 7 feet and is 21 feet high, although the current oil level is only 17 feet deep. Calculate the work required to pump all of the oil to the surface. Oil weighs 50lb/ft^3.

To calculate the work required to pump all the oil to the surface from a cylindrical tank, we will follow these steps:

1. **Set up the basic parameters**:
   - Radius of the tank: \( r = 7 \, \text{ft} \)
   - Height of the tank: \( h = 21 \, \text{ft} \), but the current oil level is only \( 17 \, \text{ft} \).
   - Oil density: \( \rho = 50 \, \text{lb/ft}^3 \)
   - The distance to pump the oil to the surface is dependent on the depth at which the oil is located. Since the top of the tank is 8 feet underground, the oil at height \( y \) in the tank must be pumped a distance of \( 8 + (21 - y) = 29 - y \) feet to the surface.

2. **Use the concept of work**:
   Work \( W \) is calculated as the force multiplied by the distance. In this context, we are integrating the weight of a differential volume of oil multiplied by the distance it has to be pumped.

   The volume of a thin slice of oil at height \( y \) with thickness \( dy \) is:
   \[
   dV = \pi r^2 \, dy = \pi (7^2) \, dy = 49\pi \, dy
   \]
   The weight of that slice of oil is:
   \[
   dF = \rho dV = 50 \times 49\pi \, dy = 2450\pi \, dy \, \text{lb}
   \]
   The distance that this slice needs to be pumped is \( 29 - y \) feet.

3. **Set up the integral**:
   To find the total work required to pump all the oil to the surface, we integrate the weight of each slice multiplied by the distance it must be pumped over the entire height of the oil in the tank (from \( y = 0 \) to \( y = 17 \)):

   \[
   W = \int_0^{17} 2450\pi (29 - y) \, dy
   \]

4. **Solve the integral**:

   First, break up the integral:
   \[
   W = 2450\pi \int_0^{17} (29 - y) \, dy
   \]
   This can be split into two simpler integrals:
   \[
   W = 2450\pi \left( \int_0^{17} 29 \, dy - \int_0^{17} y \, dy \right)
   \]

   Now solve each part:
   \[
   \int_0^{17} 29 \, dy = 29y \Big|_0^{17} = 29 \times 17 = 493
   \]
   \[
   \int_0^{17} y \, dy = \frac{y^2}{2} \Big|_0^{17} = \frac{17^2}{2} = \frac{289}{2} = 144.5
   \]

   So the total work is:
   \[
   W = 2450\pi (493 - 144.5) = 2450\pi \times 348.5
   \]

   Finally, calculate the value:
   \[
   W = 2450 \times 348.5 \times \pi \approx 2450 \times 348.5 \times 3.1416 \approx 2,683,908.17 \, \text{ft-lb}
   \]

Thus, the total work required to pump all the oil to the surface is approximately \( 2,683,908.17 \, \text{ft-lb} \).

Would you like a more detailed breakdown, or do you have any other questions?

### 5 Related Questions:
1. How does the depth of the oil affect the amount of work required?
2. What if the oil was 21 feet deep, how would the work change?
3. Can this method be applied to tanks of different shapes, such as rectangular tanks?
4. How does the density of the oil affect the work required?
5. What is the relationship between the height of the tank and the total work done?

### Tip:
Remember that work is directly related to both the force applied and the distance over which it is applied—changing either variable can have a significant impact on the result.

Explore how to calculate the work required to pump oil from a cylindrical tank 8 feet below ground level. This advanced calculus problem uses integration to determine the total work needed, considering the tank's dimensions, oil weight, and pumping distance. Ideal for Grades 11-12.

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