If I have a 110 Gallon Inductor Full Drain Conical Tank that is 30 inches in diameter and 56 inches in height with a 55 degree angled cone bottom, if the tank is full of water, how do I use calculus to determine how much work is required to pump all of the water over the top of the tank?

To find the work required to pump all the water over the top of the tank, we will integrate the work done to move each infinitesimal volume of water up to the top. The formula for work is:

W = \int_a^b F(y) \, dy

Where:
- \( F(y) \) is the force required to move the water at height \( y \)
- \( a \) and \( b \) are the limits of integration from the bottom to the top of the tank

1. **Convert dimensions to consistent units (feet):**
- Diameter = 30 inches = 2.5 feet
- Radius \( r = \frac{2.5}{2} = 1.25 \) feet
- Height \( h = 56 \) inches = 4.67 feet

2. **Volume of small element at height \( y \):**

dV = \pi (x^2) \, dy

3. **Scale height with radius:**

x = \frac{1.25}{4.67} y = \frac{y}{3.736}

4. **Force (weight) of small element of water:**
- Volume \( V = \pi \left( \frac{y}{3.736} \right)^2 \, dy \)
- \( dF = \rho g \, dV = \rho g \pi \left( \frac{y}{3.736} \right)^2 \, dy \)
- \( \rho = 62.4 \text{ lbs/ft}^3 \) (density of water)

5. **Height to lift each element \( (56 - y) \):**

dW = \rho g \pi \left( \frac{y}{3.736} \right)^2 (56 - y) \, dy

6. **Integrate from 0 to 4.67 ft (height of tank):**

W = \int_0^{4.67} 62.4 \pi \left( \frac{y}{3.736} \right)^2 (4.67 - y) \, dy

Simplified:

W = 62.4 \pi \int_0^{4.67} \left( \frac{1}{13.96} \right)^2 y^2 (4.67 - y) \, dy

W = \frac{62.4 \pi}{194.8816} \int_0^{4.67} y^2 (4.67 - y) \, dy

7. **Actual integration:**

W = \frac{62.4 \pi}{194.8816} \int_0^{4.67} (4.67 y^2 - y^3) \, dy

W = \frac{62.4 \pi}{194.8816} \left[ 4.67 \frac{y^3}{3} - \frac{y^4}{4} \right]_0^{4.67}

8. **Evaluate definite integral:**

W = \frac{62.4 \pi}{194.8816} \left( 4.67 \frac{4.67^3}{3} - \frac{4.67^4}{4} \right)

After calculation:

W = 13868.91 \text{ ft-lb}

Therefore, the work required to pump all of the water over the top of the tank is:

W = 13868.91 \text{ ft-lb}