A cylindrical tank of radius 3 ft length 8 ft is laid out horizontally .The tank is half full of oil that weighs 60 pounds per cubic foot.

a.)determine the work done in pumping out oil to the top of the tank.
b.)determine the work required to pump out the oil to leave 4 feet above the top of the tank.

Draw a cross-section. If the oil is at depth y, then the width of the thin rectangle of oil is 2x, where

(3-y)^2 + x^2 = 9
x = √(6y-y^2)

So, the weight of oil for that thin sheet is

2x*8*60 dy

and the total work to lift all the oil out of the top of the tank is thus

∫[0,3] 2*8*60(6-y)√(6y-y^2) dy = 2160(4+3π) = 28998

Now our engineer friends will just want to find out how far the center of mass is lifted, times the weight of the oil.

The weight is 1/2 * 9π * 8 * 60 = 2160π lbs

The centroid of a semicircle of radius r is at a distance 4r/(3π) from the center of the circle. That means we have to lift the centroid a distance of r + 4r/3π = r(4+3π)/3π

Multiply that by the weight of oil, and you have work of

3(4+3π)/3π * 2160π = 2160(4+3π)

as above.

To determine the work done in pumping out oil to the top of the tank, we need to find the volume of the oil and then multiply it by the weight of the oil.

a) First, let's find the volume of the oil in the tank.

The formula to calculate the volume of a cylinder is given by V = πr²h, where V is the volume, r is the radius, and h is the height.

Given:
Radius (r) = 3 ft
Length (h) = 8 ft

The tank is half full, so the height of the oil is (1/2) * 8 = 4 ft.

Now we can calculate the volume of the oil:
V = π * (3 ft)² * 4 ft
V = π * 9 ft² * 4 ft
V = 36π ft³

Next, we'll find the weight of the oil:
Weight of oil = Volume of oil * Density
Weight of oil = 36π ft³ * 60 lb/ft³

To simplify calculations, we'll approximate π to 3.14:
Weight of oil = 36 * 3.14 * 60 lb
Weight of oil = 6782.4 lb

So, the work done in pumping out oil to the top of the tank would be 6782.4 lb-ft (pound-feet).

b) To find the work required to pump out the oil to leave 4 feet above the top of the tank, we need to calculate the volume of the remaining oil.

Given:
Radius (r) = 3 ft
Length (h) = 8 ft

The height of the remaining oil is 4 ft (leaving 4 ft above the top of the tank).

Now we can calculate the volume of the remaining oil:
V_remaining = π * (3 ft)² * 4 ft
V_remaining = π * 9 ft² * 4 ft
V_remaining = 36π ft³

Again, let's approximate π to 3.14:
V_remaining = 36 * 3.14 ft³

Now we can calculate the weight of the remaining oil:
Weight_remaining = V_remaining * Density
Weight_remaining = 36π ft³ * 60 lb/ft³

Weight_remaining = 36 * 3.14 * 60 lb

So, the work required to pump out the oil to leave 4 feet above the top of the tank would be 6782.4 lb-ft (pound-feet). This is the same as the previous case because the weights are the same in both cases.

To find the work done in pumping out the oil, we need to first determine the volume of oil in the tank.

a.) To calculate the volume of the cylindrical tank, we use the formula:

V = π * r^2 * h

Where:
V = Volume of the tank
π = Pi (approximately 3.14159)
r = Radius of the tank (3 ft)
h = Length of the tank (8 ft)

Plugging in the values, we can calculate the volume of the tank:

V = π * (3 ft)^2 * 8 ft
V = π * 9 ft^2 * 8 ft
V = 72π ft^3

Since the tank is half-full, the volume of oil in the tank will be half of the volume of the tank:

Volume of oil = 1/2 * 72π ft^3
Volume of oil = 36π ft^3

Next, we need to find the weight of the oil:

Weight of the oil = Volume of oil * Density of oil

Given that the oil weighs 60 pounds per cubic foot, we can calculate the weight:

Weight of the oil = 36π ft^3 * 60 lbs/ft^3
Weight of the oil = 2160π lbs

Now, we can determine the work done in pumping out the oil to the top of the tank. Work is calculated using the formula:

Work = Force * Distance

In this case, the force is the weight of the oil and the distance is the height of the tank, which is 8 ft.

Work = Weight of the oil * Height of the tank
Work = 2160π lbs * 8 ft
Work = 17280π ft-lbs (approximately)

Therefore, the work done in pumping out the oil to the top of the tank is approximately 17280π ft-lbs.

b.) To determine the work required to pump out the oil to leave 4 feet above the top of the tank, we need to find the new height of the oil level.

The new height of the oil level would be the original height of the tank (8 ft) minus the additional 4 feet:

New height of oil = 8 ft - 4 ft
New height of oil = 4 ft

Using the same formula as before, we can calculate the volume and weight of the oil at this new height:

Volume of oil = (π * r^2 * new height) / 2
Volume of oil = (π * (3 ft)^2 * 4 ft) / 2
Volume of oil = 18π ft^3

Weight of the oil = Volume of oil * Density of oil
Weight of the oil = 18π ft^3 * 60 lbs/ft^3
Weight of the oil = 1080π lbs

Finally, we can calculate the work required in pumping out the oil to leave 4 feet above the top of the tank:

Work = Weight of the oil * Height of the tank
Work = 1080π lbs * 8 ft
Work = 8640π ft-lbs (approximately)

Therefore, the work required to pump out the oil and leave 4 feet above the top of the tank is approximately 8640π ft-lbs.