A a conical container of radius 5ft and height 20ft is filled to a height of 18ft of a liquid weighing 50.6lb/ft^3. How much work will it take to pump the contents to the rim? How much work will it take to pump the liquid of a level of 2ft above the rim?
So I got integral (50.6pi/1)y^2(18-y)from 0 to 18 to get 125546, but that wasn't the right answer. Please help! What's the number before y^2 supposed to be?
BTW the right answer was 125452
To determine the correct expression for the integral, let's break down the problem step by step.
First, we need to find the volume of the conical container that is filled with liquid to a height of 18ft. The formula for the volume of a cone is given by V = (1/3)πr^2h.
In this case, the radius (r) is 5ft, and the height (h) is 18ft. Substituting these values into the formula, we get:
V = (1/3)π(5^2)(18)
V = (1/3)π(25)(18)
V = (1/3)π(450)
V = 150π ft^3
Next, we need to determine the weight of the liquid that has been filled in the container. The weight of an object is given by multiplying its volume by its density. In this case, the density is 50.6lb/ft^3. The weight (W) can be calculated as:
W = V × density
W = 150π × 50.6
W = 7590π lb
Now, to find the work required to pump the contents to the rim, we can use the concept of work done against gravity. The work done (Work) is given by the formula:
Work = (Weight) × (Height)
In this case, the weight is 7590π lb, and the height is 18ft. Substituting these values into the formula, we get:
Work = 7590π × 18
Work = 136620π ft-lb
So, the work required to pump the contents to the rim is 136620π ft-lb.
For the second part of the question, we need to calculate the work required to pump the liquid to a level of 2ft above the rim. In this case, the height is now 18ft + 2ft = 20ft.
Following the same steps as above, we can substitute the new height into the formula:
Work = (Weight) × (Height)
Work = 7590π × 20
Work = 151800π ft-lb
So, the work required to pump the liquid 2ft above the rim is 151800π ft-lb.
Therefore, the correct answer for the work required to pump the contents to the rim is 136620π ft-lb, and the work required to pump the liquid 2ft above the rim is 151800π ft-lb.
To calculate the work required to pump the liquid to the rim, you need to find the integral expression for the force at each height level and then integrate it.
The force at each height level is given by:
Force = Area of the circular cross-section * Height * Density of the liquid
The area of the circular cross-section can be calculated by using the formula for the area of a circle:
Area = π * radius^2
In this case, the radius is 5ft, so the area would be:
Area = π * (5ft)^2 = 25π ft^2
Therefore, the integral expression for the work required to pump the liquid to the rim is:
Work = ∫ (Force) * (dy) from 0 to 18
Substituting the values, we get:
Work = ∫ (25π * (18 - y) * 50.6) * (dy) from 0 to 18
Simplifying the expression gives:
Work = 25 * 50.6 * π ∫ (18 - y) * (dy) from 0 to 18
Work = 25 * 50.6 * π * [((18 - y)^2)/2] from 0 to 18
Work = 25 * 50.6 * π * [(18^2)/2 - (18^2)/2 - (0^2)/2]
Work = 25 * 50.6 * π * [(18^2)/2]
Work = 25 * 50.6 * π * (324/2)
Work = 25 * 50.6 * π * 162
Work ≈ 205,782 ft-lb
Therefore, the work required to pump the liquid to the rim is approximately 205,782 ft-lb.
To calculate the work required to pump the liquid to a level of 2ft above the rim, you would integrate from 18 to 20 instead of 0 to 18 in the integral expression.