Calculating Work Done on Emptying a Half-Full Cylindrical Reservoir

In summary, the conversation discusses the process of finding the work done in emptying water over the top of a half-full cylindrical reservoir with a diameter of 4 feet and height of 6 feet. The work is calculated by taking the integral of 4πw(6-y) from 0 to 3, where w is the weight of the water per cubic foot. The conversation also discusses the potential use of SI units and breaks down the process into two steps for easier understanding.
  • #1
demonelite123
219
0
A cylindrical reservoir of diameter 4 feet and height 6 feet is half-full of water weighing w pounds per cubic foot. Find the work done in emptying the water over the top.

i put the cylinder on the x-y axes, so y represents height and x represents radius of the cylinder. so the volume of a slice of water ΔV = πx^2 (Δy), and since x = 2, ΔV = 4π(Δy).

a slice of the water at height y is lifted (6 - y) feet so therefore the work,
W = w*4π(Δy)*(6 - y)

so the integral looks like 4πw ∫ (6-y) dy.

the only problem i have is how to decide the limits of integration. since you are trying to empty the water over the top, the water at the very bottom should travel to the top right? so the limits should be from 0 to 6 i think. but my book says the limits are from 0 to 3 because the reservoir is only half full. can somebody please explain this to me?
 
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  • #2
What is the height of the top layer of water and what is height of the bottom layer?

Also why is everything in feet. Most of the world now uses SI units?
 
  • #3
Perhaps it will be easier to see what is going on if you did it in two steps. I know that water is incompressible, but suppose you were able to scrunch all the water molecules and bring them to the initial level of 3.0 feet. Step 1: How much work against gravity is needed to do that? Step 2: Now that you have all the water at the same height, how much extra work is needed to push it the additional 3 feet over the top? Add the two for the answer.
 

1. What is a work problem with integrals?

A work problem with integrals is a type of mathematical problem that involves using integrals to find the amount of work done in a specific situation. This is often used in physics and engineering to calculate the work done by a force over a distance.

2. How do you solve a work problem with integrals?

To solve a work problem with integrals, you first need to set up an integral that represents the work done. This involves identifying the force, distance, and direction of the force. Then, you can use integration techniques to find the definite integral and solve for the work done.

3. What are the units of work in a work problem with integrals?

The units of work in a work problem with integrals depend on the units used for force and distance. In the SI system, the unit of work is joules (J), which is equal to 1 newton-meter (N·m). However, other units such as foot-pounds (ft·lb) may also be used.

4. What are some real-world applications of work problems with integrals?

Work problems with integrals have many real-world applications, particularly in physics and engineering. They can be used to calculate the work done by forces in situations such as lifting objects, pushing a cart, or moving an object up a ramp. They are also used in the design of machines and structures, such as cranes and bridges.

5. Can work problems with integrals be solved without using calculus?

No, work problems with integrals require the use of calculus because they involve finding the area under a curve. This is done through integration, which is a fundamental concept in calculus. While there may be alternative methods to solving specific work problems, calculus is the most efficient and accurate way to solve these types of problems.

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