Math Problem Statement

Create a bucket by rotating around the y axis the curve y=4ln(x-5) from y = 0 to y = 4. If this bucket contains a liquid with density 780 kg/m3 filled to a height of 2 meters, find the work required to pump the liquid out of this bucket (over the top edge). Use 9.8 m/s2 for gravity.

Work = Joules

Solution

We are tasked with finding the work required to pump liquid out of a bucket, which is created by rotating the curve y=4ln(x5)y = 4 \ln(x - 5) around the y-axis from y=0y = 0 to y=4y = 4. The bucket is filled with a liquid of density ρ=780kg/m3\rho = 780 \, \text{kg/m}^3, filled to a height of 2 meters, and we will use gravity g=9.8m/s2g = 9.8 \, \text{m/s}^2.

Step 1: Parameterize the shape of the bucket

The given curve is y=4ln(x5)y = 4 \ln(x - 5). We need to find the radius of the bucket at any height yy.

Solve y=4ln(x5)y = 4 \ln(x - 5) for xx: x5=ey/4x - 5 = e^{y/4} x=ey/4+5x = e^{y/4} + 5 Since we are rotating the curve around the y-axis, the radius r(y)r(y) at a height yy is given by: r(y)=x5=ey/4r(y) = x - 5 = e^{y/4}

Step 2: Volume of a thin slice of liquid

Consider a thin slice of the liquid at height yy with thickness dydy. The volume of this slice, which has the shape of a thin disk, is: dV=π[r(y)]2dy=π(ey/4)2dy=πey/2dydV = \pi [r(y)]^2 \, dy = \pi (e^{y/4})^2 \, dy = \pi e^{y/2} \, dy

Step 3: Mass of a thin slice of liquid

The mass of the liquid in this thin slice is the density times the volume: dm=ρdV=780πey/2dydm = \rho \, dV = 780 \, \pi e^{y/2} \, dy

Step 4: Work to pump out a thin slice

The work required to pump this slice to the top of the bucket (which is at y=4y = 4) is the force needed to lift it (mass times gravity) times the distance it needs to be lifted (which is 4y4 - y):

dW=dmg(4y)=7809.8πey/2(4y)dydW = dm \cdot g \cdot (4 - y) = 780 \cdot 9.8 \cdot \pi e^{y/2} \cdot (4 - y) \, dy dW=7644πey/2(4y)dydW = 7644 \pi e^{y/2} \cdot (4 - y) \, dy

Step 5: Total work

To find the total work, integrate dWdW from y=0y = 0 to y=2y = 2 (since the liquid is filled to a height of 2 meters): W=027644πey/2(4y)dyW = \int_0^2 7644 \pi e^{y/2} \cdot (4 - y) \, dy

Let's compute this integral.

The total work required to pump the liquid out of the bucket is approximately 234,050Joules234,050 \, \text{Joules}.

Would you like more details on any part of this process or have any questions?

Here are five related questions to explore:

  1. How would the work change if the bucket was filled to 3 meters instead of 2 meters?
  2. What happens to the required work if the density of the liquid changes?
  3. How does the shape of the bucket affect the total work required?
  4. Can you find the volume of the bucket formed by rotating the curve up to y=4y = 4?
  5. How would the work calculation differ if the gravitational acceleration changed?

Tip: The work done in pumping liquids depends heavily on both the height to which the liquid is raised and the shape of the container.

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Math Problem Analysis

Mathematical Concepts

Calculus
Solid of Revolution
Work-Energy Principle
Integration

Formulas

Work = ∫ dm * g * (h - y)
Volume of revolution: V = π ∫[a,b] (f(y))^2 dy
dm = ρ * dV
dV = π (r(y))^2 dy

Theorems

Work-Energy Theorem
Disk Method for Solids of Revolution

Suitable Grade Level

Undergraduate (Calculus Level)