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


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

### Step 1: Parameterize the shape of the bucket

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

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

### Step 2: Volume of a thin slice of liquid

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

### Step 3: Mass of a thin slice of liquid

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

### 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 \( y = 4 \)) is the force needed to lift it (mass times gravity) times the distance it needs to be lifted (which is \( 4 - y \)):

\[
dW = dm \cdot g \cdot (4 - y) = 780 \cdot 9.8 \cdot \pi e^{y/2} \cdot (4 - y) \, dy
\]
\[
dW = 7644 \pi e^{y/2} \cdot (4 - y) \, dy
\]

### Step 5: Total work

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

Let's compute this integral.

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

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 \( y = 4 \)?
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.

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/m^3 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/s^2 for gravity.

Learn how to calculate the work required to pump liquid out of a bucket formed by rotating the curve y = 4ln(x - 5) around the y-axis. This problem uses concepts from calculus, such as solids of revolution, integration, and the work-energy principle. The liquid has a density of 780 kg/m^3 and is pumped to a height of 2 meters. Suitable for undergraduate calculus students.

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