Use cylindrical shells to find the volume of the solid generated by rotating the curve y=4ln(x) about the x-axis for 1≤x≤e

To find the volume of the solid generated by rotating the curve \( y = 4\ln(x) \) about the x-axis for \( 1 \leq x \leq e \), we use the method of **cylindrical shells**.

### Step-by-step solution:

1. **Formula for the volume using cylindrical shells:**
   The formula for the volume using cylindrical shells is:

   \[
   V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx
   \]
   
   where \( x \) is the radius of the shell, \( f(x) \) is the height of the shell, and the limits \( a \) and \( b \) are the bounds of the interval.

2. **Determine the function and limits:**
   Here, the function \( f(x) = 4\ln(x) \), and we are rotating around the x-axis over the interval \( 1 \leq x \leq e \).

3. **Set up the integral:**
   The radius of the cylindrical shell is \( x \), and the height of the shell is \( 4\ln(x) \). Therefore, the volume is given by:

   \[
   V = 2\pi \int_{1}^{e} x \cdot 4\ln(x) \, dx
   \]

4. **Simplify the integral:**

   \[
   V = 8\pi \int_{1}^{e} x \ln(x) \, dx
   \]

5. **Solve the integral:**
   To solve \( \int x \ln(x) \, dx \), we use integration by parts, where:
   
   - Let \( u = \ln(x) \), so \( du = \frac{1}{x} \, dx \),
   - Let \( dv = x \, dx \), so \( v = \frac{x^2}{2} \).

   Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get:

   \[
   \int x \ln(x) \, dx = \frac{x^2}{2} \ln(x) - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx
   \]
   \[
   = \frac{x^2}{2} \ln(x) - \frac{1}{2} \int x \, dx
   \]
   \[
   = \frac{x^2}{2} \ln(x) - \frac{x^2}{4}
   \]

6. **Evaluate the definite integral:**
   Now, evaluate this expression from \( 1 \) to \( e \):

   \[
   \int_{1}^{e} x \ln(x) \, dx = \left[ \frac{x^2}{2} \ln(x) - \frac{x^2}{4} \right]_{1}^{e}
   \]

   At \( x = e \):

   \[
   \frac{e^2}{2} \ln(e) - \frac{e^2}{4} = \frac{e^2}{2} \cdot 1 - \frac{e^2}{4} = \frac{e^2}{4}
   \]

   At \( x = 1 \):

   \[
   \frac{1^2}{2} \ln(1) - \frac{1^2}{4} = 0 - \frac{1}{4} = -\frac{1}{4}
   \]

   Therefore, the definite integral is:

   \[
   \frac{e^2}{4} - \left(-\frac{1}{4}\right) = \frac{e^2}{4} + \frac{1}{4}
   \]
   \[
   = \frac{e^2 + 1}{4}
   \]

7. **Final volume:**
   Multiply by \( 8\pi \):

   \[
   V = 8\pi \cdot \frac{e^2 + 1}{4} = 2\pi (e^2 + 1)
   \]

Thus, the volume of the solid is:

\[
V = 2\pi (e^2 + 1)
\]

---

Would you like more details on any specific part or have any questions?

Here are 5 related questions to explore:

1. How does the method of cylindrical shells compare with the disk/washer method for volume?
2. Can we apply cylindrical shells if we rotate around the y-axis instead?
3. What happens to the volume if we rotate \( y = 4\ln(x) \) around a line like \( y = 2 \)?
4. What is the geometrical interpretation of cylindrical shells in volume calculation?
5. Can cylindrical shells be used for non-logarithmic functions? What changes?

**Tip:** Integration by parts is a crucial technique, especially for solving integrals involving logarithmic and polynomial functions.

Use cylindrical shells to find the volume of the solid generated by rotating the curve y=4ln(x) about the x-axis for 1≤x≤e.

Calculate the volume of the solid generated by rotating the curve y=4ln(x) about the x-axis between x=1 and x=e using the method of cylindrical shells. This solution involves advanced calculus concepts, including integration by parts, suitable for undergraduate students.

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