Let R
be the region in the xy
-plane bounded by the curves y=ln(x), y=1
and x=1
. If S is the solid obtained by rotating R
about the y
-axis, then find the volume of S.

The problem requires finding the volume of the solid \( S \) obtained by rotating the region \( R \) bounded by the curves \( y = \ln(x) \), \( y = 1 \), and \( x = 1 \) around the \( y \)-axis.

To solve this, we will use the **method of cylindrical shells**. Here's the step-by-step solution:

### Step 1: Set up the cylindrical shell formula
The formula for the volume of a solid generated by rotating a region around the \( y \)-axis using cylindrical shells is:

\[
V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx
\]

where \( f(x) \) is the height of the shell, and \( x \) is the radius of the shell.

### Step 2: Define the boundaries and height of the region
The region is bounded by:
- \( y = \ln(x) \) (the lower boundary),
- \( y = 1 \) (the upper boundary),
- \( x = 1 \) (the right boundary).

To find the limits of integration, note that the region is bounded by \( x = e^0 = 1 \) on the right and \( x = e^1 = e \) on the left (since \( y = 1 \) corresponds to \( x = e^1 = e \)).

### Step 3: Apply the cylindrical shell method
The height of each cylindrical shell is \( f(x) = 1 - \ln(x) \) (the difference between the upper boundary and the curve \( y = \ln(x) \)).

Thus, the volume is given by:

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

### Step 4: Solve the integral
Now, break this integral into two parts:

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

#### First integral:
\[
\int_{1}^{e} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{e} = \frac{e^2}{2} - \frac{1^2}{2} = \frac{e^2 - 1}{2}
\]

#### Second integral:
To solve \( \int_{1}^{e} x \ln(x) \, dx \), use integration by parts:
- Let \( u = \ln(x) \) and \( dv = x \, dx \).
- Then \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^2}{2} \).

Applying integration by parts:
\[
\int x \ln(x) \, dx = \frac{x^2}{2} \ln(x) - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \ln(x) - \int \frac{x}{2} \, dx
\]

Now, compute both terms:
\[
\frac{x^2}{2} \ln(x) \bigg|_1^e = \frac{e^2}{2} \ln(e) - \frac{1^2}{2} \ln(1) = \frac{e^2}{2}
\]

\[
\int \frac{x}{2} \, dx = \frac{x^2}{4} \bigg|_1^e = \frac{e^2}{4} - \frac{1^2}{4} = \frac{e^2 - 1}{4}
\]

Thus, the second integral is:

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

### Step 5: Final expression for the volume
Substitute the results into the original expression:

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

Simplifying:

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

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

\[
V = 2\pi \cdot \frac{e^2 - 1}{2}
\]

Thus, the volume is:

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

### Final Answer:
The volume of the solid is \( \pi(e^2 - 1) \).

Would you like more details or further explanation?

---

Here are 5 related questions to deepen your understanding:
1. What is the method of washers, and how is it different from the method of cylindrical shells?
2. How would the volume change if the region were rotated around the x-axis instead of the y-axis?
3. Can you compute the volume using the disk method for the same problem?
4. How do you apply integration by parts, and why was it necessary here?
5. What if the boundaries were different, say \( y = \ln(x) \) and \( y = 2 \)?

**Tip:** When rotating a region around the y-axis, the cylindrical shells method often simplifies calculations, especially when the function is in terms of \( x \).

Let R be the region in the xy-plane bounded by the curves y=ln(x), y=1, and x=1. If S is the solid obtained by rotating R about the y-axis, then find the volume of S.

Find the volume of a solid obtained by rotating the region bounded by y=ln(x), y=1, and x=1 around the y-axis using the cylindrical shells method. This problem covers key calculus concepts like solids of revolution, integration by parts, and the cylindrical shells formula. Suitable for advanced high school or college-level calculus students.

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