Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the​ x-axis.
y equals e Superscript StartFraction x Over 2 EndFraction
​, y equals e Superscript negative StartFraction x Over 2 EndFraction
​, x equals ln 6​, x equals ln 12
y equals e Superscript StartFraction x Over 2 EndFraction
y equals e Superscript negative StartFraction x Over 2 EndFraction
ln 6
ln 12
x
y
1
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Part 1
Set up the integral that gives the volume of the solid. 
Integral from ln 6 to nothing left parenthesis nothing right parenthesis dx

To find the volume of the solid generated when the region \( R \) is revolved about the \( x \)-axis, we use the method of disks or washers. The region is bounded by the curves:

- \( y = e^{\frac{x}{2}} \)
- \( y = e^{-\frac{x}{2}} \)
- \( x = \ln 6 \)
- \( x = \ln 12 \)

The volume is determined by the washer method, where we subtract the volume of the inner curve from the outer curve, using the formula:

\[
V = \pi \int_{a}^{b} \left[ f_{\text{outer}}(x)^2 - f_{\text{inner}}(x)^2 \right] \, dx
\]

### Step 1: Identify the bounds and curves
- The outer curve is \( y = e^{\frac{x}{2}} \).
- The inner curve is \( y = e^{-\frac{x}{2}} \).
- The bounds are \( x = \ln 6 \) and \( x = \ln 12 \).

### Step 2: Set up the integral
The volume integral is given by:

\[
V = \pi \int_{\ln 6}^{\ln 12} \left( e^{\frac{x}{2}} \right)^2 - \left( e^{-\frac{x}{2}} \right)^2 \, dx
\]

### Step 3: Simplify the integrand
Simplify the squares of the functions:

\[
\left( e^{\frac{x}{2}} \right)^2 = e^x
\]
\[
\left( e^{-\frac{x}{2}} \right)^2 = e^{-x}
\]

Now, substitute into the integral:

\[
V = \pi \int_{\ln 6}^{\ln 12} \left( e^x - e^{-x} \right) \, dx
\]

### Step 4: Solve the integral
We can now integrate term by term:

\[
\int e^x \, dx = e^x
\]
\[
\int e^{-x} \, dx = -e^{-x}
\]

Thus, the volume integral becomes:

\[
V = \pi \left[ e^x + e^{-x} \right]_{\ln 6}^{\ln 12}
\]

### Step 5: Evaluate the integral
Now, evaluate the definite integral at the bounds \( x = \ln 6 \) and \( x = \ln 12 \):

At \( x = \ln 12 \):

\[
e^{\ln 12} + e^{-\ln 12} = 12 + \frac{1}{12}
\]

At \( x = \ln 6 \):

\[
e^{\ln 6} + e^{-\ln 6} = 6 + \frac{1}{6}
\]

Now, subtract:

\[
V = \pi \left( \left( 12 + \frac{1}{12} \right) - \left( 6 + \frac{1}{6} \right) \right)
\]

### Step 6: Simplify
Simplify the expression:

\[
V = \pi \left( \left( 12 + \frac{1}{12} \right) - \left( 6 + \frac{1}{6} \right) \right)
= \pi \left( 12 - 6 + \frac{1}{12} - \frac{1}{6} \right)
= \pi \left( 6 + \frac{1}{12} - \frac{2}{12} \right)
= \pi \left( 6 - \frac{1}{12} \right)
\]

Finally:

\[
V = \pi \left( \frac{72}{12} - \frac{1}{12} \right) = \pi \cdot \frac{71}{12}
\]

Thus, the volume is:

\[
V = \frac{71\pi}{12}
\]

### Would you like more details or have any questions?

Here are some additional questions to expand your understanding:
1. What is the geometric interpretation of the washer method?
2. How would the volume change if the region was revolved around the y-axis instead of the x-axis?
3. How would the integral change if different bounds were used, such as \( \ln 3 \) to \( \ln 12 \)?
4. What happens if the inner and outer functions are reversed in a similar problem?
5. How does the integration differ if cylindrical shells are used instead of washers?

**Tip:** Always simplify the integrand before integrating—it makes the process much easier and reduces errors!

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the​ x-axis. y = e^(x/2), y = e^(-x/2), x = ln(6), x = ln(12). Set up the integral that gives the volume of the solid.

Discover how to find the volume of a solid generated when the region bounded by the curves y = e^(x/2), y = e^(-x/2), x = ln(6), and x = ln(12) is revolved about the x-axis. This solution uses the washer method and includes detailed steps for setting up and evaluating the integral.

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