Find the volume of the solid generated by revolving the region enclosed by the graphs of 

yequals=e Superscript StartFraction x Over 2 EndFractionex2​, 

yequals=​1, 

and 

xequals=ln 17ln17 

about the​ x-axis.

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Part 1

The volume is 

enter your response here.

​(Type an exact​ answer, using 

piπ 

and 

ee 

as​ needed.)

The volume of the solid generated by revolving the given region around the x-axis can be found using the method of disks or washers. In this case, we'll use the method of washers because we are revolving the region enclosed by two curves around the x-axis.

### Step 1: Identify the boundaries

We are given:

- \( y = e^{\frac{x}{2}} \)
- \( y = 1 \)
- \( x = \ln 17 \)

We revolve the region enclosed by these curves about the x-axis.

### Step 2: Set up the volume integral

For the washer method, the formula for volume is:

\[
V = \pi \int_{a}^{b} \left( R(x)^2 - r(x)^2 \right) dx
\]

Where:
- \( R(x) \) is the outer radius (distance from the x-axis to the top curve)
- \( r(x) \) is the inner radius (if there is a lower curve, otherwise 0)
- \( a \) and \( b \) are the bounds of integration.

Here, the outer radius \( R(x) = e^{\frac{x}{2}} \), the inner radius is the line \( r(x) = 1 \), and the bounds are \( x = 0 \) to \( x = \ln 17 \).

### Step 3: Write the volume integral

The volume integral becomes:

\[
V = \pi \int_{0}^{\ln 17} \left( \left( e^{\frac{x}{2}} \right)^2 - 1^2 \right) dx
\]

Simplify the integrand:

\[
\left( e^{\frac{x}{2}} \right)^2 = e^x
\]
\[
V = \pi \int_{0}^{\ln 17} \left( e^x - 1 \right) dx
\]

### Step 4: Integrate

Now, we compute the integral:

\[
V = \pi \left[ \int_{0}^{\ln 17} e^x \, dx - \int_{0}^{\ln 17} 1 \, dx \right]
\]

- The integral of \( e^x \) is \( e^x \),
- The integral of \( 1 \) is \( x \).

Thus, the volume becomes:

\[
V = \pi \left[ e^x \Big|_0^{\ln 17} - x \Big|_0^{\ln 17} \right]
\]

### Step 5: Evaluate the definite integrals

- For \( e^x \), we have:

\[
e^{\ln 17} = 17, \quad e^0 = 1
\]

So:

\[
\int_0^{\ln 17} e^x \, dx = 17 - 1 = 16
\]

- For \( x \), we have:

\[
\ln 17 - 0 = \ln 17
\]

Thus, the volume becomes:

\[
V = \pi \left( 16 - \ln 17 \right)
\]

### Final Answer:

\[
V = \pi \left( 16 - \ln 17 \right)
\]

This is the exact volume of the solid.

---

Would you like a breakdown of any steps or have questions?

Here are 5 related questions:

1. How is the method of washers different from the method of disks in volume computation?
2. Can you find the volume if the region is revolved around the y-axis instead of the x-axis?
3. What is the significance of the boundaries \( y = 1 \) and \( x = \ln 17 \) in this problem?
4. What changes if we revolve the region around the line \( y = 1 \) instead of the x-axis?
5. How do you handle cases when both functions are below the x-axis in volume problems?

**Tip:** When solving volume problems, always sketch the region and identify the axis of revolution to help set up the correct integral!

Find the volume of the solid generated by revolving the region enclosed by the graphs of y = e^(x/2), y = 1, and x = ln 17 about the x-axis. The volume is (Type an exact answer, using pi and e as needed).

Calculate the volume of the solid formed by revolving the region enclosed by y = e^(x/2), y = 1, and x = ln 17 around the x-axis using the washer method. This problem includes concepts such as exponential functions and definite integrals, perfect for students studying calculus.

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