 Let R
 be the region bounded by y=ex
, x=0
 and y=4
. Let S
 be the solid obtained by revolving R
 around the line y=5
. If we use the shell method to find the volume of S
 

To find the volume of the solid \( S \) obtained by revolving the region \( R \) (bounded by \( y = e^x \), \( x = 0 \), and \( y = 4 \)) around the line \( y = 5 \) using the **shell method**, we follow these steps:

### Step 1: Identify the Shell Radius and Height
In the shell method, we revolve vertical slices of the region around the axis of revolution (in this case, the line \( y = 5 \)).

- The **height** of the shell at any point \( x \) is the difference between the function \( y = e^x \) and \( y = 0 \), so the height of each shell is \( e^x \).
  
- The **radius** of the shell is the vertical distance from the function \( y = e^x \) to the axis of revolution \( y = 5 \). This distance is \( 5 - e^x \).

### Step 2: Set up the Volume Integral
The formula for the volume of a solid using the shell method is:

\[
V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \, dx
\]

- The region \( R \) is bounded by \( x = 0 \) on the left and by \( y = 4 \) on the top. To find the corresponding \( x \)-value when \( y = 4 \), we solve \( e^x = 4 \), which gives \( x = \ln(4) \).
  
So, the limits of integration are from \( x = 0 \) to \( x = \ln(4) \).

Thus, the volume integral is:

\[
V = 2\pi \int_{0}^{\ln(4)} (5 - e^x)(e^x) \, dx
\]

### Step 3: Simplify the Integral
Simplifying the integrand:

\[
V = 2\pi \int_{0}^{\ln(4)} \left( 5e^x - e^{2x} \right) \, dx
\]

### Step 4: Integrate
Now, integrate term by term:

- The integral of \( 5e^x \) is \( 5e^x \).
- The integral of \( e^{2x} \) is \( \frac{e^{2x}}{2} \).

So, the volume is:

\[
V = 2\pi \left[ 5e^x - \frac{e^{2x}}{2} \right]_{0}^{\ln(4)}
\]

### Step 5: Evaluate the Integral
Now, substitute the limits of integration:

At \( x = \ln(4) \):

\[
5e^{\ln(4)} = 5 \cdot 4 = 20, \quad \frac{e^{2\ln(4)}}{2} = \frac{16}{2} = 8
\]

At \( x = 0 \):

\[
5e^0 = 5, \quad \frac{e^{2(0)}}{2} = \frac{1}{2}
\]

Thus, the volume is:

\[
V = 2\pi \left[ (20 - 8) - (5 - \frac{1}{2}) \right]
\]

Simplifying:

\[
V = 2\pi \left[ 12 - \left( 5 - \frac{1}{2} \right) \right] = 2\pi \left[ 12 - 4.5 \right] = 2\pi \cdot 7.5 = 15\pi
\]

### Final Answer:
The volume of the solid \( S \) is \( \boxed{15\pi} \).

---

Would you like more details or have any questions about the steps? Here are 5 related questions to expand your understanding:

1. What is the shell method, and how does it differ from the disk/washer method?
2. How do you set up integrals for volumes of revolution using other axes (e.g., \( y = -1 \) or \( x = 3 \))?
3. How does the choice of axis of revolution affect the integral for the volume?
4. How do you integrate more complex exponential functions in volume problems?
5. What are some common applications of the shell method in physics or engineering?

**Tip:** When setting up shell method integrals, always carefully consider the radius and height of the shells based on the axis of revolution.

Let R be the region bounded by y=ex, x=0 and y=4. Let S be the solid obtained by revolving R around the line y=5. If we use the shell method to find the volume of S.

Discover how to calculate the volume of a solid generated by revolving the region bounded by y=ex, x=0, and y=4 around the line y=5 using the shell method. This step-by-step solution covers key calculus concepts like exponential functions, shell method integration, and volumes of revolution. Suitable for college-level calculus students.

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