Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
y = 3e−x,   y = 3,   x = 4;
   about y = 6

We are tasked with finding the volume of the solid obtained by rotating the region bounded by the curves \( y = 3e^{-x} \), \( y = 3 \), and \( x = 4 \) around the line \( y = 6 \).

### Step-by-Step Approach:
We will use the **method of cylindrical shells**, since the axis of rotation is horizontal and does not coincide with the boundary of the region. The volume is calculated using the formula for the method of shells:

\[
V = 2\pi \int_a^b ( \text{radius}(x) ) \cdot ( \text{height}(x) ) \, dx
\]

- The **radius** of each cylindrical shell is the distance from the curve to the line \( y = 6 \). This distance is \( 6 - y \).
- The **height** is determined by the difference between the upper boundary \( y = 3 \) and the lower boundary \( y = 3e^{-x} \).

Thus, the volume becomes:

\[
V = 2\pi \int_0^4 (6 - y) \cdot dx
\]

However, since we are rotating around a line above the region, the radius needs to account for the distance from the curves to \( y = 6 \). Specifically:
- For \( y = 3e^{-x} \), the radius is \( 6 - 3e^{-x} \),
- For \( y = 3 \), the radius is \( 6 - 3 = 3 \).

So, the volume becomes:

\[
V = 2\pi \int_0^4 \left( (6 - 3e^{-x}) - 3 \right) \, dx
\]

Simplifying:

\[
V = 2\pi \int_0^4 \left( 3 - 3e^{-x} \right) \, dx
\]

### Solving the Integral:
The integral to solve is:

\[
V = 2\pi \int_0^4 \left( 3 - 3e^{-x} \right) \, dx
\]

We can split this into two integrals:

\[
V = 2\pi \left[ \int_0^4 3 \, dx - \int_0^4 3e^{-x} \, dx \right]
\]

- The first integral is straightforward:

\[
\int_0^4 3 \, dx = 3x \bigg|_0^4 = 12
\]

- The second integral requires the antiderivative of \( e^{-x} \), which is \( -e^{-x} \):

\[
\int_0^4 3e^{-x} \, dx = -3e^{-x} \bigg|_0^4 = -3(e^{-4} - e^{0}) = -3(e^{-4} - 1)
\]

Thus, we have:

\[
V = 2\pi \left( 12 - 3(-e^{-4} + 1) \right)
\]
\[
V = 2\pi \left( 12 - 3(1 - e^{-4}) \right)
\]
\[
V = 2\pi \left( 12 - 3 + 3e^{-4} \right) = 2\pi (9 + 3e^{-4})
\]

Finally:

\[
V = 18\pi + 6\pi e^{-4}
\]

This is the volume of the solid.

Would you like more details on any part of this process or further clarifications?

### Follow-up Questions:
1. What is the method of cylindrical shells and how does it differ from the disk/washer method?
2. How would the formula change if the axis of rotation was \( y = 3 \)?
3. Could you calculate the volume if the rotation was about the x-axis?
4. How do we apply integration by parts in volume calculation by shell method?
5. How does changing the boundary curve to \( y = e^{-x} \) impact the volume?

**Tip:** Always ensure the correct setup of radius and height before integrating to avoid errors in volume problems.

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 3e−x, y = 3, x = 4; about y = 6

Learn how to find the volume of a solid obtained by rotating the region bounded by y = 3e^(-x), y = 3, and x = 4 around the line y = 6. This calculus problem covers the method of cylindrical shells, integration, and solving for the volume using step-by-step calculations. Suitable for college-level calculus students.

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