Find the volume of

y=3e^-x, y=3, x=2, about y=6

To find the volume of the solid generated by rotating the region enclosed by \( y = 3e^{-x} \), \( y = 3 \), and \( x = 2 \) about the line \( y = 6 \), we use the **method of cylindrical shells** or the **washer method**. Let's use the washer method in this case.

### Step 1: Set up the integral for the washer method
The formula for the volume of revolution using the washer method is:
\[
V = \pi \int_{a}^{b} \left( R_{\text{outer}}^2 - R_{\text{inner}}^2 \right) dx
\]
Where:
- \( R_{\text{outer}} \) is the distance from the axis of rotation (in this case, \( y = 6 \)) to the upper function.
- \( R_{\text{inner}} \) is the distance from the axis of rotation to the lower function.
- \( a \) and \( b \) are the bounds of integration.

### Step 2: Define the radii of the washers
- The outer radius \( R_{\text{outer}} \) is the distance from \( y = 6 \) to \( y = 3 \) (the line):
\[
R_{\text{outer}} = 6 - 3 = 3
\]
- The inner radius \( R_{\text{inner}} \) is the distance from \( y = 6 \) to \( y = 3e^{-x} \) (the curve):
\[
R_{\text{inner}} = 6 - 3e^{-x}
\]

### Step 3: Set up the bounds
The bounds of integration are from \( x = 0 \) to \( x = 2 \), because the region is bounded by \( x = 2 \).

### Step 4: Set up the integral
Now, we can write the volume integral as:
\[
V = \pi \int_{0}^{2} \left( (3)^2 - (6 - 3e^{-x})^2 \right) dx
\]
Simplifying:
\[
V = \pi \int_{0}^{2} \left( 9 - (6 - 3e^{-x})^2 \right) dx
\]

### Step 5: Expand and simplify the integrand
Now, expand \( (6 - 3e^{-x})^2 \):
\[
(6 - 3e^{-x})^2 = 36 - 36e^{-x} + 9e^{-2x}
\]
So the volume integral becomes:
\[
V = \pi \int_{0}^{2} \left( 9 - (36 - 36e^{-x} + 9e^{-2x}) \right) dx
\]
\[
V = \pi \int_{0}^{2} \left( 9 - 36 + 36e^{-x} - 9e^{-2x} \right) dx
\]
\[
V = \pi \int_{0}^{2} \left( -27 + 36e^{-x} - 9e^{-2x} \right) dx
\]

### Step 6: Integrate
Now, integrate each term:
\[
V = \pi \left[ -27x + 36(-e^{-x}) - 9\left( \frac{-e^{-2x}}{2} \right) \right]_{0}^{2}
\]
Simplifying:
\[
V = \pi \left[ -27x - 36e^{-x} + \frac{9e^{-2x}}{2} \right]_{0}^{2}
\]

### Step 7: Evaluate the bounds
At \( x = 2 \):
\[
-27(2) - 36e^{-2} + \frac{9e^{-4}}{2} = -54 - 36e^{-2} + \frac{9e^{-4}}{2}
\]
At \( x = 0 \):
\[
-27(0) - 36e^{0} + \frac{9e^{0}}{2} = -36 + \frac{9}{2}
\]
Simplifying these values:
\[
V = \pi \left[ (-54 - 36e^{-2} + \frac{9e^{-4}}{2}) - (-36 + \frac{9}{2}) \right]
\]

### Step 8: Final Expression
This expression can be evaluated numerically, but this is the integral set up and the general form of the solution.

Would you like me to evaluate this numerically or go into more details?

### Follow-up Questions:
1. How do you compute the exponential terms in definite integrals?
2. What are the differences between the washer and cylindrical shell methods?
3. How does the choice of axis of rotation affect the formula?
4. Can you simplify the integrand before expanding it?
5. How do boundaries in real-world problems change the setup of volume integrals?

**Tip:** Always double-check the setup of radii in the washer method, especially when revolving around lines other than the x- or y-axis.

Find the volume of y=3e^-x, y=3, x=2, about y=6

Learn how to find the volume of the solid formed by rotating the region enclosed by y=3e^-x, y=3, and x=2 about the line y=6. This step-by-step guide covers the washer method and exponential functions, suitable for advanced calculus students in Grades 11-12.

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