y=6e^-2x and y=6+4x-2x^2 between x=0 and x=1 about the line y=-2 find volume

To find the volume of the solid generated by rotating the region bounded by the curves \( y = 6e^{-2x} \) and \( y = 6 + 4x - 2x^2 \) about the line \( y = -2 \) between \( x = 0 \) and \( x = 1 \), we can use the **method of washers**.

### Step 1: Express the functions relative to the axis of rotation \( y = -2 \)
1. **Upper function**: \( y = 6 + 4x - 2x^2 \)
   - Distance from \( y = -2 \): \( R(x) = (6 + 4x - 2x^2) - (-2) = 8 + 4x - 2x^2 \)

2. **Lower function**: \( y = 6e^{-2x} \)
   - Distance from \( y = -2 \): \( r(x) = (6e^{-2x}) - (-2) = 6e^{-2x} + 2 \)

### Step 2: Set up the washer method integral
The volume \( V \) is given by:

\[
V = \pi \int_{0}^{1} \left[ R(x)^2 - r(x)^2 \right] \, dx
\]

Substitute the expressions for \( R(x) \) and \( r(x) \):

\[
V = \pi \int_{0}^{1} \left[ (8 + 4x - 2x^2)^2 - (6e^{-2x} + 2)^2 \right] \, dx
\]

### Step 3: Expand and simplify the integrand
Expand \( (8 + 4x - 2x^2)^2 \) and \( (6e^{-2x} + 2)^2 \), then simplify:

\[
(8 + 4x - 2x^2)^2 = (64 + 64x - 32x^2 + 16x^2 - 16x^3 + 4x^4) 
\]

\[
(6e^{-2x} + 2)^2 = 36e^{-4x} + 24e^{-2x} + 4
\]

Thus, the integrand becomes:

\[
V = \pi \int_{0}^{1} \left[ (64 + 64x - 16x^2 - 16x^3 + 4x^4) - (36e^{-4x} + 24e^{-2x} + 4) \right] \, dx
\]

### Step 4: Integrate
Now, integrate each term within the bounds \( x = 0 \) and \( x = 1 \):

\[
V = \pi \left[ \int_{0}^{1} \left(60 + 64x - 16x^2 - 16x^3 + 4x^4 - 24e^{-2x} - 36e^{-4x}\right) dx \right]
\]

Each integral can be computed:

\[
\int_{0}^{1} 60 \, dx = 60x \bigg|_0^1 = 60
\]

\[
\int_{0}^{1} 64x \, dx = 32x^2 \bigg|_0^1 = 32
\]

\[
\int_{0}^{1} -16x^2 \, dx = -\frac{16}{3}x^3 \bigg|_0^1 = -\frac{16}{3}
\]

\[
\int_{0}^{1} -16x^3 \, dx = -4x^4 \bigg|_0^1 = -4
\]

\[
\int_{0}^{1} 4x^4 \, dx = \frac{4}{5}x^5 \bigg|_0^1 = \frac{4}{5}
\]

The exponential integrals can be computed, though they often require numerical methods:

\[
\int_{0}^{1} -24e^{-2x} \, dx = \frac{-12}{e^2} + 12
\]

\[
\int_{0}^{1} -36e^{-4x} \, dx = \frac{-9}{e^4} + 9
\]

### Step 5: Compute the final volume
Substitute the values back into the equation and sum them up:

\[
V = \pi \left[ 60 + 32 - \frac{16}{3} - 4 + \frac{4}{5} + \left(12 - \frac{12}{e^2}\right) + \left(9 - \frac{9}{e^4}\right) \right]
\]

This will give the exact volume.

Would you like the detailed numeric computation, or have any other questions? 

Here are five related questions for further exploration:

1. How does the volume change if the region is rotated around the x-axis instead of \( y = -2 \)?
2. What is the effect of changing the limits of integration on the volume?
3. How would the volume differ if the curves were shifted vertically by a constant?
4. Can the problem be solved using the cylindrical shells method?
5. How does the choice of axis of rotation impact the complexity of the integral?

**Tip**: When solving volumes of revolution, always sketch the region first to clearly understand the bounds and distances from the axis of rotation.

Explore how to calculate the volume of a solid formed by rotating the region between the curves y = 6e^{-2x} and y = 6 + 4x - 2x^2 about the line y = -2. Detailed step-by-step solution using the washer method and integral calculus.

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