$$ y=8 e^{x}, \quad y=8 e^{-x}, \quad x=1 ; $$ about the $$ y $$-axis

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To determine the volume of the solid obtained by rotating the region bounded by the curves $$ y = 8e^x $$, $$ y = 8e^{-x} $$, and $$ x = 1 $$ about the $$ y $$-axis, we'll use the **method of cylindrical shells**.

### Step 1: Understand the Region

1. **Intersection Point:**
   - Set $$ 8e^x = 8e^{-x} $$ to find the intersection:
     $$
     e^x = e^{-x} \implies x = 0
     $$
   - So, the curves intersect at $$ x = 0 $$, $$ y = 8 $$.

2. **Boundaries:**
   - The region is bounded between $$ x = 0 $$ and $$ x = 1 $$.
   - **Upper Function:** $$ y = 8e^x $$
   - **Lower Function:** $$ y = 8e^{-x} $$

### Step 2: Set Up the Integral Using the Shell Method

The volume $$ V $$ of the solid of revolution can be calculated using the formula:
$$
V = 2\pi \int_{a}^{b} x \cdot ( \text{Upper Function} - \text{Lower Function} ) \, dx
$$
Plugging in the functions and limits:
$$
V = 2\pi \int_{0}^{1} x \left( 8e^x - 8e^{-x} \right) dx = 16\pi \int_{0}^{1} x \left( e^x - e^{-x} \right) dx
$$

### Step 3: Evaluate the Integral

Split the integral:
$$
\int_{0}^{1} x e^x \, dx - \int_{0}^{1} x e^{-x} \, dx
$$

1. **Integral of $$ x e^x $$:**
   $$
   \int x e^x \, dx = e^x (x - 1) + C
   $$
   Evaluated from 0 to 1:
   $$
   \left[ e^1 (1 - 1) - e^0 (0 - 1) \right] = 0 - (-1) = 1
   $$

2. **Integral of $$ x e^{-x} $$:**
   $$
   \int x e^{-x} \, dx = -e^{-x} (x + 1) + C
   $$
   Evaluated from 0 to 1:
   $$
   \left[ -e^{-1} (1 + 1) - (-e^{0} (0 + 1)) \right] = -2e^{-1} + 1
   $$

### Step 4: Combine the Results

$$
V = 16\pi \left( 1 - (-2e^{-1} + 1) \right) = 16\pi (2e^{-1}) = \frac{32\pi}{e}
$$

### Final Answer

The volume of the solid is $$ \dfrac{32\,\pi}{e} $$.
The volume of the solid is $$ \dfrac{32\,\pi}{e} $$.

\( y=8 e^{x}, \quad y=8 e^{-x}, \quad x=1 ; \) about the \( y \)-axis

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