Find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis.  Sketch the region and the solid.

Y = ex; y = 0; x = -5; x = 1.


To find the volume of the solid obtained by rotating the region bounded by the curves \( y = e^x \), \( y = 0 \), \( x = -5 \), and \( x = 1 \) about the x-axis, we'll use the **method of disks** (or washers). Here's the step-by-step process:

### 1. Sketch the Region
- The curve \( y = e^x \) is an exponential function, starting near zero at \( x = -5 \) and increasing rapidly as \( x \) approaches 1.
- The line \( y = 0 \) represents the x-axis.
- The lines \( x = -5 \) and \( x = 1 \) are vertical boundaries.

### 2. Set Up the Integral
To find the volume \( V \) of the solid, we can integrate using the formula for the volume of a solid of revolution:

\[
V = \pi \int_{a}^{b} [f(x)]^2 \, dx
\]

Here, \( f(x) = e^x \), and the bounds \( a \) and \( b \) are \( -5 \) and \( 1 \), respectively. So, the volume is:

\[
V = \pi \int_{-5}^{1} (e^x)^2 \, dx
\]

### 3. Simplify the Integral
Since \( (e^x)^2 = e^{2x} \), the integral becomes:

\[
V = \pi \int_{-5}^{1} e^{2x} \, dx
\]

### 4. Evaluate the Integral
To evaluate the integral, use the substitution method. Let:

\[
u = 2x \quad \text{so that} \quad du = 2 \, dx \quad \text{and} \quad dx = \frac{du}{2}
\]

Substitute into the integral:

\[
V = \pi \int_{-10}^{2} e^{u} \cdot \frac{du}{2} = \frac{\pi}{2} \int_{-10}^{2} e^u \, du
\]

Integrating \( e^u \):

\[
V = \frac{\pi}{2} \left[ e^u \right]_{-10}^{2} = \frac{\pi}{2} \left( e^{2} - e^{-10} \right)
\]

### 5. Final Answer
Thus, the volume of the solid is:

\[
V = \frac{\pi}{2} \left( e^2 - \frac{1}{e^{10}} \right)
\]

This represents the exact volume of the solid obtained by rotating the given region about the x-axis.

### Sketch of the Region and Solid
- **Region:** The region to be rotated lies between the curve \( y = e^x \), the x-axis, and the vertical lines \( x = -5 \) and \( x = 1 \).
- **Solid:** Upon rotation about the x-axis, this region forms a solid with a shape resembling a horn or funnel, starting with a small radius at \( x = -5 \) and increasing in radius as \( x \) moves towards \( 1 \).

Would you like further explanation or a specific diagram?

### Related Questions

1. How would the volume change if the region were rotated around the y-axis instead of the x-axis?
2. What if the region were bounded by a different exponential function, like \( y = e^{-x} \)?
3. Can you find the surface area of the solid formed by this rotation?
4. How does the choice of bounds \( x = -5 \) and \( x = 1 \) affect the volume?
5. What happens if the region is rotated about a line other than the x-axis or y-axis?

### Tip
When dealing with rotation problems, visualize the solid by sketching both the region and its rotation to better understand the problem.

Calculate the volume of the solid obtained by rotating the region bounded by y = e^x, y = 0, x = -5, and x = 1 about the x-axis. Step-by-step solution using the method of disks/washers with detailed integration process. Suitable for advanced high school students and college-level calculus enthusiasts.

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