A region is defined by the functions $$ y = e^{-x} $$ for $$ x \geq 0 $$ and $$ y = 0 $$. What is the volume generated when this region is revolved around the line $$ y = 1 $$?

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UpStudy AI · Accepted Answer

To find the volume generated by revolving the region bounded by $$ y = e^{-x} $$ for $$ x \geq 0 $$ and $$ y = 0 $$ around the line $$ y = 1 $$, we can use the **washer method**. Here's a step-by-step solution:

### 1. **Understanding the Region:**
- The region is bounded by the curve $$ y = e^{-x} $$, the x-axis ($$ y = 0 $$), and extends to $$ x = \infty $$.

### 2. **Setting Up the Washer Method:**
- **Outer Radius (R):** The distance from the axis of rotation $$ y = 1 $$ to the x-axis ($$ y = 0 $$) is $$ R = 1 $$.
- **Inner Radius (r):** The distance from $$ y = 1 $$ to $$ y = e^{-x} $$ is $$ r = 1 - e^{-x} $$.

### 3. **Volume Element:**
The area of a washer at a particular $$ x $$ is:
$$
\text{Area} = \pi(R^2 - r^2) = \pi \left(1^2 - (1 - e^{-x})^2\right) = \pi \left(2e^{-x} - e^{-2x}\right)
$$
Thus, the volume $$ V $$ is:
$$
V = \pi \int_{0}^{\infty} \left(2e^{-x} - e^{-2x}\right) dx
$$

### 4. **Evaluating the Integral:**
$$
\int_{0}^{\infty} 2e^{-x} dx = 2
$$
$$
\int_{0}^{\infty} e^{-2x} dx = \frac{1}{2}
$$
So,
$$
V = \pi \left(2 - \frac{1}{2}\right) = \pi \left(\frac{3}{2}\right) = \frac{3\pi}{2}
$$

### 5. **Alternative Approach - Cylindrical Shells:**
Using the method of cylindrical shells also leads to the same result. By setting up the integral with respect to $$ y $$, you ultimately derive:
$$
V = \frac{3\pi}{2}
$$

### **Final Answer:**
$$
\boxed{\dfrac{3}{2}\pi}
$$
The volume generated is $$ \frac{3}{2}\pi $$.

A region is defined by the functions for and . What is the volume generated when this region is revolved around the line ?

Upstudy AI Solution

Answer

Solution

1. Understanding the Region:

2. Setting Up the Washer Method:

3. Volume Element:

4. Evaluating the Integral:

5. Alternative Approach - Cylindrical Shells:

Final Answer:

Beyond the Answer

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