Use the shell method to find the volume of the solid generated by revolving the region bounded by x equals StartRoot y EndRoot​, x equals negative 7 y​, and y equals 2 about the​ x-axis.
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Part 1
The volume of the solid generated by revolving the region bounded by x equals StartRoot y EndRoot​, x equals negative 7 y​, and y equals 2 about the​ x-axis is 
  
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To find the volume of the solid generated by revolving the region bounded by \( x = \sqrt{y} \), \( x = -7y \), and \( y = 2 \) about the x-axis, we will use the **shell method**. Here's the process step by step:

### Step 1: Understanding the region
The boundaries given are:
- \( x = \sqrt{y} \): A parabola opening to the right.
- \( x = -7y \): A linear equation.
- \( y = 2 \): The upper bound of the region.

We need to revolve this region about the **x-axis**.

### Step 2: Setting up the shell method
For the shell method, we revolve vertical elements of the region around the x-axis. The formula for the volume using the shell method is:

\[
V = 2\pi \int_{y_{\text{min}}}^{y_{\text{max}}} \text{(radius)} \times \text{(height)} \, dy
\]

- **Radius**: Since we're rotating about the x-axis, the radius of each shell is \( y \).
- **Height**: The height of each shell is the horizontal distance between \( x = \sqrt{y} \) and \( x = -7y \), which is the difference \( \sqrt{y} - (-7y) = \sqrt{y} + 7y \).

### Step 3: Identifying limits of integration
The region is bounded between \( y = 0 \) and \( y = 2 \).

### Step 4: Setting up the integral
Now, we set up the integral for the volume:

\[
V = 2\pi \int_0^2 y \left( \sqrt{y} + 7y \right) \, dy
\]

### Step 5: Simplifying the integrand
We simplify the expression inside the integral:

\[
V = 2\pi \int_0^2 y (\sqrt{y}) + y (7y) \, dy
\]
\[
V = 2\pi \int_0^2 y^{3/2} + 7y^2 \, dy
\]

### Step 6: Computing the integral
Now, we integrate each term:

1. \[
\int y^{3/2} \, dy = \frac{2}{5} y^{5/2}
\]
2. \[
\int 7y^2 \, dy = \frac{7}{3} y^3
\]

So the volume becomes:

\[
V = 2\pi \left[ \frac{2}{5} y^{5/2} + \frac{7}{3} y^3 \right]_0^2
\]

### Step 7: Evaluating the integral at the limits
Now, we evaluate the expression at \( y = 2 \) and \( y = 0 \):

At \( y = 2 \):

\[
\frac{2}{5} (2)^{5/2} + \frac{7}{3} (2)^3 = \frac{2}{5} \cdot 4\sqrt{2} + \frac{7}{3} \cdot 8 = \frac{8\sqrt{2}}{5} + \frac{56}{3}
\]

At \( y = 0 \), the expression is 0.

### Step 8: Final answer
Thus, the volume is:

\[
V = 2\pi \left( \frac{8\sqrt{2}}{5} + \frac{56}{3} \right)
\]

You can leave the answer in this form, or you can approximate it numerically if needed.

---

Do you have any questions or want further clarification?  
Here are five related questions to deepen your understanding:

1. How does the shell method differ from the disk/washer method for finding volumes?
2. What would the volume be if the region were revolved around the y-axis instead?
3. Can you explain the significance of the limits of integration in this problem?
4. How would the process change if the upper bound for \( y \) was \( y = 3 \) instead of \( y = 2 \)?
5. Why is the height of the shell equal to \( \sqrt{y} + 7y \) in this case?

**Tip:** When using the shell method, always ensure that you clearly understand whether you're revolving around the x-axis or y-axis as this will affect the setup of your integral.

Use the shell method to find the volume of the solid generated by revolving the region bounded by x = √y, x = -7y, and y = 2 about the x-axis.

Learn how to find the volume of a solid generated by revolving the region bounded by x = √y, x = -7y, and y = 2 about the x-axis using the shell method. This problem involves calculus concepts like volume of solids of revolution and integral setup for the shell method. Suitable for undergraduate Calculus students.

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