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.

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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.