To find the volume of the solid generated by revolving the region $R$ around the $y$-axis using the shell method, we follow the general formula for the shell method:

$V = 2\pi \int_{a}^{b} \text{(radius)} \times \text{(height)} \, dx$

Given curves:

• $y = \frac{8}{1 + 8x^2}$
• $y = 0$
• $x = 0$
• $x = 2$

We revolve this region around the $y$-axis, so:

• The radius of each cylindrical shell is simply $x$.
• The height of each shell is the function $y = \frac{8}{1 + 8x^2}$.

The limits of integration for $x$ are from 0 to 2.

Thus, the volume $V$ is given by:

$V = 2\pi \int_0^2 x \cdot \frac{8}{1 + 8x^2} \, dx$

This is the setup for the integral. Would you like me to solve the integral for the volume?

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5 Related Questions:

1. What is the geometric interpretation of the shell method in calculating volume?
2. How does the shell method differ from the disk/washer method?
3. Can the volume also be calculated using the disk method for this problem? If so, how?
4. What happens if we revolve the region around the $x$-axis instead?
5. What are the limits of integration if we were to revolve the region around a vertical line other than the $y$-axis?

Tip:

When using the shell method, the "radius" is always the distance from the axis of revolution to the point on the curve, and the "height" is the value of the function itself.