To find the volume of the solid generated by revolving the region $R$ around the y-axis, we can use the method of cylindrical shells.

Step 1: Identify the boundaries of the region $R$

The region $R$ is bounded by the following curves:

1. $y = x$
2. $y = 2x$
3. $y = 8$

We need to determine the points of intersection between these curves.

Intersection of $y = x$ and $y = 8$:

x = 8 \quad \text{(since y = x)} Thus, the point of intersection is $(8, 8)$.

Intersection of $y = 2x$ and $y = 8$:

$2x = 8 \quad \Rightarrow \quad x = 4$ Thus, the point of intersection is $(4, 8)$.

Therefore, the region $R$ is bounded between $x = 4$ and $x = 8$ along the x-axis.

Step 2: Set up the volume integral using the method of cylindrical shells

The volume $V$ generated by revolving the region $R$ around the y-axis is given by the formula for cylindrical shells: $V = 2\pi \int_{x_1}^{x_2} \text{(radius)} \times \text{(height)} \, dx$ Where:

• The radius of a shell is the distance from the y-axis, which is just $x$.
• The height of the shell is the difference between the two functions $y = 2x$ (top curve) and $y = x$ (bottom curve).

Thus, the volume integral becomes: $V = 2\pi \int_4^8 x \left( (2x) - x \right) dx$ Simplifying the integrand: $V = 2\pi \int_4^8 x \cdot x \, dx = 2\pi \int_4^8 x^2 \, dx$

Step 3: Compute the integral

Now, we calculate the integral: $V = 2\pi \int_4^8 x^2 \, dx$ The antiderivative of $x^2$ is $\frac{x^3}{3}$, so we evaluate: $V = 2\pi \left[ \frac{x^3}{3} \right]_4^8$ Substitute the limits of integration: $V = 2\pi \left( \frac{8^3}{3} - \frac{4^3}{3} \right)$ $V = 2\pi \left( \frac{512}{3} - \frac{64}{3} \right)$ $V = 2\pi \left( \frac{448}{3} \right)$ $V = \frac{896\pi}{3}$

Final Answer:

The volume of the solid generated when the region $R$ is revolved about the y-axis is $\frac{896\pi}{3}$ cubic units.

Would you like further details or clarification on any step? Here are some related questions to explore further:

1. What is the volume when the region $R$ is revolved about the x-axis?
2. How would the volume change if the region was bounded by $y = 3x$ instead of $y = 2x$?
3. Can the method of disks also be used to solve this problem?
4. How do you calculate the surface area of the solid generated by revolution around the y-axis?
5. What is the relationship between the radius and height in the method of cylindrical shells?

Tip: Always carefully analyze the points of intersection between the curves to determine the bounds of the region accurately before setting up the integral.