We are tasked with finding the volume of the solid generated when the region $R$, bounded by the curves $y = x$, $y = 9x$, and $y = 36$, is revolved about the y-axis.

Step 1: Sketch the region

The curves are:

1. $y = x$ — a line with slope 1.
2. $y = 9x$ — a line with slope 9.
3. $y = 36$ — a horizontal line at $y = 36$.

These curves intersect at the following points:

• Intersection of $y = x$ and $y = 36$: x = 36 \quad \text{(since y = x)} So the point is $(36, 36)$.

• Intersection of $y = 9x$ and $y = 36$: $9x = 36 \quad \Rightarrow \quad x = 4$ So the point is $(4, 36)$.

Step 2: Set up the volume integral

The volume of a solid generated by revolving a region around the y-axis can be computed using the shell method. The formula for the volume is:

$V = 2\pi \int_{a}^{b} x \cdot [f(x) - g(x)] \, dx$

Where:

• $a$ and $b$ are the x-values that bound the region.
• $f(x)$ is the upper function (the function on the "outside" of the region).
• $g(x)$ is the lower function (the function on the "inside" of the region).

Here, we have two parts:

1. From $x = 0$ to $x = 4$, the region is bounded by the lines $y = 9x$ (upper) and $y = x$ (lower).
2. From $x = 4$ to $x = 36$, the region is bounded by the line $y = 36$ (upper) and $y = x$ (lower).

Step 3: Volume integral setup

1. From $x = 0$ to $x = 4$: $V_1 = 2\pi \int_0^4 x \cdot [(9x) - x] \, dx = 2\pi \int_0^4 x \cdot 8x \, dx = 16\pi \int_0^4 x^2 \, dx$

2. From $x = 4$ to $x = 36$: $V_2 = 2\pi \int_4^{36} x \cdot [36 - x] \, dx = 2\pi \int_4^{36} x(36 - x) \, dx$

Step 4: Compute the integrals

For $V_1$:

$V_1 = 16\pi \int_0^4 x^2 \, dx = 16\pi \left[ \frac{x^3}{3} \right]_0^4 = 16\pi \left( \frac{4^3}{3} \right) = 16\pi \cdot \frac{64}{3} = \frac{1024\pi}{3}$

For $V_2$:

$V_2 = 2\pi \int_4^{36} (36x - x^2) \, dx = 2\pi \left[ 18x^2 - \frac{x^3}{3} \right]_4^{36}$

Now evaluate the integral at the bounds:

• At $x = 36$: $18(36)^2 - \frac{(36)^3}{3} = 18(1296) - \frac{46656}{3} = 23328 - 15552 = 7776$
• At $x = 4$: $18(4)^2 - \frac{(4)^3}{3} = 18(16) - \frac{64}{3} = 288 - \frac{64}{3} = \frac{864}{3} - \frac{64}{3} = \frac{800}{3}$

Thus: $V_2 = 2\pi \left( 7776 - \frac{800}{3} \right) = 2\pi \left( \frac{23328}{3} - \frac{800}{3} \right) = 2\pi \cdot \frac{22528}{3} = \frac{45056\pi}{3}$

Step 5: Total Volume

Now, add the two volumes $V_1$ and $V_2$:

$V = V_1 + V_2 = \frac{1024\pi}{3} + \frac{45056\pi}{3} = \frac{46080\pi}{3}$

So, the total volume is: $V = \frac{46080\pi}{3} \, \text{cubic units}.$

Would you like further clarification on any part of this process? Here are some related questions to explore:

1. How would the volume change if the region was revolved around the x-axis instead of the y-axis?
2. Can you apply the disk method to compute volumes of solids of revolution? How would that work in this case?
3. What happens if the curves $y = 9x$ and $y = x$ intersect at different points?
4. How would the volume formula change if the boundary lines were curves instead of straight lines?
5. What role does symmetry play in simplifying volume integrals?

Tip: Always check the bounds of integration carefully, as the limits depend on where the curves intersect.