To find the volume of the solid generated when R is revolved about the y-axis using the disk (washer) method, we need to integrate the area of the cross-sections perpendicular to the y-axis.
Let's break down the problem into two parts based on the given curves.
1. The region bounded by y = x and y = 3x:
First, we need to determine the limits of integration for this part of the region. The curves intersect at y = 0 and y = 6, so the limits of integration for this part will be from y = 0 to y = 6.
Now, let's consider an arbitrary height y within this range. At height y, the outer radius of the washer is given by the distance from the y-axis to the curve y = 3x, which is equal to x = y/3. The inner radius is given by the distance from the y-axis to the curve y = x, which is equal to x = y. Therefore, the radius of the washer is (y/3) - y = -2y/3.
The area of each washer is given by π((outer radius)^2 - (inner radius)^2). Substituting the values:
Area = π[((y/3) - y)^2 - (y^2)]
= π[(y^2/9) - (2y^2/3) + (4y^2/9) - (y^2)]
= π[y^2/9 - 2y^2/3 + 4y^2/9 - y^2]
= π[-(4y^2/9) + 2y^2/3]
2. The region bounded by y = 3x and y = 6:
Similarly, for this part, the limits of integration will be from y = 0 to y = 6.
At height y, the outer radius of the washer is given by x = y/3, and the inner radius is given by x = 6/3 = 2. Therefore, the radius of the washer is (y/3) - 2.
The area of each washer is given by π[((y/3) - 2)^2 - (y^2)]. Substituting the values:
Area = π[((y/3) - 2)^2 - y^2]
= π[(y^2/9) - (4y/3) + 4 - y^2]
= π[-(8y/3) + (4 - 8/9)y^2]
Now, we need to set up the integral to find the total volume V:
V = ∫[0, 6] [-(4y^2/9) + 2y^2/3] dy + ∫[0, 6] [-(8y/3) + (4 - 8/9)y^2] dy
Evaluating this integral will give us the volume of the solid generated when R is revolved about the y-axis.