Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis. (Use Disc Method)

y=3(2-x), y=0, x=0

y-intercept is 6

y = 6-3x
3x = -y + 6
x = -y/3 + 2
x^2= y^2/9 - 4y/3 + 4

vol = π[integral] y^2/9 - 4y/3 + 4 dy from 0 to 6
= π[ y^3/27 - 2y^2/3 + 4y] from 0 to 6
= π( 8 - 24 + 24 - (0) )
= 8π

better check my arithmetic

To find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis using the disc method, follow these steps:

1. Sketch the region bounded by the graphs of the equations.
- The graph y = 3(2-x) is a downward-opening line that passes through (0,6) and (2,0). The x-axis and y-axis form the boundaries of the region.

2. Determine the limits of integration.
- The region is bounded by y = 3(2-x) above and y = 0 below. To find the limits of integration, set the two equations equal to each other and solve for x:
3(2 - x) = 0
2 - x = 0
x = 2
- The limits of integration for x are from 0 to 2.

3. Set up the volume integral using the disc method.
- The volume integral for the disc method is given by:
V = π ∫[a, b] (R(x)^2 - r(x)^2) dx
where a and b are the limits of integration, R(x) is the outer radius, and r(x) is the inner radius.

4. Determine the formula for R(x) and r(x).
- Since we are revolving the region about the y-axis, the outer radius R(x) is the distance from the y-axis to the curve y = 3(2-x), which is x.
- The inner radius r(x) is the distance from the y-axis to the x-axis, which is 0.

5. Evaluate the volume integral.
- By substituting the limits of integration and the formulas for R(x) and r(x) into the volume integral, we get:
V = π ∫[0, 2] (x^2 - 0^2) dx
V = π ∫[0, 2] x^2 dx
- Integrate x^2 with respect to x:
V = π [x^3/3] from 0 to 2
V = π (2^3/3 - 0)
V = (8π/3) cubic units

Thus, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis using the disc method is (8π/3) cubic units.