Answer:
the volume of the solid formed by revolving the region bounded by y=2-x, y=0, and x=0 about the line y=3 is 8π/3 cubic units.
Explanation:
To compute the volume of the solid formed by revolving the region bounded by y=2-x, y=0, and x=0 about the line y=3, we can use the method of cylindrical shells.
First, we need to sketch the region and the line of revolution:
```
| . (3,0)
3 | .
| / |
| / |
| / |
2-y| / |
| / |
|/__________|__
| x=0 x=2
0 3
```
The line of revolution y=3 passes through the point (3,0) and is parallel to the x-axis.
Next, we need to find the equation of the curve that forms the boundary of the region when it is revolved around the line y=3. This curve is a circle with radius 3 (the distance between the line of revolution and the point (0,0)) and center (0,3).
The equation of this circle is (x-0)^2 + (y-3)^2 = 3^2, which simplifies to x^2 + (y-3)^2 = 9.
We can now set up the integral for the volume using cylindrical shells:
```
V = ∫[0,2] 2πrh dx
where r = distance between the axis of rotation and the shell
h = height of the shell
```
In this case, the radius of each shell is x (the distance between the axis of rotation and the curve) and the height of each shell is y=2-x. So we have:
```
V = ∫[0,2] 2πx(2-x) dx
= 2π ∫[0,2] (2x - x^2) dx
= 2π [x^2 - (1/3)x^3] |[0,2]
= 2π [(2^2) - (1/3)(2^3)]
= 8π/3
```
Therefore, the volume of the solid formed by revolving the region bounded by y=2-x, y=0, and x=0 about the line y=3 is 8π/3 cubic units.