The region bounded by the curve y=1÷(1+2x) , the line X=2 , the x-axis and the y axis is rotated completely about the y-axis.

Show that the volume generated is 1/2π(4-ln5)

The curve intersects x=2 at (2,1/5)

The y-intercept is at (0,1)

Using discs,

v = ∫π(2^2)(1/5) dy [0,1/5] + ∫π((1-y/2y)^2 (y-1/5) dy [1/5,1]
4π/5 + π/4 (y^2-1-2ylny)/4y [1/5,1]
= π/2 (4-ln5)

Using shells,

v = ∫2πx(1/(1+2x)) dx [0,2]
= π/2 ((2x+1)-ln(2x+1)) [0,2]
= π/2 (4-ln5)

To find the volume generated when the region bounded by a curve is rotated completely about the y-axis, we can use the method of cylindrical shells.

First, let's graph the given curve y = 1/(1 + 2x), the line x = 2, and the x and y axes. Note that the curve is only defined for x < -1/2 since the denominator cannot be zero:

```
^
| .
| .
y | . .
-axis | . . .
|___________
| | | | |
-1 0 1 2 x-axis
```

We are interested in finding the volume of the solid formed when this region is rotated about the y-axis.

To do this, we will use the formula for the volume of a solid of revolution using cylindrical shells:

V = ∫(2π * radius * height) * dx

In this case, the radius is the distance from the y-axis to an arbitrary point (x, 1/(1 + 2x)), given by x. The height is the distance between y = 0 and y = 1/(1 + 2x), given by 1/(1 + 2x). We will integrate these values with respect to x.

The limits of integration are determined by the points where the curve intersects x = 2. To find these points, we set x = 2 and solve for y:

1/(1 + 2*2) = 1/5 = 0.2

So the curve intersects x = 2 at y = 0.2.

Now let's set up and evaluate the integral:

V = ∫[from x = -1/2 to x = 2](2π * x * (1/(1 + 2x))) dx

To simplify this integral, we can use the substitution u = 1 + 2x. The differential du = 2dx.

When x = -1/2, u = 1 + 2(-1/2) = 0.
When x = 2, u = 1 + 2(2) = 5.

Substituting these values and simplifying the integral, we have:

V = ∫[from u = 0 to u = 5](π * (u - 1)/2) du

V = (1/2π) * [∫(u - 1) du] evaluated from u = 0 to u = 5

V = (1/2π) * [(1/2 * u^2 - u)] evaluated from u = 0 to u = 5

V = (1/2π) * [(1/2 * 5^2 - 5) - (1/2 * 0^2 - 0)]

V = (1/2π) * [(25/2 - 5) - 0]

V = (1/2π) * [(25 - 10)/2]

V = (1/2π) * (15/2)

V = 15/4π

Simplifying further:

V = (3/4)(5/π)

V = (15/4π)

So the volume generated when the region is rotated completely about the y-axis is 1/(2π)(4 - ln5).