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).