Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 2, and x = 1 is revolved around the line y = −2.
Ugh Steve why
To set up the integral for finding the volume, we can use the method of cylindrical shells.
First, let's sketch the region bounded by the curves y = Ln(x), y = 2, and x = 1:
2
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x=1
To find the volume of the solid when this region is revolved around the line y = -2, we can divide the region into infinitesimally thin cylindrical shells.
Consider a vertical strip of width Δx at a distance x from the y-axis. The height of this strip will be given by the difference between the upper and lower curves, which is 2 - Ln(x). The length of the strip will be Δx.
For each strip, the volume of the corresponding cylindrical shell will be approximately equal to the surface area of the shell multiplied by the thickness Δx.
The surface area of the cylindrical shell is given by 2πrh, where r is the radius of the shell and h is the height of the shell.
Since we are revolving around the line y = -2, the radius r will be x + 2.
Therefore, the volume of the cylindrical shell will be approximately equal to:
(2π(x+2)(2-Ln(x)))Δx
To find the total volume, we need to integrate this expression over the interval where x ranges from 1 to the x-coordinate of the point where the two curves intersect.
Let's denote the x-coordinate of the point where the two curves intersect as a.
The integral to find the volume is then:
V = ∫[1, a] (2π(x+2)(2-Ln(x))) dx
To set up the integral that gives the volume of the solid when the region bounded by the curves is revolved around the line y = -2, you can use the method of cylindrical shells.
First, let's find the interval over which we need to integrate. The curves y = Ln(x) and y = 2 intersect at a point. To find that point, we set the two equations equal to each other:
Ln(x) = 2
To solve this equation, we raise e to the power of both sides:
e^(Ln(x)) = e^2
x = e^2
So, the interval of integration is [1, e^2], since the region is bounded by x = 1 and x = e^2.
Next, let's determine the height of the cylindrical shell at a given x-value. The height of the shell is the difference between the upper and lower curves at that x-value. In this case, the upper curve is y = 2, and the lower curve is y = Ln(x).
So, the height of the shell is given by:
h(x) = 2 - Ln(x)
For each shell, the circumference or perimeter of the shell is given by 2πr, where r is the distance from the x-axis to the shell. In this case, since the shells are being revolved around the line y = -2, the distance is given by:
r(x) = x - (-2) = x + 2
Finally, the volume of each cylindrical shell is given by the product of the height, the circumference, and a small differential width dx:
dV = 2π * (x + 2) * (2 - Ln(x)) * dx
To get the total volume, we integrate this expression over the interval of integration:
V = ∫[1, e^2] 2π * (x + 2) * (2 - Ln(x)) dx
And that's the integral set up to find the volume when the region bounded by the given curves is revolved around y = -2.
you can do this as shells or washers.
with shells of thickness dy, remember that the volume of a shell is 2πrh. So, draw the region, and
∫[0,2] 2πrh dy
For discs with holes, of thickness dx, work with
∫[1,2] π(R^2-r^2) dx
your job is to examine the region in question and figure out R,r,h as required.