Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = -1.
I don't understand the washer thickness topic which is why im having trouble with this question
Oops the volume is
v = ∫[1,e] π(4-(lnx+1)^2) dx
A circle of radius R has area πR^2
so, a disc of thickness dx and radius R has a volume
v = πR^2 dx
Now, if it has a hole in it of radius r, then the hole has volume
v = πr^2 dx
So, the washer (a disc with hole) has volume
v = π(R^2-r^2) dx
Sketch the region. It is a curvy triangular area, which is to be revolved around the line y = -1, so
v = ∫[1,e] π(R^2-r^2) dx
where R=2 and r=y+1=ln(x)+1
v = ∫[1,e] π(4-(lnx)^2) dx
You can also consider the volume to be a set of nested cylinders (shells). If you take a sheet of material of length c and width h, and thickness dy, its volume is ch dy. Now, roll it up so that its length is the circumference (c=2πr), and you have a cylinder of volume
v = 2πrh dy
Now you can see that the volume of the rotated region is
v = ∫[0,1] 2πrh dy
where r=1+y and h=x-1 = e^y - 1
v = ∫[0,1] 2π(y+1)(e^y-1) dy
To set up the integral for the volume when the region bounded by the curves y = ln(x), y = 1, and x = 1 is revolved around the line y = -1, we can use the method of cylindrical shells instead of washers.
The method of cylindrical shells involves integrating along the axis of rotation, using cylindrical shells to represent infinitesimally thin slices of the solid.
First, let's identify the limits of integration. The region is bounded by y = ln(x), y = 1, and x = 1. To find the limits of integration for x, we set the two curves equal to each other and solve for x:
ln(x) = 1
Using exponential notation, we have:
x = e^1
x = e
So, the limits of integration for x are from x = 1 to x = e.
Next, let's express the volume element, which is the volume of an individual cylindrical shell, in terms of x. The volume of a cylindrical shell is given by:
dV = 2πrh * dx
where r represents the distance from the axis of rotation to the shell, h represents the height of the shell (which is the difference between the upper and lower bounds for y), and dx represents the infinitesimally small thickness of the shell along the x-axis.
In this case, the radius of the cylindrical shell, r, is the distance from the line y = -1 to the curve y = ln(x). So, r = ln(x) - (-1) = ln(x) + 1.
The height, h, is the difference between the upper and lower bounds of y, which is 1 - ln(x).
Plugging these values into the volume element, we get:
dV = 2π(ln(x) + 1)(1 - ln(x)) * dx
To find the total volume of the solid, we integrate this expression with respect to x, from x = 1 to x = e:
V = ∫[1 to e] 2π(ln(x) + 1)(1 - ln(x)) * dx
This integral represents the set-up for finding the volume of the solid formed by revolving the region bounded by y = ln(x), y = 1, and x = 1 around the line y = -1.
To find the volume of the solid formed by revolving the region bounded by the curves y = Ln(x), y = 1, and x = 1 around the line y = -1, we can use the method of cylindrical shells. This involves setting up an integral that represents the sum of infinitesimally thin cylindrical shells that make up the solid.
First, we need to identify the limits of integration. The region bounded by the curves lies between the x-values 0 (where y = Ln(x)) and 1 (the vertical line). Therefore, the limits of integration for x will be [0, 1].
Now, let's consider an infinitesimally thin cylindrical shell at a certain x-value within the integration limits. The radius of the shell will be the distance between the curve y = Ln(x) and the line y = -1. This distance can be calculated by subtracting the y-values of the two curves: Ln(x) - (-1) = Ln(x) + 1.
The height of the shell will be the difference between the upper and lower boundaries of the solid, which is given as 1 - Ln(x).
The thickness of the shell (Δx) represents an infinitesimally small change in x. This will allow us to sum up all the infinitesimally thin shells to determine the total volume.
Thus, the integral representing the volume of the solid formed by revolving the region is:
∫[0, 1] (2π * r * h) dx
where:
- r represents the radius of the shell, which is Ln(x) + 1
- h represents the height of the shell, which is 1 - Ln(x)
- dx represents an infinitesimally small change in x
Therefore, the integral becomes:
∫[0, 1] (2π * (Ln(x) + 1) * (1 - Ln(x))) dx