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 = −3.
The volume is just a stack of discs, each of radius y and thickness x, so
v = ∫[a,b] πr^2 dx
y(1) = 0
y(e) = 1
so,
v = ∫[1,e] πr^2 dx
so, what's r? If we were revolving around y=0, r would just be y. But the axis is 3 units farther away than that, and the area between y=0 and y=-3 is empty, so we really have washers with holes in them, so
v = ∫[1,e] π(R^2-r^2) dx
where R=y+3 and r=3.
v = ∫[1,e] π((ln(x)+3)^2-3^2) dx
= π∫[1,e] ln(x)^2 + 6 ln(x) dx
To set up the integral that gives the volume of the solid formed by revolving the region bounded by the given curves, you can use the method of cylindrical shells.
First, let's find the range of x-values that encloses the region. The curves y = Ln(x) and y = 1 intersect when Ln(x) = 1. Taking the exponential of both sides, we have x = e^1 = e. So, the region is bounded by x = 1 and x = e.
Next, consider an infinitesimally thin vertical strip of width ∆x within the region. This strip can be located at any x-value between 1 and e. Revolving this strip around the line y = -3 will create a cylindrical shell.
The height of the cylindrical shell is given by the difference between the equations of the curves that bound the region. In this case, the height h is (1 - Ln(x)).
The circumference (or perimeter) of the cylindrical shell is given by the formula for the circumference of a circle with a radius equal to the distance between the x-value and the line y = -3. The distance is (-3 - Ln(x)). So, the circumference C is 2π(-3 - Ln(x)).
The thickness or width of the cylindrical shell is just ∆x.
Setting up the integral, we integrate the product of the height, circumference, and thickness of the shell, summed up for all infinitesimally thin shells:
∫[from 1 to e] 2π(-3 - Ln(x))(1 - Ln(x)) ∆x
Now, this integral gives the setup for finding the volume of the solid generated by revolving the region bounded by the curves around the line y = -3. To evaluate the integral, you can calculate the antiderivative of the integrand and use the limits of integration, but that is not required according to the given instructions.