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.
Why did the region bounded by the curves decide to have a revolving party around the line y = -3? Because it wanted to let loose and break free from the constraints of the x-axis, of course!
To set up the integral, we can use the method of cylindrical shells. Think of the region as a bunch of stacked cylindrical shells, each with a thickness dx and a height y. The radius of each shell will be the distance from the line y = -3 to the curve y = Ln(x). So, the volume of each shell can be approximated as 2π(y+3)dx.
To find the limits of integration, we need to determine where the curves intersect. At x = 1, both y = Ln(x) and y = 1 intersect. So, our limits of integration will be from x = 1 to the x-coordinate where the two curves intersect.
Putting it all together, the integral to give the volume is:
V = ∫[1, x-coordinate of intersection] 2π(y+3)dx.
Now, let the mathematical revelry begin!
To set up the integral for finding the volume, we can use the method of cylindrical shells.
Let's first look at the region bounded by the curves y = Ln(x), y = 1, and x = 1.
To revolve this region around the line y = -3, we can imagine stacking infinitely thin cylindrical shells on top of each other, forming a solid shape. The radius of each shell will be the distance between the point on the curve and the line y = -3.
Let's denote the radius of each shell as r(x), where x is the variable. The height of each shell will be given by the difference between the upper and lower curves, which is (Ln(x) - 1).
To calculate the volume of each shell, we use the formula for the volume of a cylindrical shell:
V = 2πrhΔx
where V is the volume of each shell, r(x) is the radius, h is the height, and Δx is an infinitely small width element along the x-axis.
Integrating this formula over the interval from x = 1 to x = b (where b is the x-coordinate of the point where the two curves intersect) will give us the total volume of the solid shape.
So the integral to set up the volume is:
∫[1, b] 2π(r(x))(Ln(x) - 1) dx
Note that we don't provide the limits for the integral as we don't have the value for b. This integral will give us the setup for finding the volume, but to evaluate it, we need to find the value of b and plug it in.
To find the volume of the solid when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = -3, we can use the method of cylindrical shells.
Here is how to set up the integral:
1. Identify the limits of integration:
Since we are revolving the region around the line y = -3, the lower limit of integration will be -3. The upper limit will be the point where the two curves intersect, which is also the x-coordinate of the point (1, 1). Therefore, the limits of integration for x will be from 1 to e (since Ln(x) is the inverse function of e^x).
2. Determine the height of each cylindrical shell:
The height of each cylindrical shell will be the difference between the two curves at a particular x-value. In this case, it will be Ln(x) - 1.
3. Find the radius of each cylindrical shell:
The radius of each cylindrical shell will be the distance from the axis of rotation (y = -3) to the curve y = Ln(x). Since we are revolving around the line y = -3, the radius is given by the distance between -3 and the curve y = Ln(x), which is Ln(x) - (-3) = Ln(x) + 3.
4. Set up the integral:
The volume of the solid can be calculated using the integral:
V = ∫[1 to e] 2π(Radius)(Height) dx
Substituting in the values we found earlier, the integral becomes:
V = ∫[1 to e] 2π(Ln(x) + 3)(Ln(x) - 1) dx
This integral represents the volume of the solid when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = -3.
The region is a triangular patch with vertices at (1,0), (1,1), and (1,e)
So the volume is a stack of washers of thickness dx:
v = ∫[1,e] π(R^2-r^2) dx
where R=(1+3) and r=(lnx+3)
v = ∫[1,e] π(4^2-(lnx+3)^2) dx
Or, you can set it up as a stack of nested shells of thickness dy
v = ∫[0,1] 2πrh dy
where r=y+3 and h=x-1=e^y-1
v = ∫[0,1] 2π(y+3)(e^y-1) dy