Use the shell method to set up, but do not evaluate, an integral representing the volume of the solid generated by revolving the region bounded by the graphs of y=x^2 and y=4x-x^2 about the line x=6.
I had the shell radius as (6-x) and the shell height as (4x-2x^2).
My final integral was 2 pi * integral from 0 to 2 of [(6-x)(4x-2x^2)]. I was just wondering if I did this correct? Thanks.
Yes, your expressions for the radius and the height are correct. The expression to integrate as well. The shell method evaluates basically ∫2πr(x)h(x)dx. The limits have not been stated in the question, but I suppose they are from x=0 to x=2, as you put it.
Good work!
Yes, your setup using the shell method is correct.
To find the volume of the solid generated by revolving the region bounded by the graphs of y = x^2 and y = 4x - x^2 about the line x = 6, you need to use cylindrical shells.
The shell radius can be defined as the distance from the axis of revolution (x = 6) to the axis of the shell (x). In this case, the shell radius is (6 - x).
The shell height can be defined as the difference in y-values between the two curves at a given x-value. The top curve is y = 4x - x^2, and the bottom curve is y = x^2. So, the shell height is (4x - x^2) - x^2 = 4x - 2x^2.
The differential volume element in cylindrical coordinates is given by dV = 2πrh*dx, where dx is an infinitesimally small change in x.
Thus, the integral representing the volume becomes:
V = 2π * integral from 0 to 2 of [(6 - x)(4x - 2x^2)]dx
You have correctly set up the integral using the shell radius and shell height. However, I noticed that your limits of integration should be from 0 to 2, as that is the interval where the two curves intersect.
Remember that this integral is just the setup, and you would still need to evaluate it to find the actual volume.
To determine if your integral setup is correct, let's go through the process step by step.
The problem asks us to find the volume of the solid generated by revolving the region bounded by the graphs of y = x^2 and y = 4x - x^2 about the line x = 6.
To use the shell method, we need to consider disks or cylindrical shells that are oriented parallel to the axis of revolution. In this case, the axis of revolution is x = 6.
First, let's sketch the region bounded by the given curves to visualize it.
The curve y = x^2 is a parabola that opens upwards and passes through the origin. The curve y = 4x - x^2 is also a parabola, but it opens downwards and intersects the x-axis at (0, 0) and (4, 0). The region between these two curves is bounded by the x-values 0 and 4.
To set up the integral, we need to express the volume element as the product of the cylindrical shell's height, circumference, and thickness.
- Shell height: The difference in y-values between the curves y = 4x - x^2 and y = x^2 is (4x - x^2) - x^2 = 4x - 2x^2.
- Shell radius: The distance from the shell to the axis of revolution, x = 6, is 6 - x.
- Shell thickness: Since we want to think of the region as an infinite number of infinitely thin shells, the thickness will be dx.
The volume of each cylindrical shell is given by V = 2πrhdx, where r is the shell radius, h is the shell height, and dx is the thickness.
Therefore, the integral to find the volume is:
V = ∫[0 to 4] 2π(6 - x)(4x - 2x^2) dx
Looking at your work, you correctly identified the shell radius as (6 - x) and the shell height as (4x - 2x^2). So your final integral:
V = 2π ∫[0 to 4] (6 - x)(4x - 2x^2) dx
represents the volume using the shell method for the given region and axis of revolution.
Now, you can proceed to evaluate the integral to find the exact volume by integrating the function and substituting the limits of integration.