The discussion revolves around calculating the volume of a solid generated by revolving the region bounded by the parabola y=x^2 and the line y=1 about the line y=1, utilizing the disk method.
Participants are actively questioning the assumptions made regarding the radii and integration limits. Some have expressed confusion about the application of the disk method versus the washer method, indicating a productive exploration of the concepts involved.
There is mention of different methods (disk and washer) and how they apply to the problem, with some participants reflecting on their prior knowledge and understanding of these methods.
jsun2015 said:Homework Statement
Find the volume of the solid generated by revolving the region bounded by the parabola y=x^2 and the line y=1 about the line y=1
Homework Equations
V= integral of pi*r^2 from a to b with respect to variable "x"
The Attempt at a Solution
pi(integral of 1-(x^2-1)^2 from 0 to 1 dx)
but The answer is 15pi/16
Dick said:Why do you think the 'r^2' part in your volume equation is 1-(x^2-1)^2 and why do you think the limits to the integration are 0 to 1?
jsun2015 said:r^2
1 represents outer radius (the larger radius)
x^2-1 represents the inner radius (smaller radius)
limits to integration
the radius of the sphere is on the region x= 0 to 1
Dick said:That's confusing me. The formula you gave was actually for the method of disks, which is what I would use here. If you are rotating the region between y=x^2 and y=1 around y=1, then the inner radius is 0, isn't it? And I don't see why you are putting one of the limits to 0. y=x^2 and y=1 cross at x=1 and x=(-1), yes?
jsun2015 said:Yes. Yes. I came up with my original answer because at the outer region 1= radius so we have one and at the inner region 0 = radius as (1^2-1)^2=0
I have only studied the washer method.
Dick said:The disk method is the same as the washer method with an inner radius of 0. Isn't the outer radius ALWAYS 1-x^2? And the inner radius 0?