The discussion revolves around finding the volume of solids obtained by rotating regions bounded by specific curves about designated axes. The subject area includes calculus, specifically integral calculus and methods for calculating volumes of revolution.
Participants are actively engaging in verifying their integral setups and discussing the appropriate methods for calculating volumes. Some guidance has been offered regarding the use of washers instead of disks, and there is acknowledgment of the need for clarity in defining radii. Multiple interpretations of the problems are being explored, and there is a collaborative effort to confirm setups without reaching a definitive conclusion.
There is an emphasis on setting up integrals rather than evaluating them, and some participants express uncertainty about their approaches, indicating a learning process in understanding the concepts involved.
Did you draw a sketch? Actually, two sketches would be useful - one of the region defined by the two curves, and another that shows the solid of revolution.Jbreezy said:Homework Statement
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
Homework Equations
y = x^2 +1 , y = 9-x^2: about y = -1
The Attempt at a Solution
I used disks.
Jbreezy said:I said that r = 1 + ((9-x^2)-(x^2 +1 )) = 9 - 2x^2
So my integral all together is
V = ∏ ∫ (9 - 2x^2)^2 dx from [-2,2]
Does this set up look correct.?
Looks OK to me. It would have been helpful to indicate that your using shells.Jbreezy said:Also I have another one I would like to confirm.
This one your supposed to set up the integral not evaluate for the volume of a solid.
y = tanx, y = x, x = pi/3 : about the y axis.
I have
V = 2∏∫ x(tanx - x) dx between 0 and ∏/3
I think it is good but just to be safe.
Mark44 said:Did you draw a sketch? Actually, two sketches would be useful - one of the region defined by the two curves, and another that shows the solid of revolution.
Disks won't work, but washers will. A washer is a disk with a hole in its middle. For a typical volume element, the outer radius goes from the line y = -1 to the curve y = 9 - x2. The inner radius goes from the line y = -1 to the curve y = x2 + 1.
Your radii are correct but the integral isn't. The volume of a washer is ##\pi(R^2 - r^2)(\text{thickness})##.Jbreezy said:Yeah I always draw it. Sorry for the confusion with the other problem not indicating I was using shells.
So what about this. r = inner radius, R = outer radius
r = 1 + ( x^2 + 1) R = 1+ (9-x^2)
So V = ∏∫( x^2 +2)^2 + (10-x^2)^2 dx I think that is it but I have a little trouble with the radius sometimes.