The discussion revolves around finding the volume of a solid of revolution formed by rotating the area between the curve y=2-(1/2)x, the x-axis, and the vertical lines x=1 and x=2 about the x-axis. Participants are exploring the application of the disk method for this problem.
The discussion is active, with participants questioning each other's calculations and clarifying the integrand and bounds. Some participants have revised their answers based on feedback, indicating a productive exchange of ideas, though no consensus on the final answer has been reached.
There are indications of confusion regarding the integrand and the squaring of the radius, which may affect the final volume calculation. Participants are also reflecting on their initial assumptions and calculations.
Joe_K said:For an answer, I came up with 5pi/4. Does this seem correct? Thanks!
tylerc1991 said:I am getting a different answer. What are your bounds? What is the integrand?
Joe_K said:My bounds were from 1 to 2.
What do you mean by "with a height of 1" ?Joe_K said:Homework Statement
Find the volume obtained by rotating the solid about the specified line.
y=2-(1/2)x, y=0, x=1, x=2, about the x-axis.
Homework Equations
I used the disk method
The Attempt at a Solution
I drew a sketch and used disk method. For the radius I used 2-(1/2)x with a height of 1, and integrated. For an answer, I came up with 5pi/4. Does this seem correct? Thanks!
SammyS said:What do you mean by "with a height of 1" ?
What function did you integrate?
Joe_K said:For the integrand I just had pi*2-(1/2)x*1 dx
tylerc1991 said:The area of one of the disks is going to be \pi \cdot (2 - \frac{1}{2} x)^2. What happens when we sum those disks from x = 1 to x = 2?
Joe_K said:I think I forgot to square the radius when I originally did it. I redid the problem and came up with 19pi/12. Is this still wrong?