The discussion revolves around using the shell method to find the volume of a solid generated by revolving regions bounded by the curves \( Y = |x|/3 \) and \( Y = 1 \) about the x-axis. Participants are exploring the correct setup for the integral and addressing potential errors in their calculations.
There are multiple interpretations of the problem setup, with some participants suggesting corrections to the original attempts. Guidance has been offered regarding the need for accurate sketches of the region and the solid, as well as clarifications on the correct expressions for the radius and height of the shells. The discussion is ongoing, with no explicit consensus reached yet.
Participants are working under the constraints of homework rules, which may limit the information they can share. There is uncertainty regarding the correct limits of integration and the setup of the integral, leading to varied results in their calculations.
The above is incorrect. The upper limit of integration is not 2, and your formula for the area of the typical area element is also wrong. How did you get y/4?whatlifeforme said:Homework Statement
Use the shell method to find the volume of the solid generated by revolving the regions bounded by the curves and the lines about the x-axis.
Homework Equations
Y=|x|/3 , Y=1
The Attempt at a Solution
1. 2∏∫(0 to 2) (y) (Y/4 - (-Y/4)) dy
Using the correct setup, I get a volume of 4π as well.whatlifeforme said:2. y^3/6] (0 to 2)
Answer: 8∏/3 (which is wrong). the correct answer is 4∏ but I'm not sure how they go there.
Better, but still not right. If y = (1/3)|x|, then |x| = 3y, not y/3.whatlifeforme said:i also tried integrating just the right side and multiplying by two.
2* 2∏∫(0 to 1) (y) (Y/3) dy
No. It's 2x, which is not equal to 2y/3.whatlifeforme said:which i get the answer of 4pi/9.
1.if i draw a section parallel to the axis of revolutoin. the distance (radius) is y. thus the first part of the equation is right.
2.the height of the section would be the distance from (-y/3) on the left to (y/3).
You will if you use the correct value for the width of your shell.whatlifeforme said:however, if i take just the right section of the absolute value function (y/3), and integrate, and multiply by two (2). then i should have the right value.