Integral Calc: Volume of Solid of Revolution

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Homework Help Overview

The discussion revolves around calculating the volume of a solid of revolution formed by rotating a region in the first quadrant, specifically bounded by the curve defined by x = y - y³, the line x = 1, and the line y = 1, about the y-axis.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to set up an integral for the volume but expresses confusion about whether it is correctly configured. Some participants question the interpretation of the region being rotated and the setup of the integral.

Discussion Status

Participants are exploring different interpretations of the problem setup, particularly regarding the boundaries of the region and the method of calculating the volume. There is acknowledgment of a misunderstanding regarding the area being considered (inside vs. outside the curve). Guidance has been offered regarding the use of the washer method and the clarification of the integral setup.

Contextual Notes

There is mention of a teacher's provided answer and a discussion about the radius and height in the context of cylindrical shells, which raises questions about the geometric interpretation of the problem.

cathy
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Homework Statement



Find the volume of the first quadrant region bounded by x=y-y3, x=1 and y=1 that is revolved about the y-axis.



2. The attempt at a solution

v=∏ ∫ from 0 to 1 of (y-y^3)^2 dy
and doing this, I got the answer to be 8∏/105.

Did I set up that integral correctly? I am confused as to if I did this right or not. Please advise. Thank you in advance.
 
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What is y3?
 
the equation is x=y-y^3
 
cathy said:

Homework Statement



Find the volume of the first quadrant region bounded by x=y-y3, x=1 and y=1 that is revolved about the y-axis.
2. The attempt at a solution

v=∏ ∫ from 0 to 1 of (y-y^3)^2 dy
and doing this, I got the answer to be 8∏/105.

Did I set up that integral correctly? I am confused as to if I did this right or not. Please advise. Thank you in advance.
No, that is not set up correctly. y- y^3 is the distance from the y-axis to x= y- y^3 but you said the region to be rotated was bounded by x= y- y^3 and x=1.

What you have will give the volume from rotating the region bounded by x= 0 and x= y- y^3, the region inside the graph while you want the region outside that graph and inside x= 1.

You could use the "washer method" but equivalent, and simpler in my opinion, is to find the volume inside the rotated graph, using the integral as you have it set up, and then subtract that from the volume of the cylinder of radius 1 and height 1 leaving only the volume outside the graph.
 
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Okay, I understand what I was doing wrong. I was looking for area inside, rather than outside and within bounds. Thank you.

But, the teacher gave us the answer as V=2∏∫ from 0 to 1 of (y)(1-y+y^3) dy.

And I have a question about this. She said that r=y and h= 1-x, but if you were to draw this as shells, the "stick" would run from the line y=1 to the line x=0. The "stick" never actually touches the y-y^3 part, so how is that included? We take a vertical stick, correct? I'm confused. :(
 
Last edited:
And just to clarify, the "r" is talking about the radius of the shell formed, correct?
And since it is a vertical cylinder, why isn't the answer in terms of x?
 
Last edited:

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