Volume of Revolution By Shell Method.

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Homework Help Overview

The discussion revolves around finding the volume of a solid obtained by rotating the curves y=x^2 and x=y^2 about the line y=2. The problem involves applying the shell method for volume calculation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss their attempts to determine the radius and height for the shell method, with some expressing confusion about the appropriate coordinates to use. There are mentions of constructing graphs and evaluating integrals, as well as references to the disk method as a comparison.

Discussion Status

Some participants have shared their calculations and results, indicating a level of understanding. Guidance has been offered regarding the importance of sketching the region and solid of revolution, and there is acknowledgment of the challenges faced in identifying key elements like radius and height.

Contextual Notes

Participants note the difficulty in visualizing the problem and the importance of drawing sketches to aid in understanding the setup. There is a recognition of the time spent on the problem and the iterative nature of working through it.

OmniNewton
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Homework Statement


Find the volume of the solid obtained by rotating y=x^2 and x = y^2 about y=2

Homework Equations


V= 2πrh

The Attempt at a Solution


When I had constructed the graph I determined the following:
r= 2-(y(1/2))
h= 1-y2

after converting V into an integral I applied fundamental theorem of calculus.
2-2y^2-y^(1/2) + y^(5/2)
evaluating the expression from lower limit =0 to upper limit = 1

I obtained a value with a denominator of 21.

The final answer I should receive is 31π/30 (I proved this by disk method and this is the stated answer).
 
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OmniNewton said:

Homework Statement


Find the volume of the solid obtained by rotating y=x^2 and x = y^2 about y=2

Homework Equations


V= 2πrh

The Attempt at a Solution


When I had constructed the graph I determined the following:
r= 2-(y(1/2))
h= 1-y2
I have r = 2 - y and h = ##y^{1/2} - y^2##.
OmniNewton said:
after converting V into an integral I applied fundamental theorem of calculus.
2-2y^2-y^(1/2) + y^(5/2)
evaluating the expression from lower limit =0 to upper limit = 1

I obtained a value with a denominator of 21.

The final answer I should receive is 31π/30 (I proved this by disk method and this is the stated answer).
I get 31π/30 for the volume.
 
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Mark44 said:
I have r = 2 - y and h = ##y^{1/2} - y^2##.

I get 31π/30 for the volume.

Oh ok I see how you got that. Let me give that a try.
 
Thank you so much it worked out! I was at that problem for a good two hours. I erased so much my page nearly ripped. After taking a break to work on another problem and then coming back with your guidance I was able to figure it out.
 
OmniNewton said:
Thank you so much it worked out! I was at that problem for a good two hours. I erased so much my page nearly ripped. After taking a break to work on another problem and then coming back with your guidance I was able to figure it out.
I hope you're drawing sketches of 1) the region being revolved, with the two curves and important points labeled, and 2) the solid of revolution. Based on what you showed, I think you did, but sometimes people think they can save time by skipping that step. It's almost always a false economy...
 
Mark44 said:
I hope you're drawing sketches of 1) the region being revolved, with the two curves and important points labeled, and 2) the solid of revolution. Based on what you showed, I think you did, but sometimes people think they can save time by skipping that step. It's almost always a false economy...

Yes, my steps for these problems are.

1. draw the region
2. find necessary point of intersections
3. split the axis of rotation for disk method or split the perpendicular axis of rotation for shell method
4. draw atleast one disk or shell (more may be needed)
5. then work from there.

My problem is sometimes I still have a hard time noticing key things like radius and height (getting better with practice)
 
Very good. I think the problem you had was not realizing what the appropriate x- or y-coordinates were for writing the radius and the width of the area element. You'll get better at this with practice. Other than that, you seem to be on the right track.
 
Mark44 said:
Very good. I think the problem you had was not realizing what the appropriate x- or y-coordinates were for writing the radius and the width of the area element. You'll get better at this with practice. Other than that, you seem to be on the right track.
Yes, that makes sense. I believe that is what happened
 

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