Calculating the Shell Method for 2π∫x(9x⁴/625)dx

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Homework Help Overview

The discussion revolves around the application of the shell method for calculating volume, specifically using the integral 2π∫x(9x⁴/625)dx. Participants are examining the differences in results obtained from the shell method compared to the disk method for the same function.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand why the height in the shell method solution is expressed as 9 - (9x⁴/625) instead of simply 9x⁴/625. Some participants question the setup of the shell method in relation to the volume being calculated, particularly the distinction between the volume under the curve and the volume between two curves.

Discussion Status

Participants are exploring the reasoning behind the shell method's height calculation and discussing the implications of the volume being calculated. There is a recognition of the need to consider the difference between two heights in the shell method, and some participants express understanding of this concept.

Contextual Notes

The original poster references a specific problem from an external document, which may impose certain constraints or assumptions relevant to the discussion. The nature of the problem suggests a focus on understanding the differences between methods rather than arriving at a final solution.

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Homework Statement



http://www.catholiccentral.net/Document.Doc?id=200 problem number 5



Homework Equations


2pi intergral x(9x^4/625)


The Attempt at a Solution



When I do the shell method it doesn't give me the same answer as if I use disk. I've also noticed that when I do the shell method on the function x^4 it gives a different answer then when I use the disk method. I know what the answer is. My main question is why is there a 9-9x^4/625 in for the height in the solution instead of just a 9x^4/625.
 
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Your method would be correct if you wanted the volume underneath the rotated curve. But you don't want that. You want the volume contained between y=9 and the rotated curve y=9x^4/625. So the width of the shell is the difference of those two values.
 
Dick said:
Your method would be correct if you wanted the volume underneath the rotated curve. But you don't want that. You want the volume contained between y=9 and the rotated curve y=9x^4/625. So the width of the shell is the difference of those two values.
Oh I think I get it, so the radius would just be x and the height would be the top height 9 minus the bottom height y which is given by y=9x^4/625
 
xdrgnh said:
Oh I think I get it, so the radius would just be x and the height would be the top height 9 minus the bottom height y which is given by y=9x^4/625

Exactly.
 
Dick said:
Exactly.
Thanks a lot, have a happy new year
 

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