Calculating Volume of Rotation w/ Shell Method: Integral Difference

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Homework Help Overview

The discussion revolves around calculating the volume of a solid of revolution formed by rotating the area bounded by the curve y=e^x, the line x=1, and the line y=1 around the line x=4 using the shell method. Participants are comparing two different integrals that are supposed to yield the same volume but are producing different numerical results.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are examining two integrals: one derived from the shell method and another from the disk (washer) method. There is a focus on understanding why these integrals yield different results despite representing the same geometric object.

Discussion Status

Some participants have evaluated both integrals and found them to be approximately equal, while others are questioning the validity of their calculations and the interpretation of the integrals in computational tools. There is an ongoing exploration of the discrepancies in results and the proper use of computational resources.

Contextual Notes

There is mention of potential misinterpretation by computational tools, specifically regarding the notation used in the integrals. Participants are also discussing the implications of using different methods (shell vs. washer) for the same problem.

alingy1
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calculate the volume obtained when the area bounded by y=e^x, x=1, y=1, is rotated around x=4. use shell method. Can I see your integral?

My teacher gave me the first integral. I found the second one here: BUT: the two do not give the same numerical result:
integral of pi((4-lnx)^2-9) FROM 1 to e =/= integral of 2pi(4-x)(e^x-1) dx from 0 to 1
 
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alingy1 said:
calculate the volume obtained when the area bounded by y=e^x, x=1, y=1, is rotated around x=4. use shell method. Can I see your integral?

My teacher gave me the first integral. I found the second one here: BUT: the two do not give the same numerical result:
integral of pi((4-lnx)^2-9) FROM 1 to e =/= integral of 2pi(4-x)(e^x-1) dx from 0 to 1

This integral -- "integral of pi((4-lnx)^2-9) FROM 1 to e" -- is using disks (washers, actually), but the term you show as lnx should be ln(y). In a more nicely formated form, this integral is
$$ \pi \int_{y = 1}^e [(4 - ln(y))^2 - 9]dy$$
Since both integrals represent the volume of the same geometric object, they have to give the same result. Since you're not getting the same result, you must be doing something wrong. Please show us your work for the integral above.
 
Both integrals evaluate to roughly 14.91. Where do you get a different value?
 
That mean you shouldn't use Wolfram for this problem! At least not until you are sure you can use it properly. As you say, Wolfram is interpreting \pi(x) as a function "pi of x", which returns the number of primes less than or equal to x (or the largest integer less than or equal to x if x is not itself an integer) not as the number \pi times x.
 

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