Calculating Volume of Rotation w/ Shell Method: Integral Difference

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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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