Volumes of Revolution with e^-x

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SUMMARY

The discussion focuses on calculating the volume of the solid formed by rotating the region R, defined by the curve y=e^(-x), the line x=0, and the x-axis, around the y-axis using the shell method. The correct volume formula is established as V = 2π∫[0,∞] x e^(-x) dx, which evaluates to 2π. Confusion arose when comparing results from a calculator, which incorrectly suggested a volume of π/2, but this was clarified as being accurate only for rotation around the x-axis. The final consensus confirms that the volume of the solid of revolution is indeed 2π.

PREREQUISITES
  • Understanding of the shell method for volume calculation
  • Familiarity with the exponential function e^(-x)
  • Knowledge of integral calculus, specifically integration by parts
  • Ability to interpret and sketch graphs of functions and solids of revolution
NEXT STEPS
  • Study the shell method for calculating volumes of solids of revolution
  • Practice integration techniques, particularly integration by parts
  • Explore the differences between rotating around the x-axis and y-axis
  • Learn to use computational tools like Wolfram Alpha for verifying integral calculations
USEFUL FOR

Students studying calculus, particularly those focusing on volumes of revolution, as well as educators teaching integral calculus concepts.

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Homework Statement



Compute the region R in the first quadrant between y=e^(-x), x=0, and y=0. Compute using shells, the volume V of solid around the y-axis.

Homework Equations


Volume =integral of bounds 2pi*radius*height

The Attempt at a Solution



First I drew the graph. This graph really is just a graph of e^(-x).
I then visually rotated it around the y-axis.
This problem seems easy enough to set up:

Volume = Integral between 0 (lower limit) and infinity (upper limit) of 2*pi*x*e^-(x) dx
where x = radius
dx = width
e^(-x) = height

This problem was also easy to integrate using integral substitution first, and then integration by parts one time. The final equation was

V = 2pi[-xe^(-x)-e^(-x)] from 0 to infinity.

After calculating the simple answer is 2pi.My problem: For some reason, when I use this calculator: http://www.wolframalpha.com/widgets/view.jsp?id=1cd73be1e256a7405516501e94e892ac

I get an answer of pi/2.

Am I doing something wrong? Or is my answer/thinking correct?

Thank you.
 
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To do this by the method of shells you want to integrate dy, not dx
 
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Oh, even if it is around the y-axis? Hm. Let me try this out again. Thank you.
 
No, I think Dick misspoke there. Rotating about the y-axis you do want dx elements for shells. I also get ##2\pi##.
 
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LCKurtz said:
No, I think Dick misspoke there. Rotating about the y-axis you do want dx elements for shells. I also get ##2\pi##.

Yes, I did misspeak and 2pi is correct. pi/2 is correct if you are rotating around the x-axis.
 
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Ah, magnificent! So my answer is then verified. This makes me overjoyed.

I tried it the other way and it didn't work out too well, was getting -infinity :p.

Thank you, friends, for assisting me tonight.
 
RJLiberator said:
First I drew the graph. This graph really is just a graph of e^(-x).
I then visually rotated it around the y-axis.
It's not clear from what you wrote, but you should also draw a sketch of the solid of rotation. In your first graph you should include an incremental area element that will be rotated. In your second graph, you should include a sketch of the shell or disk or whatever. If you do that, you'll have a better chance of getting the integrand right, which in this case it seems that you did.
 
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