Volumes of Revolution with e^-x

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Homework Help Overview

The problem involves computing the volume of a solid of revolution formed by rotating the region in the first quadrant bounded by the curve y=e^(-x), the line x=0, and the x-axis around the y-axis using the method of cylindrical shells.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the integral for volume, with some questioning whether to integrate with respect to y or x. There is a focus on understanding the correct application of the shell method in this context.

Discussion Status

There is an ongoing exploration of the correct approach to setting up the integral, with some participants confirming the original poster's method while others clarify potential misunderstandings regarding the integration variable. Guidance has been provided regarding the correct interpretation of the problem.

Contextual Notes

Participants note the importance of visualizing the solid of rotation and suggest including sketches to aid in understanding the setup. There is mention of discrepancies in answers obtained from different methods or tools, highlighting the need for careful consideration of the integration process.

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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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