APC.5.2.01 AP Volume by Rotation

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    Ap Rotation Volume
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Discussion Overview

The discussion revolves around the calculation of volume generated by rotating a curve, specifically focusing on the integrand used in the volume formula. Participants explore the implications of using different exponential functions in the integral.

Discussion Character

  • Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant presents an integral involving \(e^x\) but is challenged on its validity due to the unbounded nature of the volume.
  • Another participant suggests that using \(e^{-x}\) would yield a correct answer, leading to a bounded volume calculation.
  • Multiple participants calculate the volume using \(e^{-2x}\) and arrive at the same result of \(\dfrac{\pi}{2}\), indicating a consensus on this specific calculation.
  • There is a discussion about a potential typo in the problem statement, with some participants questioning why the function is negative and suggesting that \(e^{-2x}\) is the appropriate function to use.
  • Participants express acknowledgment of each other's contributions, particularly in identifying the correct function to use in the volume calculation.

Areas of Agreement / Disagreement

There is general agreement on the use of \(e^{-x}\) and \(e^{-2x}\) for bounded volume calculations, but disagreement remains regarding the initial function presented and its implications for the volume being unbounded.

Contextual Notes

Participants note the importance of the function's behavior on the interval \([0, \infty)\) and the implications of squaring the radius in the volume formula. There is an acknowledgment of missing assumptions regarding the problem statement.

karush
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$\displaystyle\pi\int_{0}^{\infty} e^x \ dx = \pi$

Ok I looked at some of the template equations but came up with this.
 
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I think you'd have the correct answer if the integrand is \(e^{-x}\), but as given the volume is unbounded.
 
if the curve is $y=e^{-x}$ ...

$\displaystyle V = \pi \int_0^\infty e^{-2x} \, dx = \dfrac{\pi}{2}$
 
skeeter said:
if the curve is $y=e^{-x}$ ...

$\displaystyle V = \pi \int_0^\infty e^{-2x} \, dx = \dfrac{\pi}{2}$

Yes, I overlooked the squaring of the radius. :oops:
 
why is it negative?
 
karush said:
why is it negative?
It's a guess that the problem has a typo. The point is that [math]e^{-2x}[/math] is bounded on [math] [ 0, \infty )[/math] whereas [math]e^{2x}[/math] is not.

-Dan
 
karush said:
why is it negative?

compare the graphs in quadrant I ...

393FE634-9153-4CF7-981C-5CBA3AEDF622.png
393FE634-9153-4CF7-981C-5CBA3AEDF622.png
 
ok
glad you caught it
 
karush said:
ok
glad you caught it

I didn’t catch the incorrect function ... mark did. see post #2.
 

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