APC.5.2.01 AP Volume by Rotation

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    Ap Rotation Volume
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SUMMARY

The discussion centers on the calculation of the volume generated by rotating the curve defined by the function \(y=e^{-x}\) around the x-axis. The correct volume is determined using the formula \(V = \pi \int_0^\infty e^{-2x} \, dx\), resulting in \(V = \frac{\pi}{2}\). Participants noted a potential typo in the original problem statement, as the integrand \(e^{2x}\) leads to an unbounded volume, while \(e^{-2x}\) is bounded on the interval [0, ∞).

PREREQUISITES
  • Understanding of integral calculus, specifically improper integrals.
  • Familiarity with volume of revolution concepts in calculus.
  • Knowledge of exponential functions and their properties.
  • Ability to recognize and correct typographical errors in mathematical expressions.
NEXT STEPS
  • Study the method of calculating volumes of solids of revolution using the disk method.
  • Learn about improper integrals and their convergence criteria.
  • Explore the properties of exponential decay functions, particularly \(e^{-x}\).
  • Review common mistakes in calculus, such as misidentifying functions in volume calculations.
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Students and educators in calculus, mathematicians focusing on volume calculations, and anyone interested in understanding the implications of function behavior in integration.

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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