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- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,678

- Thread starter karush
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- #1

- Jan 31, 2012

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- #2

- Mar 1, 2012

- 655

if the curve is $y=e^{-x}$ ...

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

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

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- #4

Yes, I overlooked the squaring of the radius.if the curve is $y=e^{-x}$ ...

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

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- Jan 31, 2012

- 2,678

why is it negative?

- Aug 30, 2012

- 1,121

It's a guess that the problem has a typo. The point is that \(\displaystyle e^{-2x}\) is bounded on \(\displaystyle [ 0, \infty )\) whereas \(\displaystyle e^{2x}\) is not.why is it negative?

-Dan

- Mar 1, 2012

- 655

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- #8

- Jan 31, 2012

- 2,678

ok

glad you caught it

glad you caught it

- Mar 1, 2012

- 655

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

glad you caught it