APC.5.2.01 AP Volume by Rotation

[karush]

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

Ok I looked at some of the template equations but came up with this.

 

[MarkFL]

I think you'd have the correct answer if the integrand is \(e^{-x}\), but as given the volume is unbounded.

 

[skeeter]

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

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

 

[MarkFL]

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.

 

[karush]

why is it negative?

 

[topsquark]

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

 

[skeeter]

why is it negative?

compare the graphs in quadrant I ...

 

[karush]

ok
glad you caught it

 

[skeeter]

ok
glad you caught it

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