# [SOLVED]5.2.006 AP Calculus Exam Volume by Rotation

#### karush

##### Well-known member

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

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

#### MarkFL

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

#### skeeter

##### Well-known member
MHB Math Helper
if the curve is $y=e^{-x}$ ...

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

#### MarkFL

Staff member
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

##### Well-known member
why is it negative?

#### topsquark

##### Well-known member
MHB Math Helper
why is it negative?
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.

-Dan

MHB Math Helper

ok