# [SOLVED]5.2.006 AP Calculus Exam Volume by Rotation

#### MarkFL

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

• anemone and topsquark

#### skeeter

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

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

• topsquark and MarkFL

#### 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 and topsquark

#### 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

• skeeter and MarkFL

#### skeeter

##### Well-known member
MHB Math Helper
• karush and topsquark

#### karush

##### Well-known member
ok
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