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[SOLVED] 5.2.006 AP Calculus Exam Volume by Rotation

karush

Well-known member
Jan 31, 2012
2,678
006.png

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

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,774
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
Mar 1, 2012
655
if the curve is $y=e^{-x}$ ...

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,774
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:
 

karush

Well-known member
Jan 31, 2012
2,678
why is it negative?
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,121
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

Well-known member
MHB Math Helper
Mar 1, 2012
655

karush

Well-known member
Jan 31, 2012
2,678
ok
glad you caught it
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
655