# rotation to form a solid... cant figure out error

• January 25th 2007, 06:33 PM
thedoge
rotation to form a solid... cant figure out error
The region bounded by y=e^{-x^2}, y=0, x=0, and x=1 is revolved about the y-axis. Find the volume of the resulting solid.

so it should be the integral (0,1) of e^(-x^2). u=-x^2, du=-2xdx, coefficient=-1/2

therefore, (-1/2) int(0,1) e^u*du = (-1/2) e^u evaluated at |(0,1)

so, ((-e^-1)/2) - ((-e^0)/2)= -.1839397206 - (-.5) = .3160602794

multiply this by pi to get my answer, which was .9929326519, but incorrect.

what am i doing wrong?
• January 25th 2007, 07:02 PM
ThePerfectHacker
Quote:

Originally Posted by thedoge
The region bounded by y=e^{-x^2}, y=0, x=0, and x=1 is revolved about the y-axis. Find the volume of the resulting solid.

The formula is,
$2\pi \int_0^1 xe^{-x^2} dx$
To find,
$\int xe^{-x^2} dx=-\frac{1}{2}\int e^{-x^2}(-2x) dx$
Let,
$u=-x^2$ then $u'=-2x$,
Thus,
$-\frac{1}{2}\int e^{u} du=-\frac{1}{2}e^u+C=-\frac{1}{2}e^{-x^2}+C$
Evaluate,
$2\pi\left(-\frac{1}{2}e^{-1^2}+\frac{1}{2}e^0\right)$
• January 25th 2007, 07:05 PM
thedoge
Hrm, so my error was using 1*pi instead of 2*pi? Thanks again PH, you never cease to amaze me.