The region bounded by $\displaystyle y = e^{-x^{2}}, y = 0, x = 1, $ and is revolved about the y-axis. Find the volume of the resulting solid.
im not quite sure how to do this one
I think you want unbounded area.
Consinder a long interval $\displaystyle [0,N]$.
Then the volume is given by,
$\displaystyle 2\pi \int_0^N xe^{-x^2} dx$
To find the unbounded area (which is actually bounded).
Find,
$\displaystyle 2\pi \int_0^{\infty} xe^{-x^2} dx$
Hint: Use $\displaystyle t=-x^2$.
that's what i thought you meant to say, thought it probably would be cool to do it TPH's way if you didn't make a typo
We proceed by the Method of Cylindrical Shells:
$\displaystyle V = 2 \pi \int_{0}^{1} xe^{-x^2} dx$
Let $\displaystyle u = -x^2$
$\displaystyle \Rightarrow du = -2x dx$
$\displaystyle \Rightarrow - \frac {1}{2} du = x dx$
So our integral becomes:
$\displaystyle V = - 2 \pi \cdot \frac {1}{2} \int_{x=0}^{x=1} e^u du$
$\displaystyle \Rightarrow V = - \pi \left[ e^u \right]_{x=0}^{x=1} = - \pi \left[ e^{-x^2} \right]_{0}^{1}$
$\displaystyle \Rightarrow V = \pi - \frac { \pi}{e}$