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 :confused:

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- May 24th 2007, 03:47 PMvietvolume of a solid
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 :confused: - May 24th 2007, 04:12 PMJhevon
- May 24th 2007, 06:19 PMThePerfectHacker
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 $\displaystyle t=-x^2$. - May 24th 2007, 06:28 PMviet
sorry i seem to forgot something, the correct question is:

The region bounded by $\displaystyle y = e^{-x^{2}}, y = 0, x = 0, x = 1 $ and is revolved about the y-axis. Find the volume of the resulting solid. - May 24th 2007, 06:31 PMJhevon
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:

$\displaystyle V = - 2 \pi \cdot \frac {1}{2} \int_{x=0}^{x=1} e^u du$

$\displaystyle \Rightarrow V = - \pi \left[ e^u \right]_{x=0}^{x=1} = - \pi \left[ e^{-x^2} \right]_{0}^{1}$

$\displaystyle \Rightarrow V = \pi - \frac { \pi}{e}$