# Thread: volume of a solid

1. ## volume of a solid

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

im not quite sure how to do this one

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

im not quite sure how to do this one
are you sure those are the right limits? as it is, the area is not bounded. maybe it should be $x=0$ and $x = 1$ and the $x$-axis?

3. I think you want unbounded area.
Consinder a long interval $[0,N]$.

Then the volume is given by,
$2\pi \int_0^N xe^{-x^2} dx$

To find the unbounded area (which is actually bounded).

Find,
$2\pi \int_0^{\infty} xe^{-x^2} dx$

Hint: Use $t=-x^2$.

4. sorry i seem to forgot something, the correct question is:

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

5. Originally Posted by viet
sorry i seem to forgot something, the correct question is:

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

$V = 2 \pi \int_{0}^{1} xe^{-x^2} dx$

Let $u = -x^2$

$\Rightarrow du = -2x dx$

$\Rightarrow - \frac {1}{2} du = x dx$

So our integral becomes:

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

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

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