# volume of a solid

• May 24th 2007, 03:47 PM
viet
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 :confused:
• May 24th 2007, 04:12 PM
Jhevon
Quote:

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 :confused:

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?
• May 24th 2007, 06:19 PM
ThePerfectHacker
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 $t=-x^2$.
• May 24th 2007, 06:28 PM
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.
• May 24th 2007, 06:31 PM
Jhevon
Quote:

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:

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

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

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