2011-02-06, 06:19 PM
Use your own damn thread, Corn.
I still don't understand this. e^-x^2 only ever hits 0 at inf, so I don't know what you're trying to point out.
Okay, this is more helpful. But it makes me wonder. When the function is transformed to polar coordinates, it becomes this:
![[Image: msp191419e73hgg4di681gd.gif]](http://img222.imageshack.us/img222/4396/msp191419e73hgg4di681gd.gif)
r is just the function's position from zero. The volume of a cylinder is 2pi r h. Therefore, I could find the volume under the curve from r1 to r2, and since the volume of a cylinder is 2pi r h, I could divide to get the area underneath the curve?
I'm talking about applying this. How would I apply this to get the area under e^-x^2?
Noah Wrote:I meant that
and that
![[Image: 4nwchks.png]](http://mathurl.com/4nwchks.png)
![[Image: 5totwuj.png]](http://mathurl.com/5totwuj.png)
I still don't understand this. e^-x^2 only ever hits 0 at inf, so I don't know what you're trying to point out.
Shidoshi Wrote:The way you do it, you pass from a section (or the whole) x axis to a symetrical thing in 2D, which will most likely be in the form of a rectangle, which is hard to describe in polar.
Okay, this is more helpful. But it makes me wonder. When the function is transformed to polar coordinates, it becomes this:
r is just the function's position from zero. The volume of a cylinder is 2pi r h. Therefore, I could find the volume under the curve from r1 to r2, and since the volume of a cylinder is 2pi r h, I could divide to get the area underneath the curve?
I'm talking about applying this. How would I apply this to get the area under e^-x^2?