## anonymous 4 years ago what is the volume of the region bounded by y=e^-x^2, y=0, x=0, and x=2 when it is rotated around the y axis?

1. TuringTest

$y=e^{-x^2}$?

2. anonymous

yes

3. TuringTest

I don't really see how to integrate this is this how the problem was given originally? or did you get this function from somewhere?

4. anonymous

ya thats the equations we were given :/

5. anonymous

|dw:1327620023966:dw| the volume is sum of the cross-sectional areas from 0 to 1. each area is a circle with radius of x. From y=0 to e^-4, the radius is constant at 2 y = e^-x^2 lny = -x^2 -ln(y) = x^2 Area = pi*r^2 = pi*x^2 = -pi*ln(y) $\large V = \pi \int\limits_{0}^{e^{-4}} 4 dy + \pi \int\limits_{e^{-4}}^{1}-\ln (y) dy$

6. anonymous

where did you get the first integral from?

7. anonymous

$= 4\pi e^{-4} -\pi \ln(y)(y-1) {e^{-4} \to 1}$ ... from the part where radius is a constant 2 pi*2^2 = 4pi

8. anonymous

ahhh ok i see makes sense. thank you very much i was having problems with this question lol

9. TuringTest

me too, nice one dumbcow :D

10. anonymous