Question

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?

Answer 1

\[y=e^{-x^2}\]?

Answer 2

yes

Answer 3

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?

Answer 4

ya thats the equations we were given :/

Answer 5

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)

\[\large V = \pi \int\limits_{0}^{e^{-4}} 4 dy + \pi \int\limits_{e^{-4}}^{1}-\ln (y) dy\]