Question

volume of the solid under the curve y = 6e^(-x^2)

Answer 1

\[6\int\limits_{}^{}e ^{-x ^{2}}dx\]

Answer 2

need to find where y=0

Answer 3

I think I can use + and - 3

Answer 4

how is there a "solid" under a curve?

Answer 5

find the area, then multiply pi(f(x)^2

Answer 6

is this a solid of revolution? revolved around what axis?

Answer 7

volume of the solid generated by revolving the region under the curve y = 6e-x2 in the first quadrant
about the y-axis.

Answer 8

so it has to go from 0 to around 3.