Calculus of a Single Variable, Chapter 7, 7.3, Section 7.3, Problem 13

Using the shell method we can find the volume of the solid generated by the given curves,
y = e^(-x^2/2)/sqrt(2pi), y = 0 , x = 0 , x = 1

Using the shell method the volume is given as
V= 2*pi int _a^b p(x) h(x) dx
where p(x) is the function of the average radius =x
and
h(x) is the function of height = e^(-x^2/2)/sqrt(2pi)
and the range of x is given as 0 to 1
So the volume is = 2*pi int _a^b p(x) h(x) dx
= 2*pi int _0^1 (x) (e^(-x^2/2)/sqrt(2pi)) dx
=(2*pi)/(sqrt(2pi)) int _0^1 (x*e^(-x^2/2)) dx
=(2*pi)/(sqrt(2pi)) int _0^1 (x*e^(-x^2/2)) dx
let us first solve
int (x*e^(-x^2/2)) dx
let u = x^2/2
du = 2x/2 dx = xdx
so ,
int (x*e^(-x^2/2)) dx
= int (e^(-u)) du
= -e^(-u) = -e^(-x^2/2)

So, V=(2*pi)/(sqrt(2pi)) int _0^1 (x*e^(-x^2/2)) dx
=(2*pi)/(sqrt(2pi)) [-e^(-x^2/2)]_0^1
=(2*pi)/(sqrt(2pi)) [[-e^(-(1)^2/2)]-[-e^(-0^2/2)]]
=(2*pi)/(sqrt(2pi)) [[-e^(-1/2)]-[-e^(0)]]
=(sqrt(2pi)) [1-[e^(-1/2)]]
= 0.986
is the volume