Question

How do you find the volume of the solid generated by revolving the region bounded by the graphs `y=e^-x, y=0, x=0, x=1`, about the x axis?

Answer 1

`V=(pi(e^2-1))/(2e^2)`

Explanation:

At any distance x, they coordinate happens to be the radius of the revolved element
`y=e^-x`
radius is
`r=e^-x`
Circular area generated by revolving around x axis is
`A=pir^2`
volume of the elementary solid of thickness dx is
`dV=Adx`
The lower limit is given to be x=0
The upper limit is given to be x=1

`dV=Adx`

`dV=pir^2dx`
`r^2=(e^-x)^2`
`r^2=e^-(2x)`

`dV=pie^-(2x)dx`
Integrating between x=0 and x=1

`int_0^1dV=int_0^1pie^-(2x)dx`

`V=-pi/2|e^(-2x)|_0^1`

`V=-pi/2(e^(-2xx1)-e^(-2xx0))`
`V=-pi/2(e^(-2)-1)`

`V=-pi/2(1/e^2-1)`

`V=pi/2(1-1/e^2)`

`V=pi/2(e^2-1)/e^2`

`V=(pi(e^2-1))/(2e^2)`