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?

1 Answer

#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)#