What is the surface area of the solid created by revolving #f(x) = e^(x)/2 , x in [2,7]# around the x axis?

1 Answer
Nov 3, 2016

The area is #=pi/8((4(ln(e^(14)+4)+e^7)+e^7sqrt(e^(14)+4)))-pi/8(4(ln(e^(4)+4)+e^2)+e^2sqrt(e^(4)+4)))#

Explanation:

The volume of a small disc is #dS=piydr#
and #dr=sqrt((dx)^2+(dy)^2)#
#dS=piysqrt((dx)^2+(dy)^2)=piysqrt(1+(dy/dx)^2)dx#
So #y=e^x/2##=>##dy/dx=e^x/2#
so #dS=pie^x/2*sqrt(1+e^(2x)/4)dx#
#dS=pie^x/4*sqrt(4+e^(2x))dx#
so #S=pi/4inte^xsqrt(4+e^(2x))dx#
let #u=e^x# #=>##du=e^xdx#
#S=pi/4*intsqrt(4+u^2)du#
#u=2tanv##=>##du= sec^2u (dv)#
and #4tan^2v+4=4sec^2v#
#S=4pi/4intsec^3vdv=pi(1/2intsecvdv+secvtanv/2)#
#pi((ln(tanv+secv)/2)+secvtanv/2)#
#S=pi/8(4(ln(e^(2x)+4)+e^x)+e^xsqrt(e^(2x)+4)))_2^7#
#=pi/8((4(ln(e^(14)+4)+e^7)+e^7sqrt(e^(14)+4)))-pi/8(4(ln(e^(4)+4)+e^2)+e^2sqrt(e^(4)+4)))#