See the portion of the curve between #x=1# and #x=3#. Let us divide the interval #[1,3}# into say #n# equal sub-intervals each of of width #Deltax#, where #Deltax=(3-1)/n# and as #n->oo#, #Deltax->dx#.

graph{ln(4-x) [-0.99, 4.01, -0.52, 1.98]}

On each of these sub-intervals we can revolve the curve around #x#-axis to form a thin cylinder, whose widt is #dx# and radius is #r=f(x)#.

Then the surface area of this thin cylinder will be #2pirl#, where #l=dx# and #r=ln(4-x)# and adding them up will give surface area of the solid created by revolving the function #ln(4-x)# and in place of adding we can integrate it to get the surface area, which will be

#2piint_1^3ln(4-x)dx#

and as #intln(4-x)dx=-(4-x)ln(4-x)+x#

and #2piint_1^3ln(4-x)dx#

= #2pi[-(4-x)ln(4-x)+x]_1^3#

= #2pi[-ln1+3+3ln3-1]#

= #2pi(2+3ln3)#