If we consider a small strip width #dx#, it will have radius #y(x)# as it is revolved about the x axis, and thus circuference #2 pi y#.

The arc length #ds# of the tip of strip #dx# is:

#ds = sqrt ( 1 + (y')^2) dx#

With # y' = x(1 + 2 ln x)##

and so the surface area of the element is

#dS = 2 pi y ds #

#= 2 pi x^2 ln x sqrt ( 1 + ( x(1 + 2 ln x))^2) dx#

For #x in [1,3]#, the surface area #S# is therefore:

#S = 2 pi int_1^3 \ x^2 ln x sqrt ( 1 + ( x(1 + 2 ln x))^2) dx#

However because #y < 0# for #x in [1,3]#, which would generate a negative radius, we need to be sure to place a negative number on the integration.

The surface area in total is therefore

#S = 2 pi ( int_1^3 \ x^2 ln x sqrt ( 1 + ( x(1 + 2 ln x))^2) dx - int_0^1 \ x^2 ln x sqrt ( 1 + ( x(1 + 2 ln x))^2) dx )#