Calculus of a Single Variable, Chapter 7, 7.4, Section 7.4, Problem 39

Area of the surface obtained by revolving the curve y=f(x) about x-axis between a leq x leq b is given by
S_x=2pi int_a^b y sqrt(1+y'^2)dx
Let us therefore, first find y'.
y'=x^2/2-1/(2x^2)=(x^4-1)/(2x^2)
y'^2=(x^8-2x^4+1)/(4x^4)
We can now calculate the surface area.
S_x=2pi int_1^2 (x^3/6+1/(2x))sqrt(1+(x^8-2x^4+1)/(4x^4))dx=
2pi int_1^2(x^3/6+1/(2x))sqrt((x^8+2x^4+1)/(4x^4))=
2pi int_1^2(x^3/6+1/(2x))sqrt(((x^4+1)/(2x^2))^2)dx=
2pi int_1^2(x^3/6+1/(2x))(x^4+1)/(2x^2)dx=
Multiplying the terms under integral yields
2pi int_1^2 (x^5/12+x/12+x/4+1/(4x^3))dx=
2pi int_1^2(x^5/12+x/3+1/(4x^3))dx=
2pi (x^6/72+x^2/6-1/(8x^2))|_1^2=
2pi(64/72+2/3-1/32-1/72-1/6+1/8)=2pi cdot 47/32=(47pi)/16
The area of surface generated by revolving the given curve about x-axis is (47pi)/16.
Graphs of the curve and the surface can be seen in the images below.