Answer
$$S = \frac{{47}}{{16}}\pi $$
Work Step by Step
$$\eqalign{
& y = \frac{{{x^3}}}{6} + \frac{1}{{2x}},{\text{ }}1 \leqslant x \leqslant 2 \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{{x^3}}}{6} + \frac{1}{{2x}}} \right] \cr
& \frac{{dy}}{{dx}} = \frac{{3{x^2}}}{6} - \frac{1}{{2{x^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{{x^2}}}{2} - \frac{1}{{2{x^2}}} \cr
& {\text{Formula for surface area}} \cr
& S = 2\pi \int_a^b {r\left( x \right)\sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} } dx,{\text{ }}r\left( x \right) = f\left( x \right) \cr
& {\text{Then,}} \cr
& S = 2\pi \int_1^2 {\left( {\frac{{{x^3}}}{6} + \frac{1}{{2x}}} \right)\sqrt {1 + {{\left( {\frac{{{x^2}}}{2} - \frac{1}{{2{x^2}}}} \right)}^2}} } dx \cr
& S = 2\pi \int_1^2 {\left( {\frac{{{x^3}}}{6} + \frac{1}{{2x}}} \right)\sqrt {1 + \frac{{{x^4}}}{4} - \frac{1}{2} + \frac{1}{{4{x^4}}}} } dx \cr
& S = 2\pi \int_1^2 {\left( {\frac{{{x^3}}}{6} + \frac{1}{{2x}}} \right)\sqrt {\frac{{{x^4}}}{4} + \frac{1}{2} + \frac{1}{{4{x^4}}}} } dx \cr
& S = 2\pi \int_1^2 {\left( {\frac{{{x^3}}}{6} + \frac{1}{{2x}}} \right)\sqrt {{{\left( {\frac{{{x^2}}}{2} + \frac{1}{{2{x^2}}}} \right)}^2}} } dx \cr
& S = 2\pi \int_1^2 {\left( {\frac{{{x^3}}}{6} + \frac{1}{{2x}}} \right)\left( {\frac{{{x^2}}}{2} + \frac{1}{{2{x^2}}}} \right)} dx \cr
& {\text{Distribute and simplify}} \cr
& S = 2\pi \int_1^2 {\left( {\frac{{{x^5}}}{{12}} + \frac{x}{{12}} + \frac{x}{4} + \frac{1}{{4{x^3}}}} \right)} dx \cr
& S = 2\pi \int_1^2 {\left( {\frac{{{x^5}}}{{12}} + \frac{x}{3} + \frac{1}{{4{x^3}}}} \right)} dx \cr
& {\text{Integrate}} \cr
& S = 2\pi \left[ {\frac{{{x^6}}}{{72}} + \frac{{{x^2}}}{6} - \frac{1}{4}\left( {\frac{1}{{2{x^2}}}} \right)} \right]_1^2 \cr
& S = 2\pi \left[ {\frac{{{x^6}}}{{72}} + \frac{{{x^2}}}{6} - \frac{1}{{8{x^2}}}} \right]_1^2 \cr
& S = 2\pi \left[ {\frac{{{{\left( 2 \right)}^6}}}{{72}} + \frac{{{{\left( 2 \right)}^2}}}{6} - \frac{1}{{8{{\left( 2 \right)}^2}}}} \right] - 2\pi \left[ {\frac{{{{\left( 1 \right)}^6}}}{{72}} + \frac{{{{\left( 1 \right)}^2}}}{6} - \frac{1}{{8{{\left( 1 \right)}^2}}}} \right] \cr
& S = 2\pi \left( {\frac{{439}}{{288}}} \right) - 2\pi \left( {\frac{1}{{18}}} \right) \cr
& S = \frac{{47}}{{16}}\pi \cr} $$
