Answer
$$S = 12\sqrt 5 \pi $$
Work Step by Step
$$\eqalign{
& y = \frac{x}{2} + 3,{\text{ }}1 \leqslant x \leqslant 5 \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{x}{2} + 3} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{2} \cr
& {\text{Then,}} \cr
& S = 2\pi \int_a^b {r\left( x \right)\sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} } dx,{\text{ }}r\left( x \right) = x \cr
& {\text{Then,}} \cr
& S = 2\pi \int_1^5 {x\sqrt {1 + {{\left( {\frac{1}{2}} \right)}^2}} } dx \cr
& S = 2\pi \int_1^5 {x\sqrt {\frac{5}{4}} } dx \cr
& S = \sqrt 5 \pi \int_1^5 x dx \cr
& {\text{Integrate}} \cr
& S = \sqrt 5 \pi \left[ {\frac{{{x^2}}}{2}} \right]_1^5 \cr
& S = \frac{{\sqrt 5 \pi }}{2}\left( {25 - 1} \right) \cr
& S = 12\sqrt 5 \pi \cr} $$
