Answer
$$S = \frac{\pi }{{27}}\left( {145\sqrt {145} - 10\sqrt {10} } \right)$$
Work Step by Step
$$\eqalign{
& y = \root 3 \of x + 2,{\text{ }}1 \leqslant x \leqslant 8 \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\root 3 \of x + 2} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{3}{x^{ - 2/3}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{3{x^{2/3}}}} \cr
& {\text{Formula for the surface area revolved around the }}y{\text{ axis}}{\text{.}} \cr
& S = 2\pi \int_a^b {r\left( x \right)\sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} } dx,{\text{ }}r\left( x \right) = x \cr
& {\text{Then,}} \cr
& S = 2\pi \int_1^8 {x\sqrt {1 + {{\left[ {\frac{1}{{3{x^{2/3}}}}} \right]}^2}} } dx \cr
& S = 2\pi \int_1^8 {x\sqrt {1 + \frac{1}{{9{x^{4/3}}}}} } dx \cr
& S = 2\pi \int_1^8 {x\sqrt {\frac{{9{x^{4/3}} + 1}}{{9{x^{4/3}}}}} } dx \cr
& S = 2\pi \int_1^8 {\frac{x}{{3{x^{2/3}}}}\sqrt {9{x^{4/3}} + 1} } dx \cr
& S = \frac{{2\pi }}{3}\int_1^8 {{x^{1/3}}\sqrt {9{x^{4/3}} + 1} } dx \cr
& {\text{Integrate}} \cr
& u = 9{x^{4/3}} + 1,{\text{ }}du = 12{x^{1/3}}dx \cr
& {\text{The new limits of integration are:}} \cr
& x = 1 \to u = 9{\left( 1 \right)^{4/3}} + 1 = 10 \cr
& x = 8 \to u = 9{\left( 8 \right)^{4/3}} + 1 = 145 \cr
& S = \frac{{2\pi }}{3}\int_{10}^{145} {\sqrt u } \left( {\frac{1}{{12}}} \right)du \cr
& S = \frac{\pi }{{18}}\int_{10}^{145} {\sqrt u } du \cr
& S = \frac{\pi }{{18}}\left[ {\frac{{{u^{3/2}}}}{{3/2}}} \right]_{10}^{145} \cr
& S = \frac{\pi }{{27}}\left[ {{u^{3/2}}} \right]_{10}^{145} \cr
& S = \frac{\pi }{{27}}\left[ {{{\left( {145} \right)}^{3/2}} - {{\left( {10} \right)}^{3/2}}} \right] \cr
& S = \frac{\pi }{{27}}\left( {145\sqrt {145} - 10\sqrt {10} } \right) \approx 199.4804 \cr} $$
