Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.5 Area Of A Surface Of Revolution - Exercises Set 5.5 - Page 380: 6

$$S = \frac{{\left( {10\sqrt {10} - 1} \right)\pi }}{{27}}$$

$$\eqalign{ & x = {y^3},\,\,\,\,0 \leqslant y \leqslant 1 \cr & {\text{Calculate }}\frac{{dx}}{{dy}} \cr & \frac{{dx}}{{dy}} = 3{y^2} \cr & {\text{Use }}S = \int_a^b {2\pi x\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} } dy \cr & S = \int_0^1 {2\pi \left( {{y^3}} \right)\sqrt {1 + {{\left( {3{y^2}} \right)}^2}} } dy \cr & S = 2\pi \int_0^1 {{y^3}\sqrt {1 + 9{y^4}} } dy \cr & S = \frac{{2\pi }}{{36}}\int_0^1 {\left( {36{y^3}} \right)\sqrt {1 + 9{y^4}} } dy \cr & {\text{Integrate}} \cr & S = \frac{\pi }{{18}}\left[ {\frac{{2{{\left( {1 + 9{y^4}} \right)}^{3/2}}}}{3}} \right]_0^1 \cr & S = \frac{\pi }{{27}}\left[ {{{\left( {1 + 9{{\left( 1 \right)}^4}} \right)}^{3/2}} - {{\left( {1 + 9{{\left( 0 \right)}^4}} \right)}^{3/2}}} \right] \cr & S = \frac{\pi }{{27}}\left[ {{{\left( {10} \right)}^{3/2}} - {{\left( 1 \right)}^{3/2}}} \right] \cr & S = \frac{{\left( {10\sqrt {10} - 1} \right)\pi }}{{27}} \cr} $$