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: 4

$$S = \frac{{\left( {145\sqrt {145} - 10\sqrt {10} } \right)\pi }}{{27}}$$

$$\eqalign{ & x = \root 3 \of y ,\,\,\,\,1 \leqslant y \leqslant 8 \cr & x = \root 3 \of y \Rightarrow \,\,\,\,\,\,y = {x^3},\,\,\,\,\,\,\,1 \leqslant x \leqslant 2 \cr & {\text{Calculate }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = 3{x^2} \cr & {\text{Use }}S = \int_a^b {2\pi y\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx \cr & S = \int_1^2 {2\pi \left( {{x^3}} \right)\sqrt {1 + {{\left( {3{x^2}} \right)}^2}} } dx \cr & S = 2\pi \int_1^2 {{x^3}\sqrt {1 + 9{x^4}} } dx \cr & S = \frac{{2\pi }}{{36}}\int_1^2 {\left( {36{x^3}} \right)\sqrt {1 + 9{x^4}} } dx \cr & {\text{Integrate}} \cr & S = \frac{\pi }{{18}}\left[ {\frac{{2{{\left( {1 + 9{x^4}} \right)}^{3/2}}}}{3}} \right]_1^2 \cr & S = \frac{\pi }{{27}}\left[ {{{\left( {1 + 9{x^4}} \right)}^{3/2}}} \right]_1^2 \cr & S = \frac{\pi }{{27}}\left[ {{{\left( {1 + 9{{\left( 2 \right)}^4}} \right)}^{3/2}} - {{\left( {1 + 9{{\left( 1 \right)}^4}} \right)}^{3/2}}} \right] \cr & S = \frac{\pi }{{27}}\left[ {{{\left( {145} \right)}^{3/2}} - {{\left( {10} \right)}^{3/2}}} \right] \cr & S = \frac{{\left( {145\sqrt {145} - 10\sqrt {10} } \right)\pi }}{{27}} \cr} $$