Answer
$$S = \frac{\pi }{{27}}\left( {{{\left( {10} \right)}^{3/2}} - 1} \right)$$
Work Step by Step
$$\eqalign{
& y = {x^3}{\text{ on }}\left[ {0,1} \right] \cr
& {\text{use the Definition of Area of a Surface of Revolution }}\left( {{\text{see page 454}}} \right) \cr
& {\text{The area of the surface generated when the graph of }}f{\text{ on the interval }}\left[ {a,b} \right] \cr
& {\text{is revolved about the }}x - {\text{axis}} \cr
& S = \int_a^b {2\pi f\left( x \right)\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr
& {\text{then}} \cr
& y = f\left( x \right) = {x^3}{\text{ and }}\left[ {0,1} \right] \to a = 0{\text{ and }}b = 1.{\text{ then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^3}} \right] \cr
& f'\left( x \right) = 3{x^2} \cr
& {\text{substituting the values into the formula }}S = \int_a^b {2\pi f\left( x \right)\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr
& S = \int_0^1 {2\pi \left( {{x^3}} \right)\sqrt {1 + {{\left( {3{x^2}} \right)}^2}} } dx \cr
& S = 2\pi \int_0^1 {{x^3}\sqrt {1 + 9{x^4}} } dx \cr
& {\text{multiply and divide by a constant}} \cr
& S = \frac{{2\pi }}{{36}}\int_0^1 {36{x^3}\sqrt {1 + 9{x^4}} } dx \cr
& S = \frac{\pi }{{18}}\int_0^1 {{{\left( {1 + 9{x^4}} \right)}^{1/2}}} \left( {36{x^3}} \right)dx \cr
& {\text{integrate by using the power rule}} \cr
& S = \frac{\pi }{{18}}\left( {\frac{{{{\left( {1 + 9{x^4}} \right)}^{3/2}}}}{{3/2}}} \right)_0^1 \cr
& S = \frac{{2\pi }}{{54}}\left( {{{\left( {1 + 9{x^4}} \right)}^{3/2}}} \right)_0^1 \cr
& S = \frac{\pi }{{27}}\left( {{{\left( {1 + 9{{\left( 1 \right)}^4}} \right)}^{3/2}} - {{\left( {1 + 9{{\left( 0 \right)}^4}} \right)}^{3/2}}} \right) \cr
& {\text{simplifying}} \cr
& S = \frac{\pi }{{27}}\left( {{{\left( {10} \right)}^{3/2}} - {{\left( 1 \right)}^{3/2}}} \right) \cr
& S = \frac{\pi }{{27}}\left( {{{\left( {10} \right)}^{3/2}} - 1} \right) \cr} $$
