Answer
$$S = 156\sqrt {10} \pi $$
Work Step by Step
$$\eqalign{
& y = 3x + 4{\text{ on }}\left[ {0,6} \right] \cr
& {\text{use the Definition of Area of a Surface of Revolution }}\left( {{\text{see page 454}}} \right) \cr
& {\text{The area of the surface generated when the graph of }}f{\text{ on the interval }}\left[ {a,b} \right] \cr
& {\text{is revolved about the }}x - {\text{axis}} \cr
& S = \int_a^b {2\pi f\left( x \right)\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr
& {\text{then}} \cr
& {\text{Notice that }}y = f\left( x \right) = 3x + 4{\text{ and }}\left[ {0,6} \right] \to a = 0{\text{ and }}b = 6.{\text{ then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {3x + 4} \right] \cr
& f'\left( x \right) = 3 \cr
& {\text{substituting the values into the formula }}S = \int_a^b {2\pi f\left( x \right)\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr
& S = \int_0^6 {2\pi \left( {3x + 4} \right)\sqrt {1 + {{\left( 3 \right)}^2}} } dx \cr
& S = \int_0^6 {2\pi \left( {3x + 4} \right)\sqrt {10} } dx \cr
& {\text{take out the constant}} \cr
& S = 2\pi \sqrt {10} \int_0^6 {\left( {3x + 4} \right)} dx \cr
& {\text{integrate and evaluate}} \cr
& S = 2\pi \sqrt {10} \left( {\frac{{3{x^2}}}{2} + 4x} \right)_0^6 \cr
& S = 2\pi \sqrt {10} \left[ {\left( {\frac{{3{{\left( 6 \right)}^2}}}{2} + 4\left( 6 \right)} \right) - \left( {\frac{{3{{\left( 0 \right)}^2}}}{2} + 4\left( 0 \right)} \right)} \right] \cr
& {\text{simplifying}} \cr
& S = 2\pi \sqrt {10} \left[ {\left( {\frac{{108}}{2} + 24} \right) - \left( 0 \right)} \right] \cr
& S = 2\pi \sqrt {10} \left( {78} \right) \cr
& S = 156\sqrt {10} \pi \cr} $$
