Answer
$$S = 24\sqrt {10} \pi $$
Work Step by Step
$$\eqalign{
& y = 12 - 3x{\text{ on }}\left[ {1,3} \right] \cr
& {\text{use the Definition of Area of a Surface of Revolution }}\left( {{\text{see page 454}}} \right) \cr
& {\text{The area of the surface generated when the graph of }}f{\text{ on the interval }}\left[ {a,b} \right] \cr
& {\text{is revolved about the }}x - {\text{axis}} \cr
& S = \int_a^b {2\pi f\left( x \right)\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr
& {\text{then}} \cr
& y = f\left( x \right) = 12 - 3x{\text{ and }}\left[ {1,3} \right] \to a = 1{\text{ and }}b = 3.{\text{ then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {12 - 3x} \right] \cr
& f'\left( x \right) = - 3 \cr
& {\text{substituting the values into the formula }}S = \int_a^b {2\pi f\left( x \right)\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr
& S = \int_1^3 {2\pi \left( {12 - 3x} \right)\sqrt {1 + {{\left( { - 3} \right)}^2}} } dx \cr
& S = \int_1^3 {2\pi \left( {12 - 3x} \right)\sqrt {10} } dx \cr
& {\text{take out the constants}} \cr
& S = 2\pi \sqrt {10} \int_1^3 {\left( {12 - 3x} \right)} dx \cr
& {\text{integrate and evaluate}} \cr
& S = 2\pi \sqrt {10} \left( {12x - \frac{{3{x^2}}}{2}} \right)_1^3 \cr
& S = 2\pi \sqrt {10} \left[ {\left( {12\left( 3 \right) - \frac{{3{{\left( 3 \right)}^2}}}{2}} \right) - \left( {12\left( 1 \right) - \frac{{3{{\left( 1 \right)}^2}}}{2}} \right)} \right] \cr
& {\text{simplifying}} \cr
& S = 2\pi \sqrt {10} \left( {\frac{{45}}{2} - \frac{{21}}{2}} \right) \cr
& S = 2\pi \sqrt {10} \left( {12} \right) \cr
& S = 24\sqrt {10} \pi \cr} $$
