The first thing we’ll need here are the following two derivatives.

\[\frac{{dx}}{{dt}} = 2\hspace{0.25in}\hspace{0.25in}\frac{{dy}}{{dt}} = - 3\] Show Step 2 

We’ll need the \(ds\) for this problem.

\[ds = \sqrt {{{\left[ 2 \right]}^2} + {{\left[ { - 3} \right]}^2}} \space dt = \sqrt {13} \space dt\] Show Step 3 

The integral for the surface area is then,

\[SA = \int_{{}}^{{}}{{2\pi x\space ds}} = \int_{1}^{4}{{2\pi \left( {3 + 2t} \right)\sqrt {13} \space dt}} = 2\pi \sqrt {13} \int_{1}^{4}{{3 + 2t\space dt}}\]

Remember to be careful with the formula for the surface area! The formula used is dependent upon the axis we are rotating about.

Show Step 4 

This is a really simple integral…

\[SA = 2\pi \sqrt {13} \int_{1}^{4}{{3 + 2t\space dt}} = \left. {2\pi \sqrt {13} \left( {3t + {t^2}} \right)} \right|_1^4 = {{48\pi \sqrt {13} }}\]