The first thing we’ll need here are the following two derivatives.
\[\frac{{dx}}{{dt}} = \frac{{2t}}{{5 + {t^2}}}\hspace{0.25in}\hspace{0.25in}\frac{{dy}}{{dt}} = 2 - 4t\] Show Step 2
We’ll need the \(ds\) for this problem.
\[ds = \sqrt {{{\left[ {\frac{{2t}}{{5 + {t^2}}}} \right]}^2} + {{\left[ {2 - 4t} \right]}^2}} \space dt = \sqrt {\frac{{4{t^2}}}{{{{\left( {5 + {t^2}} \right)}^2}}} + {{\left( {2 - 4t} \right)}^2}} \space dt\] Show Step 3
The integral for the surface area is then,
\[SA = \int_{{}}^{{}}{{2\pi y\space ds}} = {{\int_{0}^{2}{{2\pi \left( {2t - 2{t^2}} \right)\sqrt {\frac{{4{t^2}}}{{{{\left( {5 + {t^2}} \right)}^2}}} + {{\left( {2 - 4t} \right)}^2}} \space dt}}}}\]
Remember to be careful with the formula for the surface area! The formula used is dependent upon the axis we are rotating about.