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Section 9.5 : Surface Area with Parametric Equations

4. Set up, but do not evaluate, an integral that gives the surface area of the object obtained by rotating the parametric curve about the given axis. You may assume that the curve traces out exactly once for the given range of \(t\)’s.

Rotate \(x = 1 + \ln \left( {5 + {t^2}} \right)\hspace{0.25in}y = 2t - 2{t^2}\hspace{0.25in}0 \le t \le 2\) about the \(x\)-axis.

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Start Solution

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.

\[ds = \sqrt {{{\left[ {\frac{{2t}}{{5 + {t^2}}}} \right]}^2} + {{\left[ {2 - 4t} \right]}^2}} \,dt = \sqrt {\frac{{4{t^2}}}{{{{\left( {5 + {t^2}} \right)}^2}}} + {{\left( {2 - 4t} \right)}^2}} \,dt\] Show Step 3

The integral for the surface area is then,

\[SA = \int_{{}}^{{}}{{2\pi y\,ds}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\int_{0}^{2}{{2\pi \left( {2t - 2{t^2}} \right)\sqrt {\frac{{4{t^2}}}{{{{\left( {5 + {t^2}} \right)}^2}}} + {{\left( {2 - 4t} \right)}^2}} \,dt}}}}\]

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