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Section 9.5 : Surface Area with Parametric Equations
1. Determine 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 \(x = 3 + 2t\hspace{0.25in}y = 9 - 3t\hspace{0.25in}1 \le t \le 4\) about the \(y\)-axis.
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Start SolutionThe 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 2We’ll need the \(ds\) for this problem.
\[ds = \sqrt {{{\left[ 2 \right]}^2} + {{\left[ { - 3} \right]}^2}} \,dt = \sqrt {13} \,dt\] Show Step 3The integral for the surface area is then,
\[SA = \int_{{}}^{{}}{{2\pi x\,ds}} = \int_{1}^{4}{{2\pi \left( {3 + 2t} \right)\sqrt {13} \,dt}} = 2\pi \sqrt {13} \int_{1}^{4}{{3 + 2t\,dt}}\]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 4This is a really simple integral…
\[SA = 2\pi \sqrt {13} \int_{1}^{4}{{3 + 2t\,dt}} = \left. {2\pi \sqrt {13} \left( {3t + {t^2}} \right)} \right|_1^4 = \require{bbox} \bbox[2pt,border:1px solid black]{{48\pi \sqrt {13} }}\]