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Section 9.5 : Surface Area with Parametric Equations
For all the problems in this section you should only use the given parametric equations to determine the answer.
For problems 1 – 3 determine the surface area of the object obtained by rotating the parametric curve about the given axis. For these problems 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. Solution
- Rotate \(x = 9 + 2{t^2}\hspace{0.25in}y = 4t\hspace{0.25in}0 \le t \le 2\) about the \(x\)-axis. Solution
- Rotate \(\displaystyle x = 3\cos \left( {\pi t} \right)\hspace{0.25in}y = 5t + 2\hspace{0.25in}0 \le t \le \frac{1}{2}\) about the \(y\)-axis. Solution
For problems 4 and 5 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. For these problems 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. Solution
- Rotate \(\displaystyle x = 1 + 3{t^2}\hspace{0.25in}y = \sin \left( {2t} \right)\cos \left( {\frac{1}{4}t} \right)\hspace{0.25in}0 \le t \le \frac{1}{2}\) about the \(y\)-axis. Solution