Calculus II - Surface Area with Parametric Equations

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Section 9.5 : Surface Area with Parametric Equations

2. 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 \(x = 9 + 2{t^2}\hspace{0.25in}y = 4t\hspace{0.25in}0 \le t \le 2\) about the \(x\)-axis.

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

\[\frac{{dx}}{{dt}} = 4t\hspace{0.25in}\hspace{0.25in}\frac{{dy}}{{dt}} = 4\] Show Step 2

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

\[ds = \sqrt {{{\left[ {4t} \right]}^2} + {{\left[ 4 \right]}^2}} \,dt = \sqrt {16{t^2} + 16} \,dt = 4\sqrt {{t^2} + 1} \,dt\]

Note that we factored a 16 out of the root to make the rest of the work a little simpler to deal with.

Show Step 3

The integral for the surface area is then,

\[SA = \int_{{}}^{{}}{{2\pi y\,ds}} = \int_{0}^{2}{{2\pi \left( {4t} \right)\left( {4\sqrt {{t^2} + 1} } \right)\,dt}} = 32\pi \int_{0}^{2}{{t\sqrt {{t^2} + 1} \,dt}}\]

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

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This is a simple integral to compute with a quick substitution. Here is the integral work,

\[SA = 32\pi \int_{0}^{2}{{t\sqrt {{t^2} + 1} \,dt}} = \left. {\frac{{32}}{3}\pi {{\left( {{t^2} + 1} \right)}^{\frac{3}{2}}}} \right|_0^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{32}}{3}\pi \left( {{5^{\,\frac{3}{2}}} - 1} \right) = 341.1464}}\]