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### Section 3-5 : Surface Area with Parametric Equations

In this final section of looking at calculus applications with parametric equations we will take a look at determining the surface area of a region obtained by rotating a parametric curve about the \(x\) or \(y\)-axis.

We will rotate the parametric curve given by,

\[x = f\left( t \right)\hspace{0.25in}\hspace{0.25in}y = g\left( t \right)\hspace{0.25in}\hspace{0.25in}\alpha \le t \le \beta \]about the \(x\) or \(y\)-axis. We are going to assume that the curve is traced out exactly once as \(t\) increases from \(a\) to \(b\). At this point there actually isn’t all that much to do. We know that the surface area can be found by using one of the following two formulas depending on the axis of rotation (recall the Surface Area section of the Applications of Integrals chapter).

\[\begin{align*}S & = \int{{2\pi y\,ds}}\hspace{0.25in}\hspace{0.25in}{\mbox{rotation about }}x - {\mbox{axis}}\\ S & = \int{{2\pi x\,ds}}\hspace{0.25in}\hspace{0.25in}{\mbox{rotation about }}y - {\mbox{axis}}\end{align*}\]All that we need is a formula for \(ds\) to use and from the previous section we have,

\[ds = \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} \,\,dt\hspace{0.25in}{\mbox{if }}x = f\left( t \right),y = g\left( t \right),\,\,\alpha \le t \le \beta \]which is exactly what we need.

We will need to be careful with the \(x\) or \(y\) that is in the original surface area formula. Back when we first looked at surface area we saw that sometimes we had to substitute for the variable in the integral and at other times we didn’t. This was dependent upon the \(ds\) that we used. In this case however, we will always have to substitute for the variable. The \(ds\) that we use for parametric equations introduces a \(dt\) into the integral and that means that everything needs to be in terms of \(t\). Therefore, we will need to substitute the appropriate parametric equation for \(x\) or \(y\) depending on the axis of rotation.

Let’s take a quick look at an example.

We’ll first need the derivatives of the parametric equations.

\[\frac{{dx}}{{dt}} = - 3{\cos ^2}\theta \sin \theta \hspace{0.25in}\hspace{0.25in}\frac{{dy}}{{dt}} = 3{\sin ^2}\theta \cos \theta \]Before plugging into the surface area formula let’s get the \(ds\) out of the way.

\[\begin{align*}ds & = \sqrt {9{{\cos }^4}\theta {{\sin }^2}\theta + 9{{\sin }^4}\theta {{\cos }^2}\theta } \,d\theta \\ & = 3\left| {\cos \theta \sin \theta } \right|\sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta } \\ & = 3\cos \theta \sin \theta \end{align*}\]Notice that we could drop the absolute value bars since both sine and cosine are positive in this range of \(q\) given.

Now let’s get the surface area and don’t forget to also plug in for the \(y\).

\[\begin{align*}S & = \int{{2\pi y\,ds}}\\ & = 2\pi \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{{{\sin }^3}\theta \left( {3\cos \theta \sin \theta } \right)\,\,d\theta }}\\ & = 6\pi \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{{{\sin }^4}\theta \cos \theta \,\,d\theta }}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}u = \sin \theta \\ & = 6\pi \int_{{\,0}}^{{\,1}}{{{u^4}\,du}}\\ & = \frac{{6\pi }}{5}\end{align*}\]