Surface Area Generated by a Parametric Curve

Recall that when we revolve a function [latex]y = f(x)[/latex] around the [latex]x[/latex]-axis from [latex]x = a[/latex] to [latex]x = b[/latex], the surface area is:

Now let’s find the surface area when we revolve a parametric curve around the [latex]x[/latex]-axis, as shown in the following figure.

For a parametric curve [latex]x = x(t)[/latex], [latex]y = y(t)[/latex] where [latex]a \le t \le b[/latex], the analogous formula is:

This formula requires that [latex]y(t) \ge 0[/latex] on [latex][a,b][/latex]—the curve must lie on or above the [latex]x[/latex]-axis. Notice that the expression [latex]\sqrt{[x'(t)]^2 + [y'(t)]^2}[/latex] is the arc length element we just learned, so we’re essentially multiplying the circumference [latex]2\pi y(t)[/latex] by the arc length differential.

Find the surface area of a sphere of radius [latex]r[/latex] centered at the origin.

Show Solution

We start with the curve defined by the equations

[latex]x\left(t\right)=r\cos{t},y\left(t\right)=r\sin{t},0\le t\le \pi[/latex].

 

This generates an upper semicircle of radius r centered at the origin as shown in the following graph.

When this curve is revolved around the [latex]x[/latex]-axis, it generates a sphere of radius [latex]r[/latex]. To calculate the surface area of the sphere, we use the above equation:

[latex]\begin{array}{cc}\hfill S&=2\pi {\displaystyle\int_{a}^{b}} y\left(t\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime}\left(t\right)\right)}^{2}}dt\hfill \\ &=2\pi {\displaystyle\int_{0}^{\pi}} r\sin{t}\sqrt{{\left(-r\sin{t}\right)}^{2}+{\left(r\cos{t}\right)}^{2}}dt\hfill \\ &=2\pi {\displaystyle\int }_{0}^{\pi }r\sin{t}\sqrt{{r}^{2}{\sin}^{2}t+{r}^{2}{\cos}^{2}t}dt\hfill \\ &=2\pi {\displaystyle\int }_{0}^{\pi }r\sin{t}\sqrt{{r}^{2}\left({\sin}^{2}t+{\cos}^{2}t\right)}dt\hfill \\ &=2\pi {\displaystyle\int_{0}^{\pi}{r}^{2}\sin{t}dt}\hfill \\ &=2\pi{r}^{2}\left(-\cos{t}|_{0}^{\pi}\right)\hfill \\ & =2\pi {r}^{2}\left(-\cos\pi +\cos0\right)\hfill \\ & =4\pi {r}^{2}.\hfill \end{array}[/latex]

 

This is, in fact, the formula for the surface area of a sphere.