Computing Surface Area with Surface Integrals Examples 1

Recall from the Computing Surface Area with Surface Integrals page that if a surface $\delta$ is given parametrically as $\vec{r}(u, v) = (x(u, v), y(u, v), z(u, v))$ for $(u, v) \in D$ then the surface area of $\delta$ is given by:

(1)

\begin{align} \quad \mathrm{Surface \: Area \: of \: \delta} = \iint_D \sqrt{ \left ( \frac{\partial (y, z)}{\partial (u, v)} \right )^2 + \left ( \frac{\partial (z, x)}{\partial (u, v)} \right )^2 + \left ( \frac{\partial (x, y)}{\partial (u, v)} \right )^2 } \: du \: dv \end{align}

We will now look at some more examples of computing the surface area of a surface.

Example 1

Set up an integral to find the surface area of the surface obtained by taking the function $f(x) = \sin x$ on the interval $[0, \pi]$ and rotating it $360^{\circ}$ about the $x$-axis.

We can easily parameterize this surface as follows. Let $x(x, \theta) = x$, $y(x, \theta) = f(x) \cos \theta = \sin x \cos \theta$ and $z(x, \theta) = g(x) \sin \theta = \sin x \sin \theta$. Then our surface can be parameterized as:

(2)

\begin{align} \quad \vec{r}(x, \theta) = \left\{\begin{matrix} x(x, \theta) = x \\ y(x, \theta) = \sin x \cos \theta \\ z(x, \theta) = \sin x \sin \theta \end{matrix}\right. \quad 0 ≤ x ≤ \pi, 0 ≤ \theta ≤ 2\pi \end{align}

Now let's calculate the necessary Jacobian determinants.

(3)

\begin{align} \quad \frac{\partial (y, z)}{\partial (x, \theta)} = \begin{vmatrix} \frac{\partial y}{\partial x} & \frac{\partial y}{\partial \theta} \\ \frac{\partial z}{\partial x} & \frac{\partial z}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos x \cos \theta & - \sin x \sin \theta \\ \cos x \sin \theta & \sin x \cos \theta \end{vmatrix} = \cos x \sin x \cos^2 \theta + \sin x \cos x \sin^2 \theta = \sin x \cos x \end{align}

(4)

\begin{align} \quad \frac{\partial (z, x)}{\partial (x, \theta)} = \begin{vmatrix} \frac{\partial z}{\partial x} & \frac{\partial z}{\partial \theta} \\ \frac{\partial x}{\partial x} & \frac{\partial x}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos x \sin \theta & \sin x \cos \theta \\ 1 & 0 \\ \end{vmatrix} = - \sin x \cos \theta \end{align}

(5)

\begin{align} \quad \frac{\partial (x, y)}{\partial (x, \theta)} = \begin{vmatrix} \frac{\partial x}{\partial x} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial x} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} 1 & 0 \\ \cos x \cos \theta & -\sin x \sin \theta \end{vmatrix} = -\sin x \sin \theta \end{align}

Therefore the surface area of our surface is:

(6)

\begin{align} \quad \mathrm{Surface \: Area} = \iint_D \sqrt{(\sin x \cos x)^2 + (- \sin x \cos \theta)^2 + (-\sin x \sin \theta)^2} \: dS \\ \quad \mathrm{Surface \: Area} = \iint_D \sqrt{\sin^2 x \cos^2 x + \sin^2 x \cos^2 \theta + \sin^2 x \sin^2 \theta} \: dS \\ \quad \mathrm{Surface \: Area} = \iint_D \sqrt{\sin^2 x \cos^2 x + \sin^2 x} \: dS \\ \quad \mathrm{Surface \: Area} = \iint_D \sqrt{\sin^2 x(\cos^2 x + 1)} \: dS \\ \quad \mathrm{Surface \: Area} = \iint_D \sin x \sqrt{\cos^2 x + 1} \: dS \\ \quad \mathrm{Surface \: Area} = \int_0^{2\pi} \int_0^{\pi} \sin x \sqrt{\cos^2 x + 1} \: dx \: d \theta \\ \end{align}

The integrals above can be evaluated, however, the process is cumbersome and requires a few substitutions.

The graph of this surface is lemon-shaped and is shown below: