Surface Area

Suppose that $z = f(x,y)$ is a function of the class $C^{1}$ on the closed bounded region $\Omega$. Then

$$S = \{(x,y,f(x,y) : (x,y) \in \Omega \}$$

is called smooth surface. The area of $S$ is given by

$$S = \iint_{\Omega}\sqrt{f_{x}^{2} + f_{y}^{2} + 1} dxdy.$$

where $\Omega$ is the projection of $S$ onto $xy$-plane.

NOTE

Figure 5.10: Surface Area

The vector $\boldsymbol{N}$ orthogonal to the small rectangle on the surface is given by $(f_{x},f_y,-1)$. Let $\theta$ be the angle between the vector $(f_{x},f_y,-1)$ and the vector $\boldsymbol{e}_1$ which is orthogonal to $xy$-plane. Then $\Delta S \cos{\theta}$ is equal to the $\Delta \Omega$, a small area of $xy$-plane.Thus, $\Delta S = \Delta \Omega \sec{\theta}.$ Note that

$$\boldsymbol{N}\cdot \boldsymbol{e}_1 = (f_{x},f_y,-1) \cdot (0,0,1) = \sqrt{f_x^2 + f_y^2 + 1}\cos{\theta}$$

Thus we can express the surface area $S$ as the following double integral.

$$S = \iint_{\Omega}\sqrt{f_{x}^{2} + f_{y}^{2} + 1} dxdy.$$

Example 5..7   Find the surface area of the following surface.

$$z^{2} = x^{2} + y^{2} {\rm with} 0 \leq x^{2} + y^{2} \leq 1, z \geq 0$$

SOLUTION To find the surface area, we need to find the $\Omega$ which is a projection of the surface $z = f(x,y)$. Since $z \geq 0$, the surface is $z = \sqrt{x^2 + y^2}$. Now the region $\Omega$ is given by $${\Omega = \{(x,y) : x^2 + y^2 \leq 1\}}$$. Thus, the surface area is given by the following double integral

$$S = \iint_{\Omega}\sqrt{z_{x}^2 + z_{y}^2 + 1} dx dy.$$

Since

$$z_{x} = \frac{2x}{2\sqrt{x^2 + y^2}} = \frac{x}{\sqrt{x^2 + y^2}}, z_{y} = \frac{2y}{2\sqrt{x^2 + y^2}} = \frac{y}{\sqrt{x^2 +y^2}}$$

,

$$z_{x}^2 + z_{y}^2 + 1 = \frac{x^2 + y^2 }{x^2 + y^2} + 1= 2.$$

Thus,

$$S = \iint_{\Omega}\sqrt{2} dx dy$$

Note that $\Omega$ is a circular region of the radius 1. Thus by using the polar coordinate,

$$\Gamma = \{(r,\theta) : 0 \leq \theta \leq 2\pi, 0 \leq r \leq 1\}$$

Thus the surface area is

$$S = \sqrt{2} \iint_{\Gamma}rdrd\theta = \sqrt{2} \int_{0}^{2\pi}\int_{0}^{1}r dr d\theta$$

$$= \sqrt{2} \int_{0}^{2\pi}\left[\frac{r^2}{2}\right]_{0}^1 d\theta ... ...{2}}{2}\left[\theta\right]_{0}^{2\pi} = \sqrt{2}\pi \ensuremath{ \blacksquare}$$

Exercise 5..7   Find the surface area of sphere $${x^2 + y^2 + z^2 = 1}$$ cut by cylinder $${x^2 + y^2 = x}$$.

SOLUTION Note that $x^2 + y^2 + z^2 = 1$ implies $z = \pm \sqrt{1 - x^2 - y^2}$. Then find the following surface area and double it

$$z = f(x,y) = \sqrt{1-x^2 - y^2}, \Omega = \{(x,y) : x^2 + y^2 \leq x \}$$

$$S = 2\iint_{\Omega}\sqrt{f_{x}^{2} + f_{y}^{2} + 1}dxdy$$

$$= 2\iint_{\Omega}\sqrt{(\frac{x^2}{1-x^2-y^2} + \frac{y^2}{1-x^2-y^2} + 1)}dxdy$$

$$= 2\iint_{\Omega}\frac{1}{\sqrt{1-x^2-y^2}}dxdy$$

Now using polar coordinate, $\Omega$ is mapped into

$$\Gamma = \{(r,\theta) : -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}, 0 \leq r \leq \cos{\theta} \}$$

Since $\vert J(r,\theta)\vert = r$,

$$S = 2\iint_{\Gamma}\frac{1}{\sqrt{1-r^2}}rdrd\theta = 2\int_{-\frac{\... ...frac{\pi}{2}}\big( \int_{0}^{\cos{\theta}}\frac{r}{\sqrt{1-r^2}}dr \big)d\theta$$

Let $t = 1-r^2$ Then $dt = -2rdr$, $$\begin{array}{l\vert lll} r & 0 & \to & \cos{\theta}\ \hline t & 1 & \to & 1 - \cos^{2}{\theta} = \sin^{2}{\theta} \end{array}$$. Thus,

$$S = 2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\big( \int_{1}^{\sin^{2}{\theta}}\frac{1}{-2\sqrt{t}}dt\big)d\theta$$

$$= 2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\big(\int_{1}^{\sin^{2}{\th... ...\pi}{2}}^{\frac{\pi}{2}} \big[-t^{\frac{1}{2}}\big]_1^{\sin^{2}{\theta}}d\theta$$

$$= 2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \big(-\sqrt{1-\cos^2{\thet... ...eta = 2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(1 - \vert\sin{\theta}\vert)d\theta$$

$$= 4\int_{0}^{\frac{\pi}{2}}(1 - \sin{\theta})d\theta = 4(\frac{\pi}{2} - 1) = 2\pi - 4 \ensuremath{ \blacksquare}$$