Surface Area

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

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

is called smooth surface. The area of $S$

is given by

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

where $\Omega$ is the projection of $S$

onto

-plane.

NOTE

**Figure 5.22:** Surface Area
$\includegraphics[width=6cm]{SOFTFIG-5/surfaceArea.eps}$

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

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

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

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

Dot Product
The dot product of a vector $\boldsymbol{A}=(a_1,a_2,a_3)$ and a vector $\boldsymbol{B}=(b_1,b_2,b_3)$ is given by and is written as $\boldsymbol{A} \cdot \boldsymbol{B}$ . $\boldsymbol{A}\cdot \boldsymbol{B} = \vert\boldsymbol{A}\vert\vert\boldsymbol{B}\vert\cos{\theta}$ .

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

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

$% latex2html id marker 29576 \includegraphics[width=3.5cm]{SOFTFIG-5/reidai5-7.eps}$

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 $\displaystyle{\Omega = \{(x,y) : x^2 + y^2 \leq 1\}}$ . Thus, the surface area is given by the following double integral

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

Since

$\displaystyle z_{x}$

$\displaystyle =$

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

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

Thus,

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

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

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

Thus the surface area is

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

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

Exercise5-7
$% latex2html id marker 29626 \includegraphics[width=3.5cm]{SOFTFIG-5/enshu5-7.eps}$

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

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

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

Check
implies $z = \pm \sqrt{1 - x^2 - y^2}$ . $f_x = \frac{\partial (\sqrt{1 - x^2 - y^2})}{\partial x} = \frac{-2x}{2\sqrt{1-x^2-y^2}} = -\frac{x}{\sqrt{1 - x^2 - y^2}}$ , $f_y = \frac{\partial (\sqrt{1 - x^2 - y^2})}{\partial y} = \frac{-2y}{2\sqrt{1-x^2-y^2}} = -\frac{y}{\sqrt{1 - x^2 - y^2}}$

Let $x^2 + y^2 = r^2$ , $x = r\cos{\theta}$ . Then $x^2 + y^2 \leq x$ is transformed to $0 \leq r^2 \leq r\cos{\theta}$ . Thus, $0 \leq r \leq \cos{\theta}$ . Therefore, $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$ $\sqrt{1-\cos^{2}{\theta}} = \sqrt{\sin^{2}{\theta}}$ Then for $-\frac{\pi}{2} \leq \theta \leq 0$ , $\sin{\theta} \leq 0$ . Thus $\sqrt{\sin^{2}{\theta}} = \vert\sin{\theta}\vert$ .

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

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

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

$\displaystyle S$

$\displaystyle =$

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

Let

Then

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

$\displaystyle S$	$\displaystyle =$	$\displaystyle 2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\big( \int_{1}^{\sin^{2}{\theta}}\frac{1}{-2\sqrt{t}}dt\big)d\theta$
	$\displaystyle =$	$\displaystyle 2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\big(\int_{1}^{\sin^{2}{\th... ...\pi}{2}}^{\frac{\pi}{2}} \big[-t^{\frac{1}{2}}\big]_1^{\sin^{2}{\theta}}d\theta$
	$\displaystyle =$	$\displaystyle 2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \big(-\sqrt{1-\cos^2{\thet... ...eta = 2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(1 - \vert\sin{\theta}\vert)d\theta$
	$\displaystyle =$	$\displaystyle 4\int_{0}^{\frac{\pi}{2}}(1 - \sin{\theta})d\theta = 4(\frac{\pi}{2} - 1) = 2\pi - 4 \ensuremath{ \blacksquare}$