Surface integrals

Let $f({\ rm P}) = f(x,y,z)$ be the scalar field defined for any point ${\rm P}(x,y,z)$ on the surface $S : \boldsymbol{r} = \boldsymbol{r}(u,v) = x(u,v)\:\boldsymbol{i} + y(u,v)\:\boldsymbol{j} + z(u,v)\:\boldsymbol{k}$ ．Here， is a smooth surface.

**Figure 3.3:** surface
$\begin{figure}\begin{center} \includegraphics[width=5cm]{VECANALFIG/menseki-1.eps} \end{center}\vskip -0.5cm \end{figure}$

Divide into small faces $S_{1},S_{2},\ldots,S_{n}$ , and represent this division with $\Delta$ . Next, let the area of the curved surface $S_{i}$ be $\Delta S_{i}$ , and take the point ${\ rm P}_{i}$ in $S_{i}$ and consider the following sum.

$\displaystyle S(\Delta) = \sum_{i=1}^{n}f({\rm P}_{i})\Delta S_{i}$

Here, when the surface integral is made finer and $\Delta$ is made as small as possible, if $S(\Delta)$ approaches as much as possible, this limit value is called surface integral of the scalar field and denoted by

$\displaystyle \iint_{S}f(x,y,z)dS$

**Figure 3.4:** Area element
$\begin{figure}\begin{center} \includegraphics[width=6cm]{VECANALFIG/mensekiso.eps} \end{center}\vskip -0.5cm \end{figure}$

Here the area element is approximated by the area of parallelogram with the sides $\frac{\partial \boldsymbol{r}}{\partial u}du$ and $\frac{\partial \boldsymbol{r}}{\partial v}dv$ .

$\displaystyle dS \approx \vert\frac{\partial \boldsymbol{r}}{\partial u} \times \frac{\partial \boldsymbol{r}}{\partial v}\vert du dv$

The surface integral of the scalar field on the surface is expressed as follows.

$\displaystyle \iint_{S}f(x,y,z)dS = \iint_{\Omega} f(x(u,v),y(u,v),z(u,v))\vert\boldsymbol{r}_{u} \times \boldsymbol{r}_{v}\vert du dv$

Here, $\Omega$ is the area on the plane that corresponds to .

Example 3..11

Find the surface integral of the scalar field on the paraboloid $\displaystyle{S : x^2 + y^2 + z = 4}$ where $z \geq 0$ .

$\displaystyle f(x,y,z) = \frac{2y^2 + z}{(4x^2 + 4y^2 + 1)^{1/2}}$

Answer Let $\boldsymbol{r}$ be the position vector corresponds to . Then

$\displaystyle \boldsymbol{r} = x\:\boldsymbol{i} + y\:\boldsymbol{j} + (4 - x^2 - y^2)\:\boldsymbol{k}$

Next we find the normal vecor $\boldsymbol{r}_{x} \times \boldsymbol{r}_{y}$ of .

$\displaystyle \boldsymbol{r}_{x} \times \boldsymbol{r}_{y}$	$\displaystyle =$	$\displaystyle (\boldsymbol{i} -2x\:\boldsymbol{k}) \times (\boldsymbol{j} -2y\:\boldsymbol{k})$
	$\displaystyle =$	$\displaystyle \left\vert\begin{array}{ccc} \boldsymbol{i} & \boldsymbol{j} & \b... ...nd{array}\right\vert = 2x\:\boldsymbol{i} + 2y\:\boldsymbol{j} + \boldsymbol{k}$

Thus

$\displaystyle \vert\boldsymbol{r}_{x} \times \boldsymbol{r}_{y}\vert = \sqrt{4x^2 + 4y^2 + 1}$

Here，we use the polar coordintate transformation $\displaystyle{\Omega : 4 - (x^2 + y^2) \geq 0}$ . Then

$\displaystyle \iint_{S}f(x,y,z)dS$	$\displaystyle =$	$\displaystyle \iint_{\Omega}\frac{2y^2 + z}{(4x^2 + 4y^2 + 1)^{1/2}}\vert\boldsymbol{r}_{x} \times \boldsymbol{r}_{y}\vert dx dy$
	$\displaystyle =$	$\displaystyle \iint_{\Omega}(2y^2 + z)dx dy = \int_{0}^{2\pi}\int_{0}^{2}(4 - r^2 + 2r^2 \sin^{2}{\theta}) r dr d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{2 \pi}\left[2r^2 - \frac{r^2}{4} + \frac{r^4}{2}\sin^{2... ...i} d\theta = \int_{0}^{2\pi}(4 + 8\sin^{2}\theta)d\theta = 8 \pi + 8\pi = 16\pi$

Question 3..2

Find $\displaystyle{\iint_{S}x y^2 dS}$ , where $S : x+y+z = 1, x \geq 0, y \geq 0, z \geq 0$ ．

Question 3..3

Let be the intersection of the plane and -axis，-axis, -axis and let the $\triangle$ ABC be the surface . Find $\int_{S}fdS,\ f= x^2 + 2y + z - 1$ ．

Surface integral of vector field

As in the line integral, we define surface integral of $\boldsymbol{F} = F_{1}\:\boldsymbol{i} + F_{2}\:\boldsymbol{j} + F_{3}\:\boldsymbol{k}$ on the surface using the normal vector $\boldsymbol{n}$ of or area vector ${\bf S}$ and expressed as

$\displaystyle \iint_{S} \boldsymbol{F} \cdot \boldsymbol{n} dS = \iint_{S}\boldsymbol{F} \cdot d{\bf S}$

Note that the direction of $\boldsymbol{n}$ and the direction of $\boldsymbol{r}_{u} \times \boldsymbol {r}_{v}$ are the same.

$\displaystyle \boldsymbol{n} = \frac{\boldsymbol{r}_{u} \times \boldsymbol{r}_{v}}{\vert\boldsymbol{r}_{u} \times \boldsymbol{r}_{v}\vert}$

Therefore, the surface integral of the vector field $\boldsymbol{F}$ on the surface is expressed by the double integral as follows.

$\displaystyle \iint_{S}\boldsymbol{F}\cdot\boldsymbol{n}dS$	$\displaystyle =$	$\displaystyle \iint_{\Omega}\boldsymbol{F} \cdot \frac{\boldsymbol{r}_{u} \time... ...{\Omega}\boldsymbol{F}\cdot(\boldsymbol{r}_{u} \times \boldsymbol{r}_{v}) du dv$
	$\displaystyle =$	$\displaystyle \iint_{\Omega}\left\vert\begin{array}{ccc} F_{1} & F_{2} & F_{3}\... ...ial y}{\partial v} & \frac{\partial z}{\partial v} \end{array}\right\vert du dv$

Also， using the directional cosine $\boldsymbol{n} = \cos{\alpha}\:\boldsymbol{i} + \cos{\beta}\:\boldsymbol{j} + \cos{\gamma}\:\boldsymbol{k}$ can be written as follows.

$\displaystyle \iint_{S}\boldsymbol{F}\cdot\boldsymbol{n}dS$	$\displaystyle =$	$\displaystyle \iint_{S}\boldsymbol{F}\cdot(\cos{\alpha}\:\boldsymbol{i} + \cos{... ...l{k})dS = \iint_{S}(F_{1}\cos{\alpha} + F_{2}\cos{\beta} + F_{3}\cos{\gamma})dS$
	$\displaystyle =$	$\displaystyle \iint_{S}(F_{1}dydz + F_{2}dzdx + F_{3}dxdy)$

Question 3..4

Let be the surface represented by the equation . Prove that the unit normal vector $\boldsymbol{n}$ of the surface is given by the following equation. Here $\nabla F \neq {\bf0}$ ．

$\displaystyle \boldsymbol{n} = \frac{F_{x}\boldsymbol{i} + F_{y}\boldsymbol{j} + F_{z}\boldsymbol{k}}{\sqrt{F_{x}^2 + F_{y}^2 + F_{z}^2}}$

Flux

**Figure 3.5:** flus
$\begin{figure}\begin{center} \includegraphics[width=5cm]{VECANALFIG/flux.eps} \end{center}\vskip -0.5cm \end{figure}$

Here, when the vector field $\boldsymbol{F}$ is the velocity field at a certain point when the fluid flows constantly in the flow tube, $\boldsymbol{F} \cdot \boldsymbol{n}dS$ is called flux towards ${\boldsymbol{n}}$ of $\boldsymbol{F}$ ．Therefore, the flux of $\boldsymbol{F}$ is and the surface integral is $\iint_{S}\boldsymbol{F}\cdot\boldsymbol{n}dS$ which is called flux integral and Reprresents the total flux (total flow rate)．

Example 3..12

Let the vector field be $\boldsymbol{F} = y\:\boldsymbol{j} + z\:\boldsymbol{k}$ ，surface be $\displaystyle{S : x^2 + y^2 = 4 - z , \ z \geq 0}$ ．Find the surface integral of $\displaystyle{\iint_{S} \boldsymbol{F} \cdot\boldsymbol{n} dS }$

Answer The position vector is $\boldsymbol{r} = x\:\boldsymbol{i} + y\:\boldsymbol{j} + z\:\boldsymbol{k} = x\:\boldsymbol{i} + y\:\boldsymbol{j} + (4-x^2 - y^2)\:\boldsymbol{k}$ . Then

$\displaystyle \iint_{S}\boldsymbol{F}\cdot\boldsymbol{n}dS = \iint_{\Omega}\lef... ... 1 & -2y \end{array} \right\vert dx dy = \iint_{\Omega}(4 - x^ 2 + y^2) dx dy$

Here we use the polar coordinate transformation, $\Omega : x^2 + y^2 \leq 4$ . Then $x = r\cos{\theta}, \ y = r\sin{\theta}$ implies

$\displaystyle \iint_{\Omega}(4 - x^ 2 + y^2) dx dy$	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi} \int_{0}^{2}(4 - r^2 + 2r^2 \sin^{2}{\theta})r dr... ...eft[2r^2 - \frac{r^2}{4} + \frac{r^4}{2}\sin^{2}\theta \right ]_{0}^{2} d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}(4 + 8\sin^{2}\theta)d\theta = 8 \pi + 8\pi = 16\pi$

Question 3..5

Let the vector field be $\boldsymbol{F} = x\:\boldsymbol{i} + y\:\boldsymbol{j} - 2z\:\boldsymbol{k}$ , and the surface be $\displaystyle{S : x^2 + y^2 = a^2 , \ 0 \leq z \leq 1}$ ．Then find the surface integral of $\displaystyle{\iint_{S} \boldsymbol{F} \cdot\boldsymbol{n} dS }$ .

Question 3..6

Let be the sphere of the radius with the center Ｏ．Let $\boldsymbol{r} = \overrightarrow{\rm OP}$ be the position vector of ．Then by taking the unit normal vector $\boldsymbol{n}$ of the sperical surface outward, prove the followings.

$\displaystyle \int_{S} \frac{\boldsymbol{r}}{r^3}\cdot\boldsymbol{n}\;dS = 4\pi,\ \ r = \vert\boldsymbol{r}\vert$

Answer Since ，for $z \geq 0$ ，the position vector is $\boldsymbol{r} = x\:\boldsymbol{i} + y\:\boldsymbol{j} + z\:\boldsymbol{k} = x\:\boldsymbol{i} + y\:\boldsymbol{j} + \sqrt{a^2-x^2 - y^2}\:\boldsymbol{k}$ . Thus，

$\displaystyle \iint_{S}\boldsymbol{F}\cdot\boldsymbol{n}dS = \iint_{\Omega}\lef... ...\frac{z}{r^3} \\ 1 & 0 & -2x \\ 0 & 1 & -2y \end{array} \right\vert dx dy$

Example 3..13

$\boldsymbol{F} = (2x-z)\boldsymbol{i} + x^2 y\boldsymbol{j} -x^2 z\boldsymbol{k}$ ，Let the surface be a bounded part of the plane ．Then find the surface integral $\iint_{S}\boldsymbol{F}\cdot\boldsymbol{n}dS$ .

Answer

$\includegraphics[width = 6cm]{VECANALFIG/cube.eps}$

For the plane DEFG: since , the position vector is $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k} = \boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k}$ . Here the positive direction is from the bck of the plane to the front of DEFG. Thus, the unit normal vector is

$\displaystyle \boldsymbol{n} = \frac{\boldsymbol{r}_{y} \times \boldsymbol{r}_{... ...boldsymbol{k}\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right\vert = \boldsymbol{i}$

Also，the orthogonal projection on to plane is $\Omega_{yz} = \{(y,z) : 0 \leq y \leq 1, 0 \leq z \leq 1\}$ . Then，

$\displaystyle \iint_{DEFG}\boldsymbol{F}\cdot\boldsymbol{n}dS = \int_{0}^{1}\in... ...;dz = \int_{0}^{1}(2-z)dx = \left[2z - \frac{z^2}{2}\right]_{0}^1 = \frac{3}{2}$

For the plane ABCO: since , the position vector is $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k} = y\boldsymbol{j} + z\boldsymbol{k}$ . Here，the positive direction is the direction of the back of plane ABCO to the front．Thus，the unit normal vector is

$\displaystyle \boldsymbol{n} = \frac{\boldsymbol{r}_{z} \times \boldsymbol{r}_{... ...oldsymbol{k}\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{array}\right\vert = -\boldsymbol{i}$

Also，the orthogonal projection onto plane is $\Omega_{yz} = \{(y,z) : 0 \leq y \leq 1, 0 \leq z \leq 1\}$ . Then

$\displaystyle \iint_{ABCO}\boldsymbol{F}\cdot\boldsymbol{n}dS = \int_{0}^{1}\in... ...y=0}^{1}\;dz = \int_{0}^{1}zdx = \left[\frac{z^2}{2}\right]_{0}^1 = \frac{1}{2}$

For the plane ABEF: since , the position vector is $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k} = x\boldsymbol{i} + \boldsymbol{j} + z\boldsymbol{k}$ . Here the positive direction is the direction of the back of the plane ABEF to the front. Thus, the unit normla vector is

$\displaystyle \boldsymbol{n} = \frac{\boldsymbol{r}_{z} \times \boldsymbol{r}_{... ...boldsymbol{k}\\ 0 & 0 & 1\\ 1 & 0 & 0 \end{array}\right\vert = \boldsymbol{j}$

Also，the orthogonal projection onto plane is $\Omega_{xz} = \{(x,z) : 0 \leq x \leq 1, 0 \leq z \leq 1\}$ . Then

$\displaystyle \iint_{ABEF}\boldsymbol{F}\cdot\boldsymbol{n}dS = \int_{0}^{1}\int_{0}^{1}x^2dxdz = \frac{1}{3}$

Similarly，if you do the surface integral of the remaining surface，we have

$\displaystyle \int \boldsymbol{F}\cdot\boldsymbol{n}dS = \frac{11}{6}$

Exercise3.4

1.

Let A,B,C be the intersection of the plane and -axis，-axis, and -axis, and the surface be the $\triangle$ ABC. Find the following surface integrals.

(1) $\int_{S}fdS,\ f= x^2 + 2y + z - 1$

(2) $\int_{S}{\bf A}\cdot\boldsymbol{n}dS, \ {\bf A} = x^2 \boldsymbol{i} + z \boldsymbol{k}$

2.

Let the region $x \geq 0, y \geq 0, x^2 + y^2 \leq a^2$ on the -plane be ．Find the surface integral．

$\displaystyle \int_{S}{\bf A} \times \boldsymbol{n}dS, \ {\bf A} = x\boldsymbol{i} + (x-y)\boldsymbol{j} + (\log{xy})\boldsymbol{k}$

3.

Find the following surface integrals．

$\displaystyle \iint_{S}(3x{\bf i} + 4z{\bf j} + 2y{\bf k}) \cdot {\bf n}dS, S:y... ...eq x \leq 3, y \geq 0, z \geq 0, ({\bf n}\mbox{の}z\mbox{component is positive}$

4

Find the surface integral．

$\displaystyle \iint_{S}(x{\bf i} + y{\bf j} - 2z{\bf k}) \cdot {\bf n}dS, S:x^2 + y^2 = a^2, 0 \leq z \leq 1$