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，$S$ is a smooth surface.

Figure 3.3: surface

Divide $S$ into $n$ 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.

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

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

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

Figure 3.4: Area element

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

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

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

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

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

Example 3..11  

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

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

Answer Let $\boldsymbol{r}$ be the position vector corresponds to $S : x^2 + y^2 + z = 4$. Then

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

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

$$\boldsymbol{r}_{x} \times \boldsymbol{r}_{y} = (\boldsymbol{i} -2x\:\boldsymbol{k}) \times (\boldsymbol{j} -2y\:\boldsymbol{k})$$

$$= \left\vert\begin{array}{ccc} \boldsymbol{i} & \boldsymbol{j} & \b... ...nd{array}\right\vert = 2x\:\boldsymbol{i} + 2y\:\boldsymbol{j} + \boldsymbol{k}$$

Thus

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

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

$$\iint_{S}f(x,y,z)dS = \iint_{\Omega}\frac{2y^2 + z}{(4x^2 + 4y^2 + 1)^{1/2}}\vert\boldsymbol{r}_{x} \times \boldsymbol{r}_{y}\vert dx dy$$

$$= \iint_{\Omega}(2y^2 + z)dx dy = \int_{0}^{2\pi}\int_{0}^{2}(4 - r^2 + 2r^2 \sin^{2}{\theta}) r dr d\theta$$

$$= \int_{0}^{2 \pi}\left[2r^2 - \frac{r^2}{4} + \frac{r^4}{2}\sin^{2... ...i} d\theta = \int_{0}^{2\pi}(4 + 8\sin^{2}\theta)d\theta = 8 \pi + 8\pi = 16\pi$$

Question 3..2  

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

Question 3..3  

Let $A,B,C$ be the intersection of the plane $2x + 2y + z = 2$ and $x$-axis，$y$-axis, $z$-axis and let the $\triangle$ABC be the surface $S$. 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 $S$ using the normal vector $\boldsymbol{n}$ of $S$ or area vector ${\bf S}$ and expressed as

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

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

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

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

$$\iint_{S}\boldsymbol{F}\cdot\boldsymbol{n}dS = \iint_{\Omega}\boldsymbol{F} \cdot \frac{\boldsymbol{r}_{u} \time... ...{\Omega}\boldsymbol{F}\cdot(\boldsymbol{r}_{u} \times \boldsymbol{r}_{v}) du dv$$

$$= \iint_{\Omega}\left\vert\begin{array}{ccc} F_{1} & F_{2} & F_{3}\... ...ial y}{\partial v} & \frac{\partial z}{\partial v} \end{array}\right\vert du dv$$

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

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

$$= \iint_{S}(F_{1}dydz + F_{2}dzdx + F_{3}dxdy)$$

Question 3..4  

Let $S$ be the surface represented by the equation $F(x,y,z) = 0$. Prove that the unit normal vector $\boldsymbol{n}$ of the surface $S$ is given by the following equation. Here $\nabla F \neq {\bf0}$．

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

Flux

Figure 3.5: flus

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 $dQ$ 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 $${S : x^2 + y^2 = 4 - z , \ z \geq 0}$$．Find the surface integral of $${\iint_{S} \boldsymbol{F} \cdot\boldsymbol{n} dS }$$

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

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

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

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

$$= \int_{0}^{2\pi}(4 + 8\sin^{2}\theta)d\theta = 8 \pi + 8\pi = 16\pi$$

Question 3..5  

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

Question 3..6  

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

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

Answer Since $x^2 + y^2 + z^2 = a^2$，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，

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

Example 3..13  

$\boldsymbol{F} = (2x-z)\boldsymbol{i} + x^2 y\boldsymbol{j} -x^2 z\boldsymbol{k}$，Let the surface $S$ be a bounded part of the plane $x = 0, x = 1, y = 0, y = 1, z = 0, z = 1$．Then find the surface integral $\iint_{S}\boldsymbol{F}\cdot\boldsymbol{n}dS$.

Answer

For the plane DEFG: since $x = 1$, 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

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

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

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

For the plane ABCO: since $x = 0$, 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

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

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

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

For the plane ABEF: since $y = 1$, 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

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

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

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

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

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

Exercise3.4

1.

Let A,B,C be the intersection of the plane $2x + 2y + z = 2$ and $x$-axis，$y$-axis, and $z$-axis, and the surface $S$ 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 $xy$-plane be $S$．Find the surface integral．

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

3.

Find the following surface integrals．

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

4

Find the surface integral．

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