Gauss's divergence theorem

Next, we will learn about the divergence theorem named after the German mathematician Carl Friedrich Gauss(1777-1855).

Gauss's divergence theorem

Theorem 4..1

[Gauss's divergence theorem] For the vector field $\boldsymbol{F}(x,y,z) = F_{1}\boldsymbol{i} + F_{2}\boldsymbol{j} + F_{3}\boldsymbol{k}$ ，let be the area of the space surrounded by the smooth closed surface , and let $\boldsymbol{n}$ be the normal vector. Then

$\displaystyle \iint_{S}\boldsymbol{F} \cdot\boldsymbol{n}dS = \iiint_{V} {\rm div}\boldsymbol{F} dV$

$\displaystyle \iint_{S}(F_{1}dydz + F_{2}dzdx + F_{3}dxdy) = \iiint_{V} \left(\... ...artial F_{2}}{\partial y} + \frac{\partial F_{3}}{\partial z} \right) dx dy dz$

If you think of and as variables，then

$\displaystyle \int_{S}\boldsymbol{F}\cdot\boldsymbol{n}\;dS = \int_{V} \nabla \cdot\boldsymbol{F}\;dV$

**Figure 4.1:** Gauss's theorem
$\begin{figure}\begin{center} \includegraphics[width=6cm]{CALCFIG/Fig8-6-3.eps} \end{center}\end{figure}$

Proof First, suppose is sandwiched between two curved surfaces $S_{1}, S_{2}$ from the bottom and top.．Also, $S_{1}$ is given by $z = f_{1}(x,y), \ (x,y) \in \Omega$ ， $S_{2}$ is given by $z = f_{2}(x,y) \ (x,y) \in \Omega$ . Then

$\displaystyle \iiint_{V}\frac{\partial F_{3}}{\partial z} dV$	$\displaystyle =$	$\displaystyle \iiint_{V}\frac{\partial F_{3}}{\partial z} dzdydx = \iint_{\Omega}[\int_{z=f_{1}(x,y)}^{f_{2}(x,y)}\frac{\partial F_{3}}{\partial z} dz] dydx$
	$\displaystyle =$	$\displaystyle \iint_{\Omega}\left[F_{3}\right ]_{z = f_{1}(x,y)}^{f_{2}(x,y)}dydx$
	$\displaystyle =$	$\displaystyle \iint_{\Omega}[F_{3}(x,y,f_{2}(x,y)) - F_{3}(x,y,f_{1}(x,y)) ]dy dx$

For the surface $S_{2}$ ，normal unit vector for curvilinear coordinates

$\displaystyle \frac{\boldsymbol{r}_{x} \times \boldsymbol{r}_{y}}{\Vert\boldsymbol{r}_{x} \times \boldsymbol{r}_{y}\Vert}$

is equal to $\boldsymbol{n}$ . On the surface $S_{1}$ , the unit normal vector is equal to $-\boldsymbol{n}$ . Thus,

$\displaystyle \iint_{\Omega}F_{3}(x,y,f_{2}(x,y)) dx dy = \iint_{S_{2}}F_{3} dx dy$

$\displaystyle -\iint_{\Omega}F_{3}(x,y,f_{1}(x,y)) dx dy = \iint_{S_{1}}F_{3} dx dy$

Therefore，

$\displaystyle \iiint_{V}\frac{\partial F_{3}}{\partial z} dV = \iint_{S_{2}}F_{3} dx dy + \iint_{S_{1}}F_{3} dx dy = \iint_{S}F_{3} dx dy$

Similarly, by projecting onto another plane, we can show

$\displaystyle \iiint_{V}\frac{\partial F_{1}}{\partial z} dV = \iint_{S}F_{1} dy dz$

$\displaystyle \iiint_{V}\frac{\partial F_{2}}{\partial z} dV = \iint_{S}F_{2} dz dx$

Adding each, we have

$\displaystyle \iiint_{V} \nabla \cdot\boldsymbol{F} dV = \iint_{S}\boldsymbol{F} \cdot\boldsymbol{n}dS$

If the region is common, you can prove it by dividing into subregions

Example 4..1

The surface integral

$\displaystyle \iint_{S} xz^2 dydz + (x^2 y - z^3)dzdx + (2xy + y^2 z)dxdy$

where, the surface is the upper sphere $S_{1} : z = \sqrt{a^2 - x^2 - y^2}, x^2 + y^2 \leq a^2$ と $S_{2} : z = 0, x^2 + y^2 \leq a^2$ .

(1) Evaluate this surface integral by using Gauss's divergence theorem.

(2) Evaluate this surface integral directly.

Answer (1) $\boldsymbol{F} = xz^2\boldsymbol{i} + (x^2 y - z^3)\boldsymbol{j} + (2xy + y^2 z)\boldsymbol{k}$ より $\boldsymbol{F}$ を求めると

$\displaystyle {\rm div}\boldsymbol{F} = \nabla \cdot(xz^2\boldsymbol{i} + (x^2 y - z^3)\boldsymbol{j} + (2xy + y^2 z)\boldsymbol{k}) = z^2 + x^2 + y^2$

Thus

$\displaystyle \iint_{S} xz^2 dydz + (x^2 y - z^3)dzdx + (2xy + y^2 z)dxdy = \iiint_{V}(x^2 + y^2 + z^2)dV$

Now is an upper sphere with the radius . Then by the spherical coordinate transformation $(\rho,\phi,\theta)$ , we have

$\displaystyle \int_{\theta = 0}^{2\pi}\int_{\phi = 0}^{\frac{\pi}{2}}\int_{\rho = 0}^{a}\rho^2 \rho^2 \sin{\phi}d\rho d\phi d\theta$

$\displaystyle =$

$\displaystyle \int_{\theta = 0}^{2\pi} d\theta \int_{\phi = 0}^{\frac{\pi}{2}} ... ...} \int_{\rho = 0}^{a}\rho^4 d\rho = 2\pi(1)(\frac{a^5}{5}) = \frac{2\pi a^5}{5}$

(2)

$\displaystyle \iint_{S_{1}}xz^2 dydz + (x^2 y - z^3)dzdx + (2xy + y^2 z)dxdy$	$\displaystyle =$	$\displaystyle \iint_{S_{1}}xz^2 dydz + \iint_{S_{1}}(x^2 y -z^3)dxdy$
	$\displaystyle +$	$\displaystyle \iint_{S_{1}}(2xy + y^2 z)dxdy$

Since $z = \sqrt{a^2 - x^2 - y^2}$ ，projection onto plane， implies $\Omega_{1} = \{(y,z) : y^2 + z^2 \leq a^2, z \geq 0\}$ . Here，we consier and .

$\displaystyle \iint_{S_{1}}xz^2 dydz$	$\displaystyle =$	$\displaystyle \int_{y=-a}^{a}\int_{z =0}^{\sqrt{a^2 -y^2}}z^2 \sqrt{a^2 - y^2 -... ... - \int_{y=-a}^{a}\int_{z =0}^{\sqrt{a^2 -y^2}}- z^2 \sqrt{a^2 - y^2 - z^2}dzdy$
	$\displaystyle =$	$\displaystyle 4\int_{y=0}^{a}\int_{z = 0}^{\sqrt{a^2 -y^2}}z^2 \sqrt{a^2 - y^2 - z^2}dzdy$

Next projection onto plane. Then implies $\Omega_{1} = \{(x,z) : x^2 + z^2 \leq a^2, z \geq 0\}$ ．Thus，for and ,

$\displaystyle \iint_{S_{1}}(x^2y -z^3)dzdx$

$\displaystyle =$

$\displaystyle 2\int_{x=-a}^{a}\int_{z=0}^{\sqrt{a^2 - x^2}}(x^2\sqrt{a^2 - x^2 ... ...zdx = 4\int_{x=0}^{a}\int_{z=0}^{\sqrt{a^2 - x^2}}x^2\sqrt{a^2 - x^2 - z^2}dzdx$

Lastly，projection onto plane. Then implies $\Omega_{1} = \{(x,y) : x^2 + y^2 \leq a^2\}$ ．

$\displaystyle \iint_{S_{1}}(2xy + y^2 z)dzdx$	$\displaystyle =$	$\displaystyle \int_{x=-a}^{a}\int_{y=-\sqrt{z^2 - x^2}}^{\sqrt{a^2 - x^2}}(2xy ... ...{a}\int_{y=-\sqrt{z^2 - x^2}}^{\sqrt{a^2 - x^2}} y^2\sqrt{a^2 - x^2 - y^2} dydx$
	$\displaystyle =$	$\displaystyle 4\int_{x=0}^{a}\int_{y=0}^{\sqrt{a^2 - x^2}}y^2\sqrt{a^2 - x^2 - y^2}dydx$

Next consider the surface integral for $S_{2} = \{(x,y) : x^2 + y^2 \leq a^2\}$ ．First projection onto， plane. $S_{2}$ maps to $\Omega_{2} = \{(y,z) : -a \leq y \leq a, z= 0\}$ .

$\displaystyle \iint_{S_{2}}xz^2 dydz = 0$

By Projecting onto plane， $S_{2}$ maps to $\Omega_{2} = \{(x,z) : -a \leq x \leq a, z= 0\}$ .，

$\displaystyle \iint_{S_{2}}(x^2 y - z^3) dzdx = 0$

By projecting onto plane, $S_{2}$ maps to $\Omega_{2} = \{(x,y) : x^2 + y^2 \leq a^2\}$ .

$\displaystyle \iint_{S_{2}}(2xy + y^2z)dzdy = \iint_{S_{2}}2xy dzdx = \int_{-a}^{a}\int_{-\sqrt{a^2 -x^2}}^{\sqrt{a^2 -x^2}}2xydydx = 0$

Adding these，

$\displaystyle \int_{S_{1}}\boldsymbol{F}\cdot\boldsymbol{n}\;dS$	$\displaystyle =$	$\displaystyle 4\int_{y=0}^{a}\int_{z = 0}^{\sqrt{a^2 -y^2}}z^2 \sqrt{a^2 - y^2 ... ...zdy + 4\int_{x=0}^{a}\int_{z=0}^{\sqrt{a^2 - x^2}}x^2\sqrt{a^2 - x^2 - z^2}dzdx$
	$\displaystyle +$	$\displaystyle 4\int_{x=0}^{a}\int_{y=0}^{\sqrt{a^2 - x^2}}y^2\sqrt{a^2 - x^2 - ... ...x = 12 \int_{x=0}^{a}\int_{y=0}^{\sqrt{a^2 - x^2}}y^2\sqrt{a^2 - x^2 - y^2}dydx$

Here using the polar coorinates，we have $x = r\cos{\theta}, \ y = r\sin{\theta}$ . Thus

$\displaystyle \int_{S_{1}}\boldsymbol{F}\cdot\boldsymbol{n}\;dS$	$\displaystyle =$	$\displaystyle 12\int_{\theta = 0}^{\pi/2}\int_{r = 0}^{a}r^2 \sin^{2}{\theta}\s... ...2(\int_{0}^{a}r^3 \sqrt{a^2 - r^2}dr) (\int_{0}^{\pi/2}\sin^{2}{\theta}d\theta)$
	$\displaystyle =$	$\displaystyle 12(\int_{t=a^2}^{0} \sqrt{t}(a^2-t)(\frac{dt}{-2}) ) (\frac{\pi}{4}) = (3\pi)(\frac{1}{2}\int_{0}^{a^2}(a^2 t^{1/2} - t^{3/2})dt)$
	$\displaystyle =$	$\displaystyle (\frac{3\pi}{2})\left[\frac{2}{3}a^2 t^{3/2} - \frac{2}{5}t^{5/2}... ...^{a^2} = (\frac{3\pi}{2})(\frac{2}{3}a^5 - \frac{2}{5}a^5) = \frac{2\pi a^2}{5}$

Example 4..2

Let $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z \boldsymbol{k}$ ．Show the followings for the region and the boundary surface . Here, denote the vaolume of the region by ．

$\displaystyle (1)\ \int_{S} \boldsymbol{n}dS = {\bf0}\hskip 1cm (2)\ \int_{S}\b... ...dS = 3V \hskip 1cm (3)\ \int_{S}\boldsymbol{n} \times \boldsymbol{r}dS = {\bf0}$

Answer (1) Using the arbitrary constant, we express by the surface integral and apply Gauss's divergence theorem. Then

$\displaystyle \int_{S}\boldsymbol{C} \cdot\boldsymbol{n}\;dS = \int_{V}\nabla \cdot\boldsymbol{C}\;dV = 0$

Therefore， $\int_{S}\boldsymbol{C} \cdot\boldsymbol{n}\;dS = 0$

(2) By Gauss's divergence theorem,

$\displaystyle \int_{S}\boldsymbol{r}\cdot\boldsymbol{n}\;dS = \int_{V}\nabla \c... ...al x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z})\;dV = 3V$

(3) Using a constant vector，we express by the surface integral and apply the triple scalar product and Gauss's divergance theorem,

$\displaystyle \int_{S}\boldsymbol{C} \cdot(\boldsymbol{n} \times \boldsymbol{r})\;dS$	$\displaystyle =$	$\displaystyle \int_{S}\boldsymbol{n} \cdot(\boldsymbol{r} \times \boldsymbol{C})\;dS = \int_{V}\nabla \cdot(\boldsymbol{r} \times \boldsymbol{C})\;dV$
	$\displaystyle =$	$\displaystyle \int_{V}(\boldsymbol{C} \cdot\nabla \times \boldsymbol{r} + \boldsymbol{r} \cdot\nabla \times \boldsymbol{C})\;dV = 0$

Example 4..3

Prove the following equation for the region in the scalar field $\phi$ and its boundary surface .

(1) $\displaystyle{\int_{S}\frac{\partial \phi}{\partial n}dS = \int_{V}\nabla^2 \phi dV}$

(2) If $\phi$ is a harmonic function, then

$\displaystyle \int_{S}\frac{\partial \phi}{\partial n}dS = 0$

Answer (1) The problem of surface integral is always rewrite into $\int_{S}\boldsymbol{A}\cdot\boldsymbol{n}\;dS$ ．In this case， $\frac{\partial \phi}{\partial n}$ isa directional derivative in the direction of the unit normal vector $\boldsymbol{n}$ . Then $\frac{\partial \phi}{\partial n} = \nabla \phi \cdot\boldsymbol{n}$ ．Therefore,，

$\displaystyle \int_{S}\frac{\partial \phi}{\partial n} \;dS = \int_{S}\nabla \phi \cdot\boldsymbol{n}\;dS$

Now using Gauss's divergence theorem,

$\displaystyle \int_{S}\frac{\partial \phi}{\partial n}dS$

$\displaystyle =$

$\displaystyle \int_{S}\nabla \phi \cdot\boldsymbol{n}\;dS = \int_{V}\nabla \cdot(\nabla \phi)\;dV = \int_{V}\nabla^2 \phi\;dV$

(2) $\phi$ is harmonic function. Then $\nabla^2 \phi = 0$ . Thus，(1) implies

$\displaystyle \int_{S}\frac{\partial \phi}{\partial n}\;dS = \int_{V}\nabla^2 \phi\;dV = 0$

Theorem 4..2

For any region and its bounary surface within the common definition of the scalar field $\phi$ and the vector field $\boldsymbol{A}$ , if

$\displaystyle \iiint_{V}\phi dV = \iint A\cdot\boldsymbol{n}dS$

，then $\phi = \nabla \cdot\boldsymbol{A}$ ． Proof By Gauss's divergence theorem,

$\displaystyle \iint_{S}A\cdot\boldsymbol{n}dS = \iiint_{V}A\cdot\boldsymbol{n}dV$

Then

$\displaystyle \iiint_{V}\phi\; dV = \iiint_{V}\nabla \cdot\boldsymbol{A}dV$

Thus，

$\displaystyle \iiint_{V}(\phi - \nabla \cdot A)dV = 0$

Note that $\phi - \nabla \cdot\boldsymbol{A}$ is a continuous function and for any region , the followings are true. Thus by the properties of the continuous function, we have

$\displaystyle \phi - \nabla \cdot\boldsymbol{A} = 0 \$ in other words $\displaystyle \ \phi = \nabla \cdot\boldsymbol{A}$

Theorem 4..3

For about the boundary surface of any region in $\boldsymbol{A}$ , prove that if

$\displaystyle \int_{S}\boldsymbol{A} \cdot\boldsymbol{n}dS = 0$

then $\nabla \cdot\boldsymbol{A} = 0$ .

Proof If $\phi = 0$ in the above theorem，then

$\displaystyle \int_{S}\boldsymbol{A} \cdot\boldsymbol{n}\;dS = \int_{V}\phi \;dV = 0$

implies that $\nabla \cdot A = \phi = 0$ .

Exercise4.1

1.

Let $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k},\ r = \vert\boldsymbol{r}\vert$ ．Prove that for any region and the boundary surface , the followings are true.

(1) $\int_{V}\frac{1}{r^2}\;dV = \int_{S}\frac{\boldsymbol{r}\cdot\boldsymbol{n}}{r^2}\;dS$

(2) $\int_{S}\boldsymbol{r} \times \boldsymbol{n}\; dS = 0$

(3) $\int_{S}r^{n}\boldsymbol{r} \cdot\boldsymbol{n}\; dS = (n+3)\int_{V} r^{n}\;dV$

(4) $\int_{S}r^{n} \boldsymbol{n}\; dS = n\int_{V} {\bf r}r^{n-2}\;dV$

(5) $\int_{S}r^{n}\boldsymbol{r} \times \boldsymbol{n}\; dS = 0$

(6) $\int_{V} r\;dV = \frac{1}{2}\int_{S}r^{2}\boldsymbol{n}\;dS$

(7) $\int_{S}F(r)\boldsymbol{n}\; dS = \int_{V} \frac{dF}{dr}\frac{\boldsymbol{r}}{r}\;dV$

2.

Prove the following is true for the boundary surface of any region in the scalar field $\phi$ .

$\displaystyle \int_{S}\boldsymbol{n} \times (\nabla \phi)\;dS = {\bf0}$

3.

Suppose that $\boldsymbol{A}$ satisfies $\nabla \cdot\boldsymbol{A} = 0$ ．Take the closed curve that is the boundary of the curved surface in this vector field．At this time, the surface integral $\Phi = \int_{S}\boldsymbol{A} \cdot \boldsymbol{n}\; dS$ is always the same value for any curved surface whose boundary is . And its value is determined by the closed surface ．Prove the above．

4.

The scalar fiel $\phi$ and the vector field $\boldsymbol{A}, \boldsymbol{B}$ are within the comon domain. Prove the following equation for any region and its bounary surface .

(1) $\int_{V}\boldsymbol{A} \cdot\nabla \phi \;dV = \int_{S}\phi \boldsymbol{A}\cdot\boldsymbol{n}\;dS - \int_{V}\phi \nabla\cdot\boldsymbol{A}\;dV$

(2) $\int_{V}\boldsymbol{A} \cdot(\nabla \times \boldsymbol{B}) \;dV = \int_{S} (\bo... ...oldsymbol{n}dS + \int_{V}\boldsymbol{B} \cdot(\nabla \times \boldsymbol{A})\;dV$

(3) $\int_{V}(\nabla \phi) \cdot(\nabla \times \boldsymbol{A})\;dV= - \int_{S}((\nabla \phi) \times \boldsymbol{A})\cdot\boldsymbol{n}\;dS$

(4) $\boldsymbol{A} = \nabla \phi, \nabla^2 \phi = 0$ ならば， $\int_{V} \vert\boldsymbol{A}\vert^2\;dV = \int_{S}\phi \boldsymbol{A}\cdot\boldsymbol{n}\;dS$

5.

Prove the following equation for any region and its boundary surface within the common definition of the scalar fields $\phi$ and $\psi$ .

(1) $\int_{S}\psi\frac{\partial \phi}{\partial n} \;dS = \int_{V}\{\psi \nabla^2 \phi + (\nabla \psi)\cdot(\nabla \phi)\}\;dV$

(2) $\int_{S}\phi\frac{\partial \phi}{\partial n} \;dS = \int_{V}\{\phi \nabla^2 \phi + \vert\nabla \phi\vert^2\}\;dV$

(3) $\int_{S}\left(\psi\frac{\partial \phi}{\partial n} - \phi \frac{\partial \psi}{\partial n}\right)\;dS = \int_{V}\{\psi \nabla^2 \phi -\phi \nabla^2 \psi\}\;dV$ Green's formula

(4) If $\phi$ is harmonic function, then

$\displaystyle \int_{S}\phi\frac{\partial \phi}{\partial n}\;dS = \int_{V}\vert\nabla \phi\vert^2\;dV$

(5) If $\phi, \psi$ are harmonic functions, then

$\displaystyle \int_{S}\left(\psi\frac{\partial \phi}{\partial n} - \phi \frac{\partial \psi}{\partial n}\right)\;dS = 0$

(6) If $\phi = 0$ on ，then the harmonic function $\phi$ is 0 in ．

6.

Suppose that the vector field $\boldsymbol{A}$ is defined in all space. Prove that if $\int_{S}\boldsymbol{A}\cdot\boldsymbol{n}dS = 0$ for any boundary surface ，then $\boldsymbol{A}$ has a vector potential.

7.

Suppose that $\boldsymbol{A}$ is defined for all space. Prove that if $\int_{S}\boldsymbol{A}\times \boldsymbol{n}dS = 0$ for any boundary surface ，then $\boldsymbol{A}$ has a scalar potential.