Gauss's divergence theorem

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.

8.

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.

9.

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 ．

(1)