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演習問題詳解

演習問題詳解1.1 1.

1.1 (1) $2\boldsymbol{A} = 2(-\boldsymbol{i} + 2\:\boldsymbol{j} + 3\:\boldsymbol{k}) = -2\:\boldsymbol{i} + 4\:\boldsymbol{j} + 6\:\boldsymbol{k}$. (2) $3\boldsymbol{B} - 2\boldsymbol{A} = 3(-3\:\boldsymbol{j} + 2\:\boldsymbol{k}) -... ...+ 2\:\boldsymbol{j} + 3\:\boldsymbol{k} = 2\:\boldsymbol{i} -13\:\boldsymbol{j}$. (3) $\vert\boldsymbol{A}\vert = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{14}.$ (4)

$\displaystyle \vert\boldsymbol{A} + \boldsymbol{B}\vert$ $\displaystyle =$ $\displaystyle \vert-\boldsymbol{i} + 2\:\boldsymbol{j} + 3\:\boldsymbol{k} -3\:\boldsymbol{j} + 2\:\boldsymbol{k}\vert$  
  $\displaystyle =$ $\displaystyle \vert-\boldsymbol{i} - \:\boldsymbol{j} + 5\:\boldsymbol{k}\vert = \sqrt{1+1+25} = \sqrt{27} = 3\sqrt{3}$  

1.2 (1) $\boldsymbol{A} + \boldsymbol{B} = -\boldsymbol{i} + 2\:\boldsymbol{j} + 3\:\bol... ...bol{j} + \:\boldsymbol{k} = -2\boldsymbol{i} +4\boldsymbol{j} + 4\boldsymbol{k}$ (2)

$\displaystyle \boldsymbol{A} \cdot \boldsymbol{B}$ $\displaystyle =$ $\displaystyle \vert\boldsymbol{A}\vert\vert\boldsymbol{B}\vert\cos{\theta} = \sqrt{1+4+9}\sqrt{4+16+16}\cos{\theta}$  
  $\displaystyle =$ $\displaystyle 6\sqrt{14}\cos{\theta}$  

Note that $\boldsymbol{A} \cdot \boldsymbol{B} = -2+8+12 = 18$. よって,

$\displaystyle \cos{\theta} = \frac{18}{6\sqrt{14}} = \frac{3}{\sqrt{14}}$

したがって, $\theta = \cos^{-1}{\frac{3}{\sqrt{14}}}$.

演習問題詳解1.2

(1)

$\displaystyle \boldsymbol{A} \cdot \boldsymbol{B}$ $\displaystyle =$ $\displaystyle \vert\boldsymbol{A}\vert\vert\boldsymbol{B}\vert\cos{\theta} = \sqrt{1+9+1}\sqrt{4+16+4}\cos{\theta}$  
  $\displaystyle =$ $\displaystyle \sqrt{11}\sqrt{24}\cos{\theta}$  

$\boldsymbol{A} \cdot \boldsymbol{B} = -2+12-2 = 8$に注意すると，

$\displaystyle \cos{\theta} = \frac{8}{\sqrt{11}\sqrt{24}} = \frac{4}{\sqrt{66}}$

したがって, $\theta = \cos^{-1}{\frac{4}{\sqrt{66}}}$.

(2) $\boldsymbol{A}$方向の単位ベクトル${\bf u}$は

$\displaystyle {\bf u} = \frac{-\:\boldsymbol{i} + 3\:\boldsymbol{j} + \:\boldsy... ...}} = \frac{-\:\boldsymbol{i} + 3\:\boldsymbol{j} + \:\boldsymbol{k}}{\sqrt{11}}$

2 (1)

$\displaystyle \boldsymbol{A} \cdot \boldsymbol{B}$ $\displaystyle =$ $\displaystyle 2(-1) + 2(-4) -(-1) = -9$  

(2)

$\displaystyle \boldsymbol{A} \cdot \boldsymbol{B} = \vert\boldsymbol{A}\vert \vert\boldsymbol{B}\vert \cos{\theta} $

に注意すると，

$\displaystyle \cos{\theta} = \frac{-9}{\sqrt{9}\sqrt{18}} = \frac{1}{\sqrt{2}}$

演習問題詳解1.3

1.3 1.

$\displaystyle {}^t[1 2 1] \times {}^t[2 -1 -2]$ $\displaystyle =$ $\displaystyle \Vert\begin{array}{lll}{\bf i} & {\bf j} &{\bf k}\\ 1 & 2 & 1\\ 2 & -1 & -2 \end{array}\Vert$  
  $\displaystyle =$ $\displaystyle -3{\bf i}+4{\bf j} -5{\bf k}$  

2. (1)

$\displaystyle \left\vert\begin{array}{ccc} {\bf i} & {\bf j} & {\bf k}\\ 2 & -3 & -1\\ 1 & 4& -2 \end{array}\right\vert = 10{\bf i} + 3{\bf j} + 11{\bf k}$

(2)

$\displaystyle (2{\bf A} - 3{\bf B}) \times ({\bf A} + 2{\vert bf B})$ $\displaystyle =$ $\displaystyle ({\bf i} - 18{\bf j} + 4{\bf k}) \times (4{\bf i} + 5{\bf j} - 5{\bf k})$  
  $\displaystyle =$ $\displaystyle \left\vert\begin{array}{ccc} {\bf i} & {\bf j} & {\bf k}\\ 1 & -18 & 4\\ 4 & 5 & -5 \end{array}\right\vert = 70{\bf i} + 21{\bf j} + 77{\bf k}$  

(1) 四辺形の面積は

$\displaystyle \vert{\bf A} \times {\bf B}\vert = \left\vert\begin{array}{rrr} \... ...ht\vert = \vert-3{\bf i} +4 {\bf j} -5{\bf k}\vert = \sqrt{9+16+25} = \sqrt{50}$

(2) $\left\vert\begin{array}{rrr} \:\boldsymbol{i} & \:\boldsymbol{j} & \:\boldsymbo... ...= -3\:\boldsymbol{i} + 4\:\boldsymbol{j} + -5\:\boldsymbol{k} = {}^t[-3 4 -5]$

演習問題詳解1.4 1. $\boldsymbol{A}\cdot(\boldsymbol{B}\times\boldsymbol{C}) = \left\vert\begin{arra... ...}1 & 2 & -1\\ 2 & -1 & -1\\ -1 & 3 & 4 \end{array}\right\vert = -20 \neq 0$．Thus they are not coplanar．

2.

$\displaystyle {\bf B} \times {\bf C}$ $\displaystyle =$ $\displaystyle \left\vert\begin{array}{ccc} {\bf i} & {\bf j} & {\bf k}\\ 2 & -... ... -1 & 1 & 2 \end{array}\right\vert = (-2-1){\bf i} -(4+1){\bf j} + (2-1){\bf k}$  
  $\displaystyle =$ $\displaystyle -3{\bf i} - 5{\bf j} + {\bf k}$  

Then

$\displaystyle {\bf A} \cdot ({\bf B} \times {\bf C}) = ({\bf i} + 2{\bf j} + {\bf k})\cdot (-3{\bf i} - 5{\bf j} + {\bf k}) = -12$

3. Note that $\vert{\bf A} \times {\bf B}\vert^2 = ({\bf A} \times {\bf B}) \cdot ({\bf A} \times {\bf B})$. Let ${\bf C} = {\bf A} \times {\bf B}$. Then

$\displaystyle {\bf C} = (A_2 B_3 - A_3B_2){\bf i} - (A_1 B_3 - A_3 B_1){\bf j} + (A_1B_2 0 A_2 B_1){\bf k}$

Thus,
$\displaystyle \vert{\bf A} \times {\bf B}\vert^2$ $\displaystyle =$ $\displaystyle (A_2 B_3 - A_3B_2)^2 + (A_1 B_3 - A_3B_1)^2 + (A_1 B_2 - A_2 B_1)^2$  
  $\displaystyle =$ $\displaystyle A_2^2 B_3^2 + A_3^2 B_2^2 + A_1^2B_3^2 + A_3^2B_1^2+A_1^2 B_2^2 + A_2^2 B_1^2 - (2A_2A_3B_2B_3 + 2A_1 A_3B_1B_3 + 2A_1 A_2 B_1B_2)$  
  $\displaystyle =$ $\displaystyle (A_1^2 + A_2^2 + A_3^2)(B_1^2 + B_2^2 + B_3^2) - (A_1^2B_1^2 + A_2^2B_2^2 + A_3^2 B_3^2 + 2A_2A_3B_2B_3 + 2A_1 A_3B_1B_3 + 2A_1 A_2 B_1B_2)$  
  $\displaystyle =$ $\displaystyle ({\bf A} \cdot {\bf A})({\bf B} \cdot {\bf B}) - ({\bf A} \cdot {\bf B})^2$  

演習問題詳解2.1 1. $\boldsymbol{F}(t)$の成分は $x = t\cos{t}, y = t\sin{t}, z = t^2$. よって， $z = x^2 + y^2$でその軌跡 $\boldsymbol{F}(t)$, $(x(t),y(t),z(t))$は放物面 $z = x^2 + y^2$である.

2. それぞれの成分を微分することにより

$\displaystyle \boldsymbol{F}^{\prime}(t) = 2t\:\boldsymbol{i} + \:\boldsymbol{j} + 3t^{2}\:\boldsymbol{k}$

3. $F' = (5t^2)'\:\boldsymbol{i} + (t)'\:\boldsymbol{j} + - (t^2)' \:\boldsymbol{k}... ... (\cos{t})'\:\boldsymbol{j} = \cos{t}\:\boldsymbol{i} + \sin{t}\:\boldsymbol{j}$.

(1)

$\displaystyle (\boldsymbol{F}\cdot$$\displaystyle \mbox{\boldmath$G$}$$\displaystyle )'$ $\displaystyle =$ $\displaystyle \boldsymbol{F}'\cdot$$\displaystyle \mbox{\boldmath$G$}$$\displaystyle + \boldsymbol{F}\cdot$$\displaystyle \mbox{\boldmath$G$}$$\displaystyle '$  
  $\displaystyle =$ $\displaystyle (10t \:\boldsymbol{i} + \:\boldsymbol{j} - 2t \:\boldsymbol{k})\c... ... - t^2\:\boldsymbol{k})\cdot(\cos{t}\:\boldsymbol{i} + \sin{t}\:\boldsymbol{j})$  
  $\displaystyle =$ $\displaystyle (5t^2 - 1)\cos{t} + 11\sin{t}$  

(2)
$\displaystyle (\boldsymbol{F}\times$   $\displaystyle \mbox{\boldmath$G$}$$\displaystyle )'$ $\displaystyle =$ $\displaystyle \boldsymbol{F}'\times$   $\displaystyle \mbox{\boldmath$G$}$$\displaystyle + \boldsymbol{F}\times$   $\displaystyle \mbox{\boldmath$G$}$$\displaystyle '$  
  $\displaystyle =$ $\displaystyle (10t \:\boldsymbol{i} + \:\boldsymbol{j} - 2t \:\boldsymbol{k})\t... ... t^2\:\boldsymbol{k})\times (\cos{t}\:\boldsymbol{i} + \sin{t}\:\boldsymbol{j})$  
  $\displaystyle =$ $\displaystyle \left\vert\begin{array}{ccc} \boldsymbol{i} & \boldsymbol{j} & \b... ...oldsymbol{k}\\ 5t^2 & t & -t^2\\ \cos{t} & \sin{t} & 0 \end{array}\right\vert$  
  $\displaystyle =$ $\displaystyle (t^2 \sin{t} - 2t \cos{t})\:\boldsymbol{i} - (t^2 \cos{t} + 2t \sin{t})\:\boldsymbol{j} + (5t^2 \sin{t} - \sin{t} - 11t\cos{t})\:\boldsymbol{k}$  

4. $\vert\boldsymbol{F}(t)\vert = c$とすると， $\vert\boldsymbol{F}(t)\vert^2 = \boldsymbol{F}(t) \cdot\boldsymbol{F}(t) = c^{2}$. よって，ベクトル関数の導関数より

$\displaystyle (\boldsymbol{F}(t) \cdot\boldsymbol{F}(t))^{\prime} = 2\boldsymbol{F}^{\prime}(t) \cdot\boldsymbol{F}(t) = 0 $

これより内積は 0 ， $\boldsymbol{F}(t)$ と $\boldsymbol{F}^{\prime}(t)$は直交

5. $\int \boldsymbol{F}\cdot\boldsymbol{F}'\;dt = \boldsymbol{F}\cdot\boldsymbol{F} - \int\boldsymbol{F}'\cdot\boldsymbol{F}\;dt$．したがって， $\int \boldsymbol{F}\cdot\boldsymbol{F}'\;dt = \frac{1}{2}\boldsymbol{F}\cdot\boldsymbol{F}$.

6. $(\boldsymbol{F}\cdot \boldsymbol{F})' = 2\boldsymbol{F}' \cdot \boldsymbol{F}$とすると，

$\displaystyle \int \boldsymbol{F}\cdot \boldsymbol{F'} dt = \frac{1}{2}\int \boldsymbol{R} \cdot \boldsymbol{R} dt = \frac{1}{2}\vert\boldsymbol{F}\vert^2$

よって，

$\displaystyle \int_{2}^{3}\boldsymbol{F} \cdot \boldsymbol{F'}dt = \frac{1}{2}\... ...boldsymbol{F} \mid_{2}^{3} = \frac{1}{2}(16+4+9 - (4+1 +4)) = \frac{20}{2} = 10$

演習問題詳解2.2 1. 点 $(-1,0,2)$を通り方向が ${}^t [1,4,3]-{}^t [-1,0,2] = {}^t [2,4,1]$の直線を求める.直線上の任意の点を $(x,y,z)$とすると

$\displaystyle (x,y,z) = (-1,0,2) + (2,4,1)t, t \in R $

この式を$t$について解くと

$\displaystyle \frac{x+1}{2} = \frac{y}{4} = \frac{z-2}{1} $

3. 求める直線は始点を $(-1,0,2)$ にもち，方向が ${}^t [1,4,3]-{}^t [-1,0,2] = {}^t [2,4,1]$ と考えられるので，直線上の任意の点を $(x,y,z)$ とすると，

$\displaystyle (x,y,z) = (-1,0,2) + (2,4,1)t, t \in R $

またはこの式を $t$ について解くと

$\displaystyle \frac{x+1}{2} = \frac{y}{4} = \frac{z-2}{1} $

が得られます．

4. $x = \cos{t}, y = \sin{t}, z = t$ より， $x^2 + y^2 = 1$ となり， $\boldsymbol{r}(t)$ は半径1の円柱の回りをらせん状に回転する滑らかな曲線だということが分かります．そこで

$\displaystyle \frac{d \boldsymbol{r}}{dt} = -\sin{t}\:\boldsymbol{i} + \cos{t}\:\boldsymbol{j} + \:\boldsymbol{k} $

より

$\displaystyle \vert\frac{d \boldsymbol{r}}{dt}\vert = \sqrt{(-\sin{t})^{2} + (\cos{t})^{2} + 1} = \sqrt{1 + 1} = \sqrt{2}$

よって

$\displaystyle s = \int_{0}^{2\pi}\sqrt{2}dt = 2\sqrt{2}\pi $

となります．

5.

$\displaystyle \boldsymbol{r}(t) = (\cos\pi t, \sin\pi t, t)$

$\displaystyle {\bf v}(t) = \boldsymbol{r}'(t) = (-\pi\sin\pi t, \pi\cos\pi t, 1)$

$\displaystyle \boldsymbol{A}(t) = \boldsymbol{r}''(t) = (-\pi^{2}\cos\pi t, -\pi^{2}\sin\pi t, 0)$

より $t = 1$ のとき
$\displaystyle {\bf v}(1)$ $\displaystyle =$ $\displaystyle (0, \pi, 1)$  
$\displaystyle \boldsymbol{A}(1)$ $\displaystyle =$ $\displaystyle (\pi^{2}, 0, 0)$  

となるので

$\displaystyle v = \vert{\bf v}(1)\vert = \sqrt{\pi^{2} + 1}$

$\displaystyle {\bf t} = \frac{{\bf v}(1)}{\vert{\bf v}(1)\vert} = \frac{(0, \pi, 1)}{\sqrt{\pi^{2} + 1}}$

ここで $\boldsymbol{n}$ を求めるには色々な方法があります．ここでは計算が簡単な方法を考えます．

$\displaystyle \boldsymbol{A} = a_{{\bf t}} + a_{\boldsymbol{n}},a_{{\bf t}} = (\boldsymbol{A}\cdot{\bf t}){\bf t} = 0 $

より

$\displaystyle a_{\boldsymbol{n}} = \boldsymbol{A} - \boldsymbol{A}_{{\bf t}} = (\pi^{2}, 0, 0) $

したがって，

$\displaystyle \boldsymbol{n} = \frac{a_{\boldsymbol{n}}}{\vert a_{\boldsymbol{n}}\vert} = (1, 0, 0)．$

他にも

$\displaystyle v = \vert{\bf v}(t)\vert = \sqrt{\pi^{2}\sin^{2}\pi t + \pi^{2}\cos^{2}\pi t + 1} = \sqrt{\pi^{2} + 1} より$

$\displaystyle a_{{\bf t}} = \frac{dv}{dt} = 0$

よって $\displaystyle{\boldsymbol{n} = \frac{a_{\boldsymbol{n}}}{\vert a_{\boldsymbol{n}}\vert} = (1, 0, 0)}$ と求めることができます．

(a) $\displaystyle{\frac{d \boldsymbol{r}}{dt} = 4a \sqrt{1 - t^2}\:\boldsymbol{i} + 4at\:\boldsymbol{j} + 4a\:\boldsymbol{k}}$より

$\displaystyle \vert\frac{d \boldsymbol{r}}{dt}\vert = \sqrt{16a^2 (1- t^2) + 16a^2 t^2 + 16a^2} = \sqrt{32 a^2} = 4\sqrt{2} a $

よって，

$\displaystyle s = \int_{t_{1}}^{t_{2}} \vert \frac{d \boldsymbol{r}}{dt}\vert dt = 4\sqrt{2} a (t_{2} - t_{1}) $

(b)

$\displaystyle {\bf t} = \frac{d \boldsymbol{r}}{ds} = \frac{d \boldsymbol{r}/dt... ...-t^2}\:\boldsymbol{i} + 4at\:\boldsymbol{j} + 4a\:\boldsymbol{k}}{4\sqrt{2} a} $

(c) $\displaystyle{\frac{d {\bf t}}{dt} = \frac{1}{\sqrt{2}}(\frac{-t}{\sqrt{1 - t^2}}\:\boldsymbol{i} + \boldsymbol{j}), \frac{ds}{dt} = 4\sqrt{2} a}$ より

$\displaystyle \kappa = \vert\frac{d {\bf t}}{ds}\vert = \vert\frac{d {\bf t}/dt... ...vert = \frac{1}{8a}\sqrt{\frac{t^2}{1 - t^2} + 1} = \frac{1}{8a \sqrt{1 - t^2}}$

また， $\displaystyle{\boldsymbol{n} = \frac{1}{\kappa}\frac{d {\bf t}}{ds}}$ より

$\displaystyle \boldsymbol{n} = -t\:\boldsymbol{i} + \sqrt{1 - t^2}\:\boldsymbol{j}$

(d)

$\displaystyle \boldsymbol{B} = {\bf t} \times \boldsymbol{n} = \frac{1}{\sqrt{2... ...{\sqrt{2}}(-\sqrt{1 - t^2}\:\boldsymbol{i} -t\:\boldsymbol{j} + \boldsymbol{k} $

また，

$\displaystyle \frac{d \boldsymbol{B}}{ds} = \frac{d \boldsymbol{B}/dt}{ds/dt} =... ...frac{t}{\sqrt{1- t^2}}\:\boldsymbol{i} -\boldsymbol{j}) = - \tau \boldsymbol{n}$

より $\displaystyle{\tau = \frac{1}{8a \sqrt{1 - t^2}}}$

8. まず， $\displaystyle{\vert\frac{d\phi}{ds}\vert}$ を求めてみましょう． $\tan{\phi}$ は接線の傾きなので， $\tan{\phi} = y^{\prime}(x)$ となります．よって

$\displaystyle \phi = \tan^{-1}(y^{\prime}) $

これを $x$ について微分すると

$\displaystyle \frac{d\phi}{dx} = \frac{1}{1 + (y^{\prime})^{2}} \cdot\frac{d}{dx}(y^{\prime}) = \frac{y^{\prime\prime}}{1 + (y^{\prime})^{2}} $

ここで，

$\displaystyle \frac{d \phi}{dx} = \frac{d \phi}{ds} \frac{ds}{dx} = \frac{d \phi}{ds} \sqrt{1 + (y^{\prime})^2} $

に注意すると

$\displaystyle \frac{d \phi}{ds} \sqrt{1 + (y^{\prime})^2} = \frac{y^{\prime\prime}}{1 + (y^{\prime})^2} $

よって


$\displaystyle \kappa = \vert\frac{d \phi}{ds}\vert = \frac{\vert y^{\prime\prime}\vert}{[1 + (y^{\prime})^2]^{3/2}} $


となります．

演習問題詳解3.1

1. 点 $(x_{0},y_{0},z_{0})$ を通る等位面を

$\displaystyle f(x,y,z) = c $

とし，この等位面上で点 $(x_{0},y_{0},z_{0})$ を通る任意の曲線を

$\displaystyle \boldsymbol{r} = \boldsymbol{r}(t) = x(t)\:\boldsymbol{i} + y(t)\:\boldsymbol{j} + z(t)\:\boldsymbol{k} $

とすると， $f(\boldsymbol{r}(t)) = f(x(t),y(t),z(t)) = c$ が成り立つので，この両辺を $t$ について微分すると，

$\displaystyle f_{x}x_{t} + f_{y}y_{t} + f_{z}z_{t} = (f_{x},f_{y},f_{z})\cdot(x_{t},y_{t},z_{t}) = \nabla f \cdot\frac{d \boldsymbol{r}}{dt} = 0 $

よって，すべての等位面上の曲線の接線に直交するので，勾配は等位面に直交します．

2.

$\displaystyle f(x,y,z)=x^{2}+2y^{2}-z^{2} とおくと，$

等位面 $f(x,y,z)=0$ はもとの曲面と同じ．また， 法線ベクトルは $\nabla f$ で与えられるので，

$\displaystyle \nabla f = (2x, 4y, -2z), \nabla f \mid_{(1,2,3)} = (2,8,-6)$

よって

$\displaystyle \boldsymbol{n} = \frac{\nabla f}{\vert\nabla f\vert} = \frac{(2,8,-6)}{\sqrt{4+64 +36}} = \frac{(2,8,-6)}{\sqrt{104}} $

次に点 $(1,2,3)$ での $(1,3,-1)$方向の方向微分係数を求めるため方向単位ベクトルを求めると $\displaystyle{{\bf u} = \frac{(1,3,-1)}{\sqrt{11}}}$．よって方向微分係数は

$\displaystyle \frac{\partial f}{\partial u} = \nabla f \cdot{\bf u} = (2,8,-6) \cdot\frac{(1,3,-1)}{\sqrt{11}} = \frac{32}{\sqrt{11}} $

また，椄平面の方程式は

$\displaystyle 2(x-1) + 8(y-2) - 6(z - 3) = 0 すなわち 2x + 8y - 6z = 0 $

となります．

3. $f(x,y) = c$ を流線の方程式とすると $\nabla f(x,y) = f_{x}\:\boldsymbol{i} + f_{y}\:\boldsymbol{j}$ は $f(x,y) = c$ の法線ベクトルを表わすので，

$\displaystyle \nabla f(x,y) \cdot\boldsymbol{F}(x,y) = 0 よって (f_{x}\:\bolds... ...{i} + f_{y}\:\boldsymbol{j}) \cdot(-y\:\boldsymbol{i} + x\:\boldsymbol{j}) = 0 $

これより， $-y f_{x} + xf_{y} = 0$．ここで $f(x,y) = c$ の接線の傾きは

$\displaystyle f_{x}dx + f_{y} dy = 0より\frac{dy}{dx} = - \frac{f_{x}}{f_{y}} $

に注意すると

$\displaystyle \frac{dy}{dx} = - \frac{f_{x}}{f_{y}} = - \frac{x}{y} $

より $x dx + y dy = 0$．これより $f_{x} = x, f_{y} = y$ となるので，

$\displaystyle f(x,y) = \int f_{x}dx = \frac{x^2}{2} + c(y) $ (5.1)

次に式5.1を $y$ で偏微分すると

$\displaystyle f_{y} = c'(y) = y よりc(y) = \frac{y^2}{2} $

よって求める流線は

$\displaystyle f(x,y) = \frac{x^2}{2} + \frac{y^2}{2} = c$

4.

$\displaystyle f_{x}$ $\displaystyle =$ $\displaystyle \frac{\partial}{\partial x}(x^{2}+y^{2}+x^{2})^{\frac{1}{2}} = \f... ...}{2}(x^{2}+y^{2}+z^{2})^{-\frac{1}{2}}(2x) = \frac{x}{\vert\boldsymbol{r}\vert}$  
$\displaystyle f_{y}$ $\displaystyle =$ $\displaystyle \frac{y}{\vert\boldsymbol{r}\vert}$  
$\displaystyle f_{z}$ $\displaystyle =$ $\displaystyle \frac{z}{\vert\boldsymbol{r}\vert}$  

より $-\nabla f=(f_{x},f_{y},f_{z}) = \boldsymbol{F}$． よって， $\boldsymbol{F}$ は保存場であり

$\displaystyle f(x,y,z) = {\vert\boldsymbol{r}\vert} = (x^{2}+y^{2}+z^{2})^{1/2}$

は $\boldsymbol{F}$ のスカラーポテンシャルとなります．

(1)

$\displaystyle \nabla r = (\frac{\partial}{\partial x}\:\boldsymbol{i} + \frac{\... ...ol{j} + 2z\:\boldsymbol{k}}{2\sqrt{x^2 + y^2 + z^2}} = \frac{\boldsymbol{r}}{r}$

(2)

$\displaystyle \nabla r^{n}$ $\displaystyle =$ $\displaystyle \frac{\partial}{\partial x}(r^{n})\: \boldsymbol{i} + \frac{\part... ...y}(r^{n})\: \boldsymbol{j} + \frac{\partial}{\partial z}(r^{n})\:\boldsymbol{k}$  
  $\displaystyle =$ $\displaystyle \frac{\partial}{\partial r}(r^{n})\frac{\partial r}{\partial x}\:... ...frac{\partial }{\partial r}(r^{n})\frac{\partial r}{\partial x}\:\boldsymbol{k}$  
  $\displaystyle =$ $\displaystyle nr^{n-1} \left(\frac{\partial r}{\partial x}\boldsymbol{i} + \fra... ...= nr^{n-1}\nabla r = nr^{n-1} \frac{\boldsymbol{r}}{r} = nr^{n-2}\boldsymbol{r}$  

(1)

$\displaystyle {\bf A} = \nabla \left(2r^2 - 4\sqrt{r} + \frac{6}{3\sqrt{r}}\rig... ...oldsymbol{r}) = (4 - \frac{2}{r\sqrt{r}} - \frac{1}{r^2\sqrt{r}})\boldsymbol{r}$

(2)

$\displaystyle {\bf B} = \nabla (r^2 e^{-r}) = (\nabla r^{2})e^{-r} + r^{2} (\na... ...boldsymbol{r} -r^2 e^{-r}\frac{\boldsymbol{r}}{r} = (2 - r)e^{-r}\boldsymbol{r}$

Alternate詳解 Using $\nabla f(\phi) = \frac{d f}{d\phi} \nabla \phi$, we have

$\displaystyle {\bf B} = \nabla (r^2 e^{-r}) = \frac{d (r^2 e^{-r})}{dr} \nabla r = (2re^{-r} -r^2 e^{-r})\frac{\boldsymbol{r}}{r} = (2 - r)e^{-r}\boldsymbol{r}$

7. Note that $\nabla \phi$ of $\phi(x,y,z) = x^2 y + 2xz $ is orthogonal to $\phi(x,y,z) = 4$．Therefore the unit normal vector $\boldsymbol{n}$ is

$\displaystyle \boldsymbol{n} = \frac{(\nabla \phi)_{P}}{\vert\nabla \phi\vert _{P}}$

Here， $(\nabla \phi)_{P} = (2xy +2z)\boldsymbol{i} + x^2 \boldsymbol{j} + 2x\boldsymbol{k})\mid_{(2,-2,3)} = -2\boldsymbol{i} + 4\boldsymbol{j} + 4\boldsymbol{k}$ implies

$\displaystyle \boldsymbol{n} = \frac{-2\boldsymbol{i} + 4\boldsymbol{j} + 4\bol... ... + 16 + 16}} = \frac{1}{3}(-\boldsymbol{i} + 2\boldsymbol{j} + 2\boldsymbol{k})$

8. $\nabla = \boldsymbol{i}\frac{\partial}{\partial x} + \boldsymbol{j}\frac{\partial}{\partial y} + \boldsymbol{k}\frac{\partial }{\partial z}$ implies

$\displaystyle \boldsymbol{i}\frac{\partial}{\partial x} (\frac{\phi}{\psi})$ $\displaystyle =$ $\displaystyle \boldsymbol{i}\frac{\phi_{x}\psi - \phi \psi_{x}}{\psi^2} = \frac{\psi \phi_{x}\boldsymbol{i} - \phi \psi_{x}\boldsymbol{i}}{\psi^2}$  
$\displaystyle \boldsymbol{j}\frac{\partial}{\partial y} (\frac{\phi}{\psi})$ $\displaystyle =$ $\displaystyle \boldsymbol{j}\frac{\phi_{y}\psi - \phi \psi_{y}}{\psi^2} = \frac{\psi \phi_{y}\boldsymbol{j} - \phi \psi_{y}\boldsymbol{j}}{\psi^2}$  
$\displaystyle \boldsymbol{k}\frac{\partial}{\partial z} (\frac{\phi}{\psi})$ $\displaystyle =$ $\displaystyle \boldsymbol{k}\frac{\phi_{z}\psi - \phi \psi_{z}}{\psi^2} = \frac{\psi \phi_{z}\boldsymbol{k} - \phi \psi_{z}\boldsymbol{k}}{\psi^2}$  

Here， $\phi_{x}\boldsymbol{i} + \phi_{y}\boldsymbol{j} + \phi_{z}\boldsymbol{k} = \nabla \phi$, $\psi_{x}\boldsymbol{i} + \psi_{y}\boldsymbol{j} + \psi_{z}\boldsymbol{k} = \nabla \psi$. Thus，

$\displaystyle \nabla \left(\frac{\phi}{\psi}\right) = \frac{\phi \nabla \psi - \psi \nabla \phi}{\phi^2}$

演習問題詳解3.2

1.

(1) $\phi = x^2 z + e^{y/x}$より，

$\displaystyle \nabla \phi = \frac{\partial \phi}{\partial x}\boldsymbol{i} + \f... ...x^2}))\boldsymbol{i} + (e^{y/x}(\frac{1}{x}))\boldsymbol{j} + x^2\boldsymbol{k}$

$\psi = 2z^2 y - xy^2$より，

$\displaystyle \nabla \psi = \frac{\partial \psi}{\partial x}\boldsymbol{i} + \f... ...symbol{k} = -y^2\boldsymbol{i} + (2z^2 - 2xy)\boldsymbol{j} + 4zy\boldsymbol{k}$

(2) $\nabla (\phi \psi) = (\nabla \phi)\psi + \phi \nabla \psi$. (1)より，

$\displaystyle \nabla (\phi \psi)$ $\displaystyle =$ $\displaystyle ((2xz + e^{y/x}(-\frac{y}{x^2}))\boldsymbol{i} + (e^{y/x}(\frac{1}{x}))\boldsymbol{j} + x^2\boldsymbol{k})(2z^2 y - xy^2)$  
  $\displaystyle +$ $\displaystyle (x^2 z + e^{y/x})(-y^2\boldsymbol{i} + (2z^2 - 2xy)\boldsymbol{j} + 4zy\boldsymbol{k})$  

より，点P$(1,0,-2)$での値は

$\displaystyle \nabla (\phi \psi)_{P} = (-4\boldsymbol{i} + \boldsymbol{j} + \boldsymbol{k})(0) + (-2+1)(8)\boldsymbol{j} = -8\boldsymbol{j}$

2. $\phi$の点P$(2,-1,2)$における${\bf u}$方向への方向微分係数は，

$\displaystyle \frac{\partial \phi(2,-1,2)}{\partial u} = \nabla \phi(1,0,-2) \cdot{\bf u}$

ここで，
$\displaystyle \nabla \phi(2,-1,2)$ $\displaystyle =$ $\displaystyle (4z^3 - 6xyz)\boldsymbol{i} + (-3x^2 z)\boldsymbol{j} + (12xz^2 - 3x^2y)\boldsymbol{k}\mid_{(2,-1,2)}$  
  $\displaystyle =$ $\displaystyle (32 + 24)\boldsymbol{i} + (-24)\boldsymbol{j} +(96 + 12))\boldsymbol{k}$  

より，

$\displaystyle \frac{\partial \phi(2,-1,2)}{\partial u} = (56\boldsymbol{i} - 24... ...ldsymbol{j} + 6\boldsymbol{k}}{7} = \frac{1}{7}(112 + 72 + 648) = \frac{832}{7}$

3. $\nabla r^{n} = nr^{n-1}\nabla r = nr^{n-1}\frac{\boldsymbol{r}}{r} = nr^{n-2}\boldsymbol{r}$を用いると簡単である．

(1)

$\displaystyle {\bf A} = \nabla \left(2r^2 - 4\sqrt{r} + \frac{6}{3\sqrt{r}}\rig... ...oldsymbol{r}) = (4 - \frac{2}{r\sqrt{r}} - \frac{1}{r^2\sqrt{r}})\boldsymbol{r}$

(2)

$\displaystyle {\bf B} = \nabla (r^2 e^{-r}) = (\nabla r^{2})e^{-r} + r^{2} (\na... ...boldsymbol{r} -r^2 e^{-r}\frac{\boldsymbol{r}}{r} = (2 - r)e^{-r}\boldsymbol{r}$

別解 $\nabla f(\phi) = \frac{d f}{d\phi} \nabla \phi$を用いると，

$\displaystyle {\bf B} = \nabla (r^2 e^{-r}) = \frac{d (r^2 e^{-r})}{dr} \nabla r = (2re^{-r} -r^2 e^{-r})\frac{\boldsymbol{r}}{r} = (2 - r)e^{-r}\boldsymbol{r}$

4. $\phi(x,y,z) = x^2 y + 2xz $の勾配 $\nabla \phi$は点Pでこの曲面 $\phi(x,y,z) = 4$に垂直である．したがって， 単位法ベクトル $\boldsymbol{n}$は

$\displaystyle \boldsymbol{n} = \frac{(\nabla \phi)_{P}}{\vert\nabla \phi\vert _{P}}$

ここで， $(\nabla \phi)_{P} = (2xy +2z)\boldsymbol{i} + x^2 \boldsymbol{j} + 2x\boldsymbol{k})\mid_{(2,-2,3)} = -2\boldsymbol{i} + 4\boldsymbol{j} + 4\boldsymbol{k}$より，

$\displaystyle \boldsymbol{n} = \frac{-2\boldsymbol{i} + 4\boldsymbol{j} + 4\bol... ... + 16 + 16}} = \frac{1}{3}(-\boldsymbol{i} + 2\boldsymbol{j} + 2\boldsymbol{k})$

5. $\nabla = \boldsymbol{i}\frac{\partial}{\partial x} + \boldsymbol{j}\frac{\partial}{\partial y} + \boldsymbol{k}\frac{\partial }{\partial z}$より，

$\displaystyle \boldsymbol{i}\frac{\partial}{\partial x} (\frac{\phi}{\psi})$ $\displaystyle =$ $\displaystyle \boldsymbol{i}\frac{\phi_{x}\psi - \phi \psi_{x}}{\psi^2} = \frac{\psi \phi_{x}\boldsymbol{i} - \phi \psi_{x}\boldsymbol{i}}{\psi^2}$  
$\displaystyle \boldsymbol{j}\frac{\partial}{\partial y} (\frac{\phi}{\psi})$ $\displaystyle =$ $\displaystyle \boldsymbol{j}\frac{\phi_{y}\psi - \phi \psi_{y}}{\psi^2} = \frac{\psi \phi_{y}\boldsymbol{j} - \phi \psi_{y}\boldsymbol{j}}{\psi^2}$  
$\displaystyle \boldsymbol{k}\frac{\partial}{\partial z} (\frac{\phi}{\psi})$ $\displaystyle =$ $\displaystyle \boldsymbol{k}\frac{\phi_{z}\psi - \phi \psi_{z}}{\psi^2} = \frac{\psi \phi_{z}\boldsymbol{k} - \phi \psi_{z}\boldsymbol{k}}{\psi^2}$  

ここで， $\phi_{x}\boldsymbol{i} + \phi_{y}\boldsymbol{j} + \phi_{z}\boldsymbol{k} = \nabla \phi$, $\psi_{x}\boldsymbol{i} + \psi_{y}\boldsymbol{j} + \psi_{z}\boldsymbol{k} = \nabla \psi$であるから，

$\displaystyle \nabla \left(\frac{\phi}{\psi}\right) = \frac{\phi \nabla \psi - \psi \nabla \phi}{\phi^2}$

6.

(1) $r = \sqrt{(\xi -x)^2 + (\eta - y)^2 + (\zeta - z)^2}$， $\nabla_{Q} = \boldsymbol{i}\frac{\partial}{\partial \xi} + \boldsymbol{j}\frac{\partial}{\partial \eta} + \boldsymbol{k}\frac{\partial}{\partial \zeta}$, $\nabla_{P} = \boldsymbol{i}\frac{\partial}{\partial x} + \boldsymbol{j}\frac{\partial}{\partial y} + \boldsymbol{k}\frac{\partial}{\partial z}$とすると，

$\displaystyle \nabla_{Q}r$ $\displaystyle =$ $\displaystyle \boldsymbol{i}\frac{\partial r}{\partial \xi} + \boldsymbol{j}\frac{\partial r}{\partial \eta} + \boldsymbol{k}\frac{\partial r}{\partial \zeta}$  
  $\displaystyle =$ $\displaystyle \boldsymbol{i}\frac{2(\xi-x)}{2r} + \boldsymbol{j}\frac{2(\eta -y)}{2r} + \boldsymbol{k}\frac{2(\zeta - z)}{2r}$  
$\displaystyle \nabla_{P}r$ $\displaystyle =$ $\displaystyle \boldsymbol{i}\frac{\partial r}{\partial x} + \boldsymbol{j}\frac{\partial r}{\partial y} + \boldsymbol{k}\frac{\partial r}{\partial z}$  
  $\displaystyle =$ $\displaystyle \boldsymbol{i}\frac{-2(\xi-x)}{2r} + \boldsymbol{j}\frac{-2(\eta -y)}{2r} + \boldsymbol{k}\frac{-2(\zeta - z)}{2r} = -\nabla_{Q}r$  

(2)

$\displaystyle \nabla_{Q}(\frac{1}{r})$ $\displaystyle =$ $\displaystyle \boldsymbol{i}\frac{\partial r^{-1}}{\partial \xi} + \boldsymbol{... ...l r^{-1}}{\partial \eta} + \boldsymbol{k}\frac{\partial r^{-1}}{\partial \zeta}$  
  $\displaystyle =$ $\displaystyle \boldsymbol{i}(-r^{-2}\frac{(\xi-x)}{r}) + \boldsymbol{j}(-r^{-2}\frac{(\eta-y)}{r}) + \boldsymbol{k}(-r^{-2}\frac{(\zeta-z)}{r})$  
$\displaystyle \nabla_{P}(\frac{1}{r})$ $\displaystyle =$ $\displaystyle \boldsymbol{i}\frac{\partial r^{-1}}{\partial x} + \boldsymbol{j}... ...\partial r^{-1}}{\partial y} + \boldsymbol{k}\frac{\partial r^{-1}}{\partial z}$  
  $\displaystyle =$ $\displaystyle \boldsymbol{i}(r^{-2}\frac{(\xi-x)}{r}) + \boldsymbol{j}(r^{-2}\f... ...-y)}{r}) + \boldsymbol{k}(r^{-2}\frac{(\zeta-z)}{r}) = -\nabla_{Q}(\frac{1}{r})$  

演習問題詳解3.3 1. $\boldsymbol{r} = x\boldsymbol{i} + \boldsymbol{j} + \boldsymbol{k} = (t^2 + 1)\boldsymbol{i} + 2t^2\boldsymbol{j} + t^3 \boldsymbol{k}$より，

$\displaystyle d\boldsymbol{r} = ( 2t\boldsymbol{i} + 4t\boldsymbol{j} + 3t^2 \boldsymbol{k})dt$

また， ${\mathbf F} = 3xy\boldsymbol{i} - 5z\boldsymbol{j} + 10x \boldsymbol{k} = 3(t^2+1)(2t^2)\boldsymbol{i} - 5(t^3)\boldsymbol{j} + 10(t^2 + 1)\boldsymbol{k}$． したがって，
$\displaystyle W$ $\displaystyle =$ $\displaystyle \int_{C}{\mathbf F}\cdot d\boldsymbol{r} = \int_{1}^{2}\left(12t^3(t^2+1) - 20t^4 + 30(t^2+1)t^2\right)dt$  
  $\displaystyle =$ $\displaystyle \int_{1}^{2} (12t^5 + 10t^4 + 12t^3 + 30t^2 )dt = \left[2t^6 + 2t^5 + 3t^4 + 10t^3\right]_{1}^{2}$  
  $\displaystyle =$ $\displaystyle 2(64 -1) + 2(32 - 1) + 3(16-1) + 10(8-1) = 303$  

2.

(1) $\boldsymbol{r} = x\boldsymbol{i} + \boldsymbol{j} + \boldsymbol{k} = t^2\boldsymbol{i} + 2t\boldsymbol{j} + t^3\boldsymbol{k} (0 \leq t \leq 1)$より，

$\displaystyle d\boldsymbol{r} = 2t\boldsymbol{i} + 2\boldsymbol{j} + 3t^2\boldsymbol{k}$

また， $\phi = 2xyz^2 = 2t^2(2t)(t^3)^2 = 4t^9$．したがって，
$\displaystyle \int_{C}\phi d\boldsymbol{r}$ $\displaystyle =$ $\displaystyle \int_{0}^{1}4t^{9} (2t\boldsymbol{i} + 2\boldsymbol{j} + 3t^2\boldsymbol{k})dt$  
  $\displaystyle =$ $\displaystyle \int_{0}^{1}(8t^{10}\boldsymbol{i} + 8t^{9}\boldsymbol{j} + 12t^{... ...bol{i} + \frac{8}{10}t^{10}\boldsymbol{j} + t^{12}\boldsymbol{k}\right]_{0}^{1}$  
  $\displaystyle =$ $\displaystyle \frac{8}{11}\boldsymbol{i} + \frac{4}{5}\boldsymbol{j} + \boldsymbol{k}$  

(2) $\boldsymbol{r} = x\boldsymbol{i} + \boldsymbol{j} + \boldsymbol{k} = t^2\boldsymbol{i} + 2t\boldsymbol{j} + t^3\boldsymbol{k} (0 \leq t \leq 1)$より，

$\displaystyle d\boldsymbol{r} = 2t\boldsymbol{i} + 2\boldsymbol{j} + 3t^2\boldsymbol{k}$

また， ${\mathbf F} = xy\boldsymbol{i} - z\boldsymbol{j} + x^2 \boldsymbol{k} = 2t^3\boldsymbol{i} - t^3\boldsymbol{j} + t^4\boldsymbol{k}$．したがって，
$\displaystyle \int_{C}{\mathbf F} \times d\boldsymbol{r}$ $\displaystyle =$ $\displaystyle \int_{0}^{1}\left\vert\begin{array}{ccc} \boldsymbol{i} & \boldsy... ...dsymbol{i} - (6t^5 - 2t^5)\boldsymbol{j} + (4t^3 + 2t^4)\boldsymbol{k}\right)dt$  
  $\displaystyle =$ $\displaystyle \left[(-\frac{t^6}{2} - \frac{2t^5}{5})\boldsymbol{i} - \frac{4t^6}{6}\boldsymbol{j} + (t^4 + \frac{2t^5}{5})\boldsymbol{k}\right]_{0}^{1}$  
  $\displaystyle =$ $\displaystyle (-\frac{1}{2} - \frac{2}{5})\boldsymbol{i} - \frac{2}{3}\boldsymb... ...ac{9}{10}\boldsymbol{i} - \frac{2}{3}\boldsymbol{j} + \frac{7}{5}\boldsymbol{k}$  

3. ベクトル場${\bf A}$は ${\bf A} = -\nabla \phi$のとき，がスカラー・ポテンシャルを持つといい，そのとき， $\int_{C}{\bf A}\cdot d\boldsymbol{r} = 0$である. そこで， $\boldsymbol{r} = \nabla \phi$であるような$\phi$を求める．

$\displaystyle \nabla (\boldsymbol{r} \cdot\boldsymbol{r}) = \nabla r^2 = 2 \boldsymbol{r}$

より， $\boldsymbol{r} = \frac{1}{2} \nabla (\boldsymbol{r} \cdot\boldsymbol{r}) $．したがって，

$\displaystyle \int_{C}\boldsymbol{r}\cdot d\boldsymbol{r} = \frac{1}{2}\int_{C}\nabla (\boldsymbol{r} \cdot\boldsymbol{r})d\boldsymbol{r} = 0$

4. 力の場 ${\mathbf F}$がポテンシャル$U$をもつことより， ${\mathbf F} = -\nabla U$．これより，この質点の運動方程式は

$\displaystyle m \frac{d{\bf v}}{dt} = {\mathbf F} = -\nabla U$

また， ${\bf v} = \frac{d\boldsymbol{r}}{dt}$．そこで，

$\displaystyle m \frac{d^2{\bf v}}{dt^2} \cdot{\bf v} = -\nabla U \cdot\frac{d\boldsymbol{r}}{dt}$

これより，

$\displaystyle m{\bf v}\cdot\frac{d{\bf v}}{dt} + \frac{d\boldsymbol{r}}{dt} \cdot\nabla U = 0$

ここで， $\frac{d\boldsymbol{r}}{dt} \cdot\nabla U$を計算すると，

$\displaystyle \frac{d\boldsymbol{r}}{dt} \cdot\nabla U = (\frac{dx}{dt}\boldsym... ...al y}\frac{dy}{dt} + \frac{\partial U}{\partial z}\frac{dz}{dt} = \frac{dU}{dt}$

よって，

$\displaystyle m{\bf v}\cdot\frac{d{\bf v}}{dt} + \frac{dU}{dt} = \frac{d}{dt}(\frac{1}{2}m{\bf v}\cdot{\bf v} + U)= 0$

これより，

$\displaystyle \frac{1}{2}mv^2 + U = C (C 定数)$

したがって，

$\displaystyle \frac{1}{2}mv_{A}^2 + U(A) = \frac{1}{2}mv_{B}^2 + U(B)$

5. $xy$平面上で原点Oを中心とし，半径$a$の円を$C$とすると， $C : x = a\cos{t}, y = a\sin{t}, 0 \leq t \leq 2\pi$とパラメター化できる．これより，

$\displaystyle \nabla \phi = \frac{-\frac{y}{x^2}}{1 + (\frac{y}{x})^2}\boldsymb... ...dsymbol{j} = \frac{-\sin{t}}{a}\boldsymbol{i} + \frac{\cos{t}}{a}\boldsymbol{j}$

また，

$\displaystyle d\boldsymbol{r} = (-a\sin{t}\boldsymbol{i} + a\cos{t}\boldsymbol{j})dt$

したがって，

$\displaystyle \int_{C}(\nabla \phi) \cdot d\boldsymbol{r} = \int_{0}^{2\pi}(-\s... ...} +\cos{t}\boldsymbol{j})\cdot(-\sin{t} + \cos{t})dt = \int_{0}^{2\pi}dt = 2\pi$

演習問題詳解3.4

1.

(1) 曲面$S$を$xy$平面に正射影すると，$S$は $\Omega = \{(x,y): 2x + 2y \leq 2, x \geq 0, y \geq 0\}$に移る．また，曲面 $S : 2x + 2y + z = 2$より，対応する $\boldsymbol{r}$を位置ベクトルとすると，

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

これより，曲面$S$の法線ベクトル $\boldsymbol{r}_{x} \times \boldsymbol{r}_{y}$を求めると，

$\displaystyle \boldsymbol{r}_{x} \times \boldsymbol{r}_{y} = \left\vert\begin{a... ... -2 \end{array}\right\vert = 2\boldsymbol{i} + 2\boldsymbol{j} + \boldsymbol{k}$

ここで， $dS = \vert\boldsymbol{r}_{x} \times \boldsymbol{r}_{y}\vert dx dy = \sqrt{4 + 4 + 1}dxdy = 3dxdy$に注意すると，


$\displaystyle \int_{S}fdS$ $\displaystyle =$ $\displaystyle 3\int_{\Omega}(x^2 + 2y + z - 1)dxdy = 3\int_{0}^{1}\int_{y=0}^{1-x}(x^2 + 2y + (2-2x-2y) -1)dydx$  
  $\displaystyle =$ $\displaystyle 3\int_{0}^{1}\int_{0}^{1-x}(x^2 - 2x + 1)dydx = 3\int_{0}^{1}\int_{0}^{1-x}(x-1)^2\;dydx$  
  $\displaystyle =$ $\displaystyle 3\int_{0}^{1}\left[(x-1)^2y\right]_{0}^{1-x}dx = 3\int_{0}^{1}(1-x)^3\;dx = -3\left[\frac{1}{4}(1-x)^{4}\right]_{0}^{1} = \frac{3}{4}$  

(2) $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + (2-2x-2y)\boldsymbol{k}$とすると， $\boldsymbol{n} = \frac{\boldsymbol{r}_{x} \times \boldsymbol{r}_{y}}{\vert\bold... ...bol{r}_{y}\vert} = \frac{2\boldsymbol{i} + 2\boldsymbol{j} + \boldsymbol{k}}{3}$. よって，

$\displaystyle {\bf A} \cdot\boldsymbol{n} = (x^2 \boldsymbol{i} + z\boldsymbol{... ...rac{2\boldsymbol{i} + 2\boldsymbol{j} + \boldsymbol{k}}{3} = \frac{2x^2 + z}{3}$

したがって，
$\displaystyle \int_{S}{\bf A} \cdot\boldsymbol{n}dS$ $\displaystyle =$ $\displaystyle \int_{\Omega}\frac{2x^2 + z}{3}3dxdy = \int_{\Omega}(2x^2 + (2-2x-2y))dxdy$  
  $\displaystyle =$ $\displaystyle \int_{0}^{1}\int_{0}^{1-x}(2x^2 - 2x - 2y + 2)dy dx = \int_{0}^{1}\left[2x^2 y - 2xy -y^2 + 2y\right]_{0}^{1-x}dx$  
  $\displaystyle =$ $\displaystyle \int_{0}^{1}(2x^2(1-x) - 2x(1-x) - (1-x)^2 + 2(1-x))dx$  
  $\displaystyle =$ $\displaystyle \int_{0}^{1}(1-x)(2x^2 - 2x - (1-x) + 2)dx = \int_{0}^{1}(1-x)(2x^2 - x + 1)dx$  
  $\displaystyle =$ $\displaystyle \int_{0}^{1}(-2x^3 +3x^2 - 2x + 1)dx = \left[-\frac{x^4}{2} + x^3 - x^2 + x\right]_{0}^{1}$  
  $\displaystyle =$ $\displaystyle -\frac{1}{2} + 1 - 1 + 1 = \frac{1}{2}$  

2. 曲面$S$は$xy$平面上領域より，法線単位ベクトルは $\boldsymbol{n} = \boldsymbol{k}$となる．

$\displaystyle {\bf A} \times \boldsymbol{n} = (x\boldsymbol{i} + (x-y)\boldsymb... ...{xy}\\ 0 & 0& 1 \end{array}\right\vert = (x-y)\boldsymbol{i} - x\boldsymbol{j}$

また，$S$は$xy$平面上にあるので， $S = \Omega$，$dS = dxdy$．ここで，$\Omega$は円板より，極座標を用いると， $x = r\cos{\theta}, y = r\sin{\theta}, 0 \leq \theta \leq \frac{\pi}{2}, 0 \leq r \leq a$となり,
$\displaystyle \int_{S}{\bf A} \times \boldsymbol{n}dS$ $\displaystyle =$ $\displaystyle \int_{0}^{\pi/2}\int_{0}^{a}((r\cos{\theta} - r\sin{\theta})\boldsymbol{i} - r\cos{\theta}\boldsymbol{j})rdr d\theta$  
  $\displaystyle =$ $\displaystyle \int_{0}^{\pi/2}\left[\frac{r^3}{3}\right]_{0}^{a}\left((\cos{\theta} - \sin{\theta})\boldsymbol{i} - \cos{\theta}\boldsymbol{j}\right)d\theta$  
  $\displaystyle =$ $\displaystyle \frac{a^3}{3}\left[(\sin{\theta} + \cos{\theta}\boldsymbol{i} - \... ...3}{3}((1-1)\boldsymbol{i} - \boldsymbol{j}) = -\frac{a^3}{3}\boldsymbol{j} F3-4$  

3. 曲面$S$を$xy$平面に正射影すると$S$は $\Omega = \{(x,y):y^2 = 4, 0 \leq x \leq 3, y \geq 0\}$. 次に，${\bf r}$を位置ベクトルとすると ${\bf r} =x{\bf i} + y{\bf j} + \sqrt{4-y^2}{\bf k}$. これより，曲面$S$の法線ベクトル ${\bf r}_{x} \times {\bf r}_{y} = \frac{y}{\sqrt{4-y^2}}{\bf j} +{\bf k}$． これより，

$\displaystyle \iint_{S}(3x{\bf i} + 4z{\bf j} +2y{\bf k}) \cdot {\bf n} dS = \i... ...bf j} +2y{\bf k})\cdot ({\bf r}_{x} \times {\bf r}_{y})dxdy = \iint_{S}(6y)dxdy$

ここで，$\Omega$を縦線重合で表すと，

$\displaystyle \iint_{S}(6y)dxdy = \int_{0}^{3}\int_{0}^{2}6y dy dx = \int_{0}^{3}3y^2\mid_{0}^{2} = \int_{0}^{3}12dx = 36$

4. 曲面$S$を$yz$平面に正射影すると$S$は $\Omega = \{(y,z):y^2 = a^2, 0 \leq z \leq 1\}$. 次に，${\bf r}$を位置ベクトルとすると ${\bf r} =x{\bf i} + \sqrt{a^2 -x^2}{\bf j} + z{\bf k}$. これより，曲面$S$の法線ベクトル ${\bf r}_{z} \times {\bf r}_{x} = \frac{y}{\sqrt{4-y^2}}{\bf i} +{\bf k}$．よって，

$\displaystyle {\bf n}$ $\displaystyle =$ $\displaystyle \frac{{\bf r}_{z} \times {\bf r}_{x}}{\vert{\bf r}_{z} \times {\b... ...} = \frac{\frac{y}{\sqrt{4-y^2}}{\bf i} +{\bf k}}{\sqrt{\frac{y^2}{4-y^2} + 1}}$  
  $\displaystyle =$ $\displaystyle \frac{\frac{y}{\sqrt{4-y^2}}{\bf i} +{\bf k}}{\sqrt{\frac{4}{4-y^2}}}$  

これより，

$\displaystyle \iint_{S}(3x{\bf i} + 4z{\bf j} +2y{\bf k}) \cdot {\bf n} dS = \iint_{S}{\bf r}_{x} \times {\bf r}_{y}dxdy = \iint_{S}(6y)dxdy$

ここで，$\Omega$を縦線重合で表すと，

$\displaystyle \iint_{S}(6y)dxdy = \int_{0}^{3}\int_{0}^{2}6y dy dx = \int_{0}^{3}3y^2\mid_{0}^{2} = \int_{0}^{3}12dx = 36$

演習問題詳解3.5 基本公式 $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k}, r = \sqrt{x^2 + y^2 + z^2}$とすると，(1) $\nabla r = \frac{\boldsymbol{r}}{r} (2) \nabla r^{n} = nr^{n-1}\nabla r = nr^... ...dsymbol{A}= (\nabla \phi) \cdot\boldsymbol{A} + \phi \nabla \cdot\boldsymbol{A}$

1.

(1)

$\displaystyle \nabla \cdot(2x^2 z\boldsymbol{i} - xy^2 z \boldsymbol{j} + 3yz^2... ...ial (-xy^2)}{\partial y} + \frac{\partial (3yz^2)}{\partial z} = 4xz -2xy + 6yz$

(2)

$\displaystyle \nabla^2(3x^2 z - y^2 z^3 + 4x^2 y)$ $\displaystyle =$ $\displaystyle \frac{\partial^2 (3x^2 z - y^2 z^3 + 4x^2 y)}{\partial x^2} + \fr... ...2 y)}{\partial y^2} + \frac{\partial (3x^2 z - y^2 z^3 + 4x^2 y)}{\partial z^2}$  
  $\displaystyle =$ $\displaystyle 6z + 8y - 2z^3 -6y^2 z$  

(3)

$\displaystyle \nabla(\nabla \cdot\boldsymbol{F}) = \nabla (6xy + 4y^3 -4x^2z) = (6y-8xz)\boldsymbol{i} + (6x+12y^2)\boldsymbol{j} - 4x^2\boldsymbol{k}$

2.

(1) $\nabla r^{-3} = \frac{d r^{-3}}{dr}\nabla r = -3r^{-4}\frac{\boldsymbol{r}}{r} = -3r^{-5}\boldsymbol{r}$より， $r\nabla r^{-3} = -3r^{-4}\boldsymbol{r}$. したがって，

$\displaystyle \nabla \cdot(r \nabla r^{-3}) = \nabla \cdot(-3r^{-4}\boldsymbol{... ...oldsymbol{r}}{r} \cdot\boldsymbol{r} -3r^{-4}(3) = 12r^{-4} - 9r^{-4} = 3r^{-4}$

(2) $\nabla\cdot\left(\frac{\boldsymbol{r}}{r^2}\right) = \nabla \cdot r^{-2}\boldsy... ...boldsymbol{r}}{r} \cdot\boldsymbol{r} + r^{-2}(3) = -2r^{-2} + 3r^{-2} = r^{-2}$. したがって，

$\displaystyle \nabla^{2}\left\{\nabla\cdot\left(\frac{\boldsymbol{r}}{r^2}\right)\right\}$ $\displaystyle =$ $\displaystyle \nabla \cdot(\nabla(r^{-2})) = \nabla \cdot(-2r^{-3}\frac{\boldsymbol{r}}{r} = \nabla \cdot(-2r^{-4}\boldsymbol{r}$  
  $\displaystyle =$ $\displaystyle \nabla(-2r^{-4}) \cdot\boldsymbol{r} + -2r^{-4} \nabla \cdot\bold... ...oldsymbol{r}}{r} \cdot\boldsymbol{r} - 2r^{-4}(3) = 8r^{-4} - 6r^{-4} = 2r^{-4}$  

3. ${\bf w} = w_{1}\boldsymbol{i} + w_{2}\boldsymbol{j} + w_{3}\boldsymbol{k}$ とすると，

$\displaystyle {\bf w} \times \boldsymbol{r} = \left\vert\begin{array}{ccc} \bol... ...oldsymbol{i} + (w_{3}x -w_{1}z)\boldsymbol{j} + (w_{1}y - w_{2}x)\boldsymbol{k}$

したがって，

$\displaystyle \nabla \cdot({\bf w} \times \boldsymbol{r}) = 0$

4.

(1) 基本公式より，

$\displaystyle \nabla^2(\phi \psi)$ $\displaystyle =$ $\displaystyle \nabla \cdot\nabla(\phi \psi) = \nabla \cdot((\nabla \phi)\psi + ... ...(\nabla \psi) + (\nabla \phi)\cdot(\nabla\psi) + \phi \nabla \cdot(\nabla \psi)$  
  $\displaystyle =$ $\displaystyle \psi + (\nabla^2 \phi) + 2(\nabla \phi)\cdot(\nabla \psi) + \phi \nabla^2 \psi$  

(1) 基本公式より，

$\displaystyle \nabla \cdot(\phi \nabla \psi) = (\nabla \phi) \cdot(\nabla \psi)... ...abla \cdot(\nabla \psi) = (\nabla \phi) \cdot(\nabla \psi) + \phi \nabla^2 \psi$      

(3) (2)より， $\nabla \cdot(\phi \nabla \psi) = (\nabla \phi) \cdot(\nabla \psi) + \phi \nabla^2 \psi$. 対称性より， $\nabla \cdot(\psi \nabla \phi) = (\nabla \psi) \cdot(\nabla \phi) + \psi \nabla^2 \phi$. したがって，

$\displaystyle \nabla \cdot(\phi \nabla \psi - \psi \nabla \phi) = \phi \nabla^2 \psi - \psi \nabla^2 \phi$      

5. $\nabla \phi = \frac{\partial \phi}{\partial x}\boldsymbol{i} + \frac{\partial \... ...l z} = 2xyz^3\boldsymbol{i} + x^2 z^3 \boldsymbol{j} + 3x^2 y z^2\boldsymbol{k}$より

$\displaystyle \frac{\partial \phi}{\partial x} = 2xyz^3, \frac{\partial \phi}{\partial y} = x^2 z^3, \frac{\partial \phi}{\partial z} = 3x^2 yz^2$

これより， $\phi(x,y,z) = x^2 yz^3 + c(y,z)$．ここで， $\frac{\partial \phi}{\partial y} = x^2 z^3$を用いると， $\frac{\partial c(y,z)}{\partial y} = 0$より， $c(y,z) = c(z)$．最後に， $\frac{\partial \phi}{\partial z} = 3x^2 yz^2$より， $\frac{\partial c(z)}{\partial z} = 0$となり，$c(z) = c$．よって， $\phi(x,y,z) = x^2 yz^3 + c$. ここで，初期値 $\phi(1,-2,2) =4$より，$-16+c = 4$となり，$c = 20$.

6.

$\displaystyle (\nabla U) \times (\nabla V) = \left\vert\begin{array}{ccc} \bold... ..._{z}V_{x} - U_{x}V_{z})\boldsymbol{j} + (U_{z}V_{y} - U_{y}V_{z})\boldsymbol{k}$

したがって，
$\displaystyle \nabla \cdot((\nabla U) \times (\nabla V))$ $\displaystyle =$ $\displaystyle \frac{\partial(U_{y}V_{z} - U_{z}V_{y})}{\partial x} + \frac{\par... ...U_{x}V_{z})}{\partial y} + \frac{\partial(U_{x}V_{y} - U_{y}V_{x})}{\partial z}$  
  $\displaystyle =$ $\displaystyle U_{yx}V_{z} + U_{y}V_{zx} - U_{zx}V_{y} - U_{z}V_{yx} + U_{zy}V_{x} + U_{z}V_{xy} - U_{xy}V_{z} - U_{x}V_{zy}$  
  $\displaystyle +$ $\displaystyle U_{xz}V_{y} + U_{x}V_{yz} - U_{yz}V_{x} - U_{y}V_{xz} = 0$  

7.

$\displaystyle \frac{d\boldsymbol{r}}{dt}\cdot\nabla \phi = (\frac{dx}{dt}\bolds... ...artial \phi}{\partial z}\frac{\partial z}{\partial t} = \frac{d}{dt}\phi(x,y,z)$

8.

\includegraphics[width = 4cm]{VECANALFIG/3-5-8.eps}
合成関数の微分法より，

$\displaystyle \frac{d \phi}{dt} = \frac{\partial \phi}{\partial x}\frac{dx}{dt}... ...\nabla \phi \cdot\frac{d \boldsymbol{r}}{dt} + \frac{\partial \phi}{\partial t}$

演習問題詳解3.6

1.

(1)

$\displaystyle \nabla \times \boldsymbol{A} = \left\vert\begin{array}{ccc} \bold... ...j}(4xz-3z^3) + \boldsymbol{k}(0) = y\boldsymbol{i} + (4xz - 3z^3)\boldsymbol{j}$

(2)

$\displaystyle \nabla \times (\phi \boldsymbol{A})$ $\displaystyle =$ $\displaystyle (\nabla \phi) \times \boldsymbol{A} + \phi \nabla \times \boldsymbol{A}$  
  $\displaystyle =$ $\displaystyle \left\vert\begin{array}{ccc} \boldsymbol{i} & \boldsymbol{j} & \b... ... y} & \frac{\partial}{\partial z}\\ 2xz^2 & -yz & 3xz^3 \end{array}\right\vert$  
  $\displaystyle =$ $\displaystyle (3x^3z^4 + x^2 y^2z)\boldsymbol{i} + (2x^3yz^2 -6x^2yz^4)\boldsym... ...^2-2x^3z^3)\boldsymbol{k} + x^2yz(y\boldsymbol{i} + (4xz - 3z^3)\boldsymbol{j})$  
  $\displaystyle =$ $\displaystyle (3x^3z^4 + 2x^2y^2)\boldsymbol{i} + (6x^3 yz^2 -9x^2 yz^4)\boldsymbol{i} - (2xy^2z^2 + 2x^3z^3)\boldsymbol{k}$  

(3) (1)より， $\nabla \times \boldsymbol{A} = y\boldsymbol{i} + (4xz - 3z^3)\boldsymbol{j}$.

$\displaystyle \nabla \times (\nabla \times \boldsymbol{A})$ $\displaystyle =$ $\displaystyle \left\vert\begin{array}{ccc} \boldsymbol{i} & \boldsymbol{j} & \b... ...^3 & 0 \end{array}\right\vert = (-4x+9z^2)\boldsymbol{i} + (4z-1)\boldsymbol{k}$  

2.

$\displaystyle \nabla \times {\bf v}$ $\displaystyle =$ $\displaystyle \left\vert\begin{array}{ccc} \boldsymbol{i} & \boldsymbol{j} & \b... ...vert = (c+1)\boldsymbol{i} + (a-4)\boldsymbol{j} + (b-2)\boldsymbol{k} = {\bf0}$  

より， $a = 4, b = 2, c = -1$

3.

$\displaystyle \nabla \cdot(\boldsymbol{A} \times \boldsymbol{B})$ $\displaystyle =$ $\displaystyle (\frac{\partial}{\partial x}\boldsymbol{i} + \frac{\partial}{\par... ...\partial}{\partial z}\boldsymbol{k})\cdot(\boldsymbol{A} \times \boldsymbol{B})$  
  $\displaystyle =$ $\displaystyle \boldsymbol{i}\cdot\frac{\partial}{\partial x}(\boldsymbol{A}\tim... ...dsymbol{k}\cdot\frac{\partial}{\partial z}(\boldsymbol{A}\times \boldsymbol{B})$  
  $\displaystyle =$ $\displaystyle \boldsymbol{i}\cdot(\frac{\partial \boldsymbol{A}}{\partial x} \t... ...ldsymbol{B} + \boldsymbol{A} \times \frac{\partial \boldsymbol{B}}{\partial z})$  
  $\displaystyle =$ $\displaystyle \boldsymbol{B}\cdot(\boldsymbol{i} \times \frac{\partial \boldsym... ...\frac{\partial \boldsymbol{A}}{\partial z}) \hskip 1cm ([A B C] = [C A B])$  
  $\displaystyle -$ $\displaystyle \boldsymbol{A}\cdot(\boldsymbol{i} \times \frac{\partial \boldsym... ...ymbol{B}\cdot(\boldsymbol{k} \times \frac{\partial \boldsymbol{B}}{\partial z})$  
  $\displaystyle =$ $\displaystyle \boldsymbol{B}\cdot(\nabla \times \boldsymbol{A}) - \boldsymbol{A}\cdot(\nabla \times \boldsymbol{B}) = 0$  

(1)

$\displaystyle \nabla(\boldsymbol{A} \cdot\boldsymbol{B}) = (\boldsymbol{B}\cdot... ...la \times \boldsymbol{A}) + \boldsymbol{A}\times (\nabla \times \boldsymbol{B})$

を用いると，

$\displaystyle \nabla(\boldsymbol{C} \cdot\boldsymbol{A}) = (\boldsymbol{A}\cdot... ...la \times \boldsymbol{C}) + \boldsymbol{C}\times (\nabla \times \boldsymbol{A})$

ここで， $(\boldsymbol{A}\cdot\nabla)\boldsymbol{C} = 0, \boldsymbol{A} \times(\nabla \times \boldsymbol{C})= {\bf0}$より，

$\displaystyle \nabla(\boldsymbol{C}\cdot\boldsymbol{A}) = (\boldsymbol{C}\cdot\nabla)\boldsymbol{A} + \boldsymbol{C} \times (\nabla \times \boldsymbol{A})$

(2)

$\displaystyle \nabla \cdot(\boldsymbol{C} \times \boldsymbol{A})$ $\displaystyle =$ $\displaystyle (\frac{\partial }{\partial x}\boldsymbol{i} + \frac{\partial}{\pa... ...partial}{\partial z}\boldsymbol{k}) \cdot(\boldsymbol{C} \times \boldsymbol{A})$  
  $\displaystyle =$ $\displaystyle \boldsymbol{i}\cdot\frac{\partial}{\partial x}(\boldsymbol{C} \ti... ...symbol{k}\cdot\frac{\partial }{\partial z}\boldsymbol{C} \times \boldsymbol{A})$  
  $\displaystyle =$ $\displaystyle \boldsymbol{i}\cdot(\frac{\partial C}{\partial x} \times \boldsym... ...ldsymbol{A} + \boldsymbol{C} \times \frac{\partial \boldsymbol{A}}{\partial z})$  
  $\displaystyle =$ $\displaystyle -\boldsymbol{C} \cdot(\boldsymbol{i} \times \frac{\partial \bolds... ...mbol{C} \cdot(\boldsymbol{k} \times \frac{\partial \boldsymbol{A}}{\partial z})$  
  $\displaystyle =$ $\displaystyle -\boldsymbol{C} \cdot(\nabla \times \boldsymbol{A})$  

(3)

$\displaystyle \nabla \times (\boldsymbol{C} \times \boldsymbol{A})$ $\displaystyle =$ $\displaystyle (\boldsymbol{i}\frac{\partial }{\partial x} + \boldsymbol{j}\frac... ...l{k}\frac{\partial }{\partial z}) \times (\boldsymbol{C} \times \boldsymbol{A})$  
  $\displaystyle =$ $\displaystyle \boldsymbol{i} \times \frac{\partial (\boldsymbol{C}\times\boldsy... ...ymbol{C}\times\boldsymbol{A})}{\partial z} \hskip 0.5cm (\boldsymbol{C}は定数ベクトル)$  
  $\displaystyle =$ $\displaystyle \boldsymbol{i} \times (\boldsymbol{C} \times \frac{\partial \bold... ...ldsymbol{C})\boldsymbol{B} - (\boldsymbol{A}\cdot\boldsymbol{B})\boldsymbol{C})$  
  $\displaystyle =$ $\displaystyle (\boldsymbol{i} \cdot\frac{\partial \boldsymbol{A}}{\partial x})\... ... (\boldsymbol{k} \cdot\boldsymbol{C})\frac{\partial \boldsymbol{A}}{\partial z}$  
  $\displaystyle =$ $\displaystyle \boldsymbol{C}(\nabla \cdot\boldsymbol{A}) - (\boldsymbol{C} \cdot\nabla)\boldsymbol{A}$  

5. 公式

$\displaystyle \nabla(\boldsymbol{A} \cdot\boldsymbol{B}) = (\boldsymbol{B}\cdot... ...la \times \boldsymbol{A}) + \boldsymbol{A}\times (\nabla \times \boldsymbol{B})$

を用いると，
$\displaystyle \nabla(\boldsymbol{A} \cdot\boldsymbol{A})$ $\displaystyle =$ $\displaystyle (\boldsymbol{A}\cdot\nabla)\boldsymbol{A} + (\boldsymbol{A}\cdot\... ...la \times \boldsymbol{A}) + \boldsymbol{A}\times (\nabla \times \boldsymbol{A})$  
  $\displaystyle =$ $\displaystyle 2(\boldsymbol{A}\cdot\nabla)\boldsymbol{A} + 2\boldsymbol{A} \times(\nabla \times \boldsymbol{A})$  

よって，

$\displaystyle (\boldsymbol{A}\cdot\nabla)\boldsymbol{A} = \frac{1}{2}\nabla \vert\boldsymbol{A}\vert^2 - \boldsymbol{A} \times (\nabla \times \boldsymbol{A})$

6. $\nabla \times (\rho \boldsymbol{F}) = (\nabla \rho)\times \boldsymbol{F} + \rho(\nabla \times \boldsymbol{F})$ また， $\nabla \times (\rho \boldsymbol{F}) = \nabla \times (\nabla p) = 0$より， $\nabla \times \boldsymbol{F} = -\frac{\nabla \rho}{\rho} \times \boldsymbol{F}$．したがって，

$\displaystyle \boldsymbol{F} \cdot(\nabla \times \boldsymbol{F}) = \boldsymbol{... ...l{F}) = \frac{\nabla \rho}{\rho}\cdot(\boldsymbol{F} \times \boldsymbol{F}) = 0$

7.

$\displaystyle \nabla \times \boldsymbol{A} = \left\vert\begin{array}{ccc} \bold... ...1)\boldsymbol{i} - (3z^2 - 3z^2)\boldsymbol{j} + (6x-6x)\boldsymbol{k} = {\bf0}$

ここで， $\nabla \times \boldsymbol{A} = {\bf0}$より， $\boldsymbol{A} = -\nabla \phi$となる$\phi$が存在することに注意する．

$\displaystyle \boldsymbol{A} = (6xy + z^3)\boldsymbol{i} + (3x^2 - z)\boldsymbo... ...hi}{\partial y}\boldsymbol{j} + \frac{\partial \phi}{\partial z}\boldsymbol{k})$

より，

$\displaystyle -\frac{\partial \phi}{\partial x} = 6xy + z^3, -\frac{\partial \phi}{\partial y} = 3x^2 -z, -\frac{\partial \phi}{\partial z} = 3xz^2 - y$

まず，- $\frac{\partial \phi}{\partial x} = 6xy + z^3$より， $-\phi(x,y,z) = 3x^2y + xz^3 + f(y,z)$．ここで，$\phi$を$y$について偏微分すると，

$\displaystyle -\frac{\partial \phi}{\partial y} = 3x^2 + f_{y}(y,z)$

一方，

$\displaystyle -\frac{\partial \phi}{\partial y} = 3x^2 -z$

より，

$\displaystyle f_{y}(y,z) = -z$

これより， $f(y,z) = -yz + g(z)$となる．よって， $-\phi(x,y,z) = 3x^2 y + xz^3 - yz + g(z)$. ここで， $\phi(x,y,z)$を$z$について偏微分すると，

$\displaystyle -\frac{\partial \phi}{\partial z} = 3xz^2 - y + g'(z)$

一方，

$\displaystyle -\frac{\partial \phi}{\partial z} = 3x^2 -y$

より，

$\displaystyle g'(z) = 0$

よって，$g(z) = c$となり， $\phi(x,y,z) = -(3x^2 y + xz^3 - yz + c)$

8.

演習問題詳解4.1

1.

(1) Gaussの発散定理より，

$\displaystyle \int_{S} \frac{\boldsymbol{r}}{r^2}\cdot\boldsymbol{n}\;dS = \int_{V}\nabla \cdot\frac{\boldsymbol{r}}{r^2}\;dV$

ここで，

$\displaystyle \nabla \cdot\frac{\boldsymbol{r}}{r^2} = \nabla \cdot(r^{-2}\bold... ...c{\boldsymbol{r}}{r}\cdot\boldsymbol{r} + 3r^{-2} = -2r^{-2} + 3r^{-2} = r^{-2}$

したがって，

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

(2) $\int_{S}\boldsymbol{r} \times \boldsymbol{n}\;dS$を面積分の形に直す．任意の定ベクトル $\boldsymbol{C}$とスカラー3重積の性質を用いると

$\displaystyle \int_{S}\boldsymbol{C} \cdot(\boldsymbol{r} \times \boldsymbol{n})\;dS$ $\displaystyle =$ $\displaystyle \int_{S}\boldsymbol{n} \cdot(\boldsymbol{C} \times \boldsymbol{r})$  
  $\displaystyle =$ $\displaystyle \int_{V}\nabla \cdot(\boldsymbol{C} \times \boldsymbol{r})\;dV = ... ...rtial y} & \frac{\partial }{\partial z}\\ x & y & z \end{array}\right\vert = 0$  

ここで， $\boldsymbol{C}$は任意の定ベクトルより，

$\displaystyle \int_{S}\boldsymbol{r} \times \boldsymbol{n}\;dS = 0$

(3) Gaussの発散定理より，

$\displaystyle \int_{S} r^{n} \boldsymbol{r}\cdot\boldsymbol{n}\;dS$ $\displaystyle =$ $\displaystyle \int_{V}\nabla \cdot r^{n}\boldsymbol{r}\;dV = \int_{V} \left((\nabla r^{n})\cdot\boldsymbol{r} + r^{n}\nabla \cdot\boldsymbol{r}\right)\;dV$  
  $\displaystyle =$ $\displaystyle \int_{V} (nr^{n-1}\nabla r \cdot\boldsymbol{r} + 3r^{n})\;dV = \int_{V}(nr^{n} + 3r^{n})\;dV$  
  $\displaystyle =$ $\displaystyle (3+n)\int_{V}r^{n}\;dV$  

(4) $\int_{S}r^{n}\boldsymbol{n}\;dS$を面積分の形に直す．任意の定ベクトル $\boldsymbol{C}$を用いると

$\displaystyle \int_{S}\boldsymbol{C} \cdot(r^{n}\boldsymbol{n})\;dS$ $\displaystyle =$ $\displaystyle \int_{S}r^{n}\boldsymbol{C}\cdot\boldsymbol{n}\;dS = \int_{V}\nab... ...\int_{V}(\nabla r^{n}\cdot\boldsymbol{C} + r^{n}\nabla \cdot\boldsymbol{C})\;dV$  
  $\displaystyle =$ $\displaystyle \int_{V}nr^{n-1}\frac{\boldsymbol{r}}{r}\cdot\boldsymbol{C}\;dV = \int_{V}\boldsymbol{C} \cdot nr^{n-2}\boldsymbol{r}\;dV$  

これより，

$\displaystyle \int_{S}r^{n}\boldsymbol{n} = \int_{V}nr^{n-2}\boldsymbol{r}\;dV$

(5) $\int_{S}r^{n}\boldsymbol{r} \times \boldsymbol{n}\;dS$を面積分の形に直す．任意の定ベクトル $\boldsymbol{C}$とスカラー3重積を用いると

$\displaystyle \int_{S}\boldsymbol{C}\cdot(r^{n}\boldsymbol{r} \times \boldsymbol{n})\;dS$ $\displaystyle =$ $\displaystyle \int_{S}\boldsymbol{n} \cdot\boldsymbol{C} \times (r^{n}\boldsymb... ...l{r})\;dS = \int_{V}\nabla \cdot(\boldsymbol{C} \times r^{n}\boldsymbol{r})\;dV$  
  $\displaystyle =$ $\displaystyle \int_{V}(-\boldsymbol{C} \cdot(\nabla \times r^{n}\boldsymbol{r})... ...ot(\nabla r^{n} \times \boldsymbol{r} + r^{n} \nabla \times \boldsymbol{r})\;dV$  
  $\displaystyle =$ $\displaystyle -\int_{V}\boldsymbol{C} \cdot(nr^{n-1}\frac{\boldsymbol{r}}{r} \times \boldsymbol{r} + 0)\;dV = 0$  

(6) $\int_{S}r^{2}\boldsymbol{n}\;dS$を面積分の形に直す．任意の定ベクトル $\boldsymbol{C}$を用いると，

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

したがって，

$\displaystyle \int_{S}r^{2}\boldsymbol{n}\;dS = 2\int_{V}\boldsymbol{r}\;dV$

(7) $\int_{S}F(r)\boldsymbol{n}\;dS$を面積分の形に直す．任意の定ベクトル $\boldsymbol{C}$を用いると，

$\displaystyle \int_{S}\boldsymbol{C}\cdot(F(r)\boldsymbol{n})\;dS$ $\displaystyle =$ $\displaystyle \int_{S}F(r)\boldsymbol{C} \cdot\boldsymbol{n}\;dS = \int_{V}\nab... ...\int_{V} (\nabla F(r) \cdot\boldsymbol{C} + F(r)\nabla \cdot\boldsymbol{C})\;dV$  
  $\displaystyle =$ $\displaystyle \int_{V} \frac{dF}{dr}\frac{\boldsymbol{r}}{r} \cdot\boldsymbol{C})\;dV = \boldsymbol{C}\cdot\int_{V}\frac{dF}{dr}\frac{\boldsymbol{r}}{r}\;dV$  

したがって，

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

2. 任意の定ベクトル $\boldsymbol{C}$を用いて，面積分の形に直すと，

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

ここで， $(\nabla \times \nabla \phi = 0, \nabla \times \boldsymbol{C} = {\bf0})$に注意すると，

$\displaystyle \int_{S}\boldsymbol{C} \cdot(\boldsymbol{n}\times (\nabla \phi))\;dS = 0$

3. 曲面$S$の境界線を$C$とするので，境界線で分けられた曲面を $S_{1},S_{2}$とする．また，曲面$S_{1}$の法単位ベクトルを $\boldsymbol{n}_{1}$，$S_{2}$の法単位ベクトルを $\boldsymbol{n}_{2}$とする．このとき，曲面$S$の法単位ベクトルを $\boldsymbol{n}$とすると $\boldsymbol{n}_{1} = \boldsymbol{n}, \boldsymbol{n}_{2} = -\boldsymbol{n}$または， $\boldsymbol{n}_{1} = -\boldsymbol{n}, \boldsymbol{n}_{2} = \boldsymbol{n}$．ここで，

$\displaystyle \int_{S_{1}}\boldsymbol{A}\cdot\boldsymbol{n}_{1}\;dS + \int_{S_{... ...mbol{A}\cdot\boldsymbol{n}_{2}\;dS = \int_{V}\nabla \cdot\boldsymbol{A}\;dV = 0$

より，

$\displaystyle \int_{S_{1}}\boldsymbol{A}\cdot\boldsymbol{n}_{1}\;dS = -\int_{S_{2}}\boldsymbol{A}\cdot\boldsymbol{n}_{2}\;dS$

4.

(1) Gaussの発散定理より，

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

ここで， $\nabla \cdot(\phi\boldsymbol{A})= (\nabla \phi) \cdot\boldsymbol{A} + \phi \nabla \cdot\boldsymbol{A}$であることに注意すると，

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

(2) Gaussの発散定理より，

$\displaystyle \int_{S}(\boldsymbol{B} \times \boldsymbol{A})\cdot\boldsymbol{n}dS = \int_{V}\nabla \cdot(\boldsymbol{B} \times \boldsymbol{A})\;dV$

ここで，演習問題#よ#>り， $\nabla \cdot(\boldsymbol{B} \times \boldsymbol{A}) = \boldsymbol{A}\cdot(\nabla \times \boldsymbol{B}) - \boldsymbol{B}\cdot(\nabla \times \boldsymbol{A})$．よって，

$\displaystyle \int_{S}(\boldsymbol{B} \times \boldsymbol{A})\cdot\boldsymbol{n}... ...symbol{B})\;dV - \int_{V}\boldsymbol{B} \cdot(\nabla \times \boldsymbol{A})\;dV$

(3) Gaussの発散定理より，

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

ここで，

$\displaystyle \nabla \cdot((\nabla \phi) \times \boldsymbol{A}) = \boldsymbol{A... ...bla \times \boldsymbol{A}) = -(\nabla \phi) \cdot(\nabla \times \boldsymbol{A})$

したがって，

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

(4) Gaussの発散定理より，

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

ここで， $\boldsymbol{A} = \nabla \phi, \nabla^2 \phi = 0$より，

$\displaystyle \nabla \cdot(\phi \boldsymbol{A}) = (\nabla \phi)\cdot\boldsymbol... ... = \vert\boldsymbol{A}\vert^2 + \phi \nabla^2 \phi = \vert\boldsymbol{A}\vert^2$

したがって，

$\displaystyle \int_{S}\phi \boldsymbol{A} \cdot\boldsymbol{n}dS = \int_{V}\vert\boldsymbol{A}\vert^2\;dV$

5.

(1)
labelenshu:4-1-5-1 例題4.3のように，面積分の形に書き直す． $\frac{\partial \phi}{\partial n}$は法線単位ベクトル $\boldsymbol{n}$方向での方向微分係数より， $\frac{\partial \phi}{\partial n} = \nabla \phi \cdot\boldsymbol{n}$．よって，

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

ここで，Gaussの発散定理を用いると，

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

なお， $\nabla \cdot(\psi \nabla \phi) = (\nabla \psi)\cdot(\nabla \phi) + \psi \nabla \cdot(\nabla \phi) = (\nabla \psi)\cdot(\nabla \phi) + \psi \nabla^2 \phi$より，

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

(2) 例題4.3のように，面積分の形に書き直す． $\frac{\partial \phi}{\partial n}$は法線単位ベクトル $\boldsymbol{n}$方向での方向微分係数より， $\frac{\partial \phi}{\partial n} = \nabla \phi \cdot\boldsymbol{n}$．よって，

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

ここで，Gaussの発散定理を用いると，

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

なお， $\nabla \cdot(\phi \nabla \phi) = (\nabla \phi)\cdot(\nabla \phi) + \phi \nabla \cdot(\nabla \phi) = \vert\nabla \phi\vert^2 + \phi \nabla^2 \phi$より，

$\displaystyle \int_{S}\phi \nabla \phi \cdot\boldsymbol{n}dS = \int_{V}\{\phi \nabla^2 \phi + \vert\nabla \psi\vert^2\}\;dV$

(3) (1)の結果から(2)の結果を引くと，

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

(4) $\phi$が調和関数とは， $\nabla^2 \phi = 0$となることである．したがって，(2)を用いると,

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

(5) $\phi, \psi$が調和関数とは， $\nabla^2 \phi = \nabla^2 \psi = 0$となることである．したがって，(3)を用いると,

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

(6) $\phi$が調和関数とすると，(4)より，

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

$S$上で$\phi = 0$とすると， $\int_{V}\vert\nabla \phi\vert^2\;dV = 0$となり， $\vert\nabla \phi\vert^2 = 0$．よって， $\vert\nabla \phi\vert = 0$. すなわち $\nabla \phi = 0$．したがって，$\phi$は定数．

6. $\int_{S}{\bf A}\cdot\boldsymbol{n}dS = 0$ならば，Gaussの発散定理より，

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

したがって， $\nabla \cdot{\bf A} = 0$. 定理3.4より，${\bf A}$はベクトル・ポテンシャルを持つ．

7. $\int_{S}{\bf A} \times \boldsymbol{n}dS = {\bf0}$．ここで，任意の定ベクトル${\bf C}$を用いて，面積分の形に直すと，

$\displaystyle \int_{S}{\bf C} \cdot({\bf A} \times \boldsymbol{n})\;dS = \int_{S} \boldsymbol{n} \cdot({\bf C} \times {\bf A})\;dS$

Gaussの発散定理より，

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

ここで，，

$\displaystyle 0 = \nabla \cdot({\bf C} \times {\bf A}) = {\bf A} \cdot\nabla \t... ...C} - {\bf C} \cdot\nabla \times {\bf A} = -{\bf C} \cdot(\nabla \times {\bf A})$

したがって， $\nabla \times {\bf A} = 0$となり，${\bf A}$はスカラー・ポテンシャルをもつ．

演習問題詳解4.2

1.

1. Stokesの定理より，

$\displaystyle \int_{C}(\nabla \phi \psi)\cdot d\boldsymbol{r} = \int_{S}(\nabla \times (\nabla \phi \psi))\cdot\boldsymbol{n}dS = 0$

ここで，

$\displaystyle \nabla \phi \psi = \psi (\nabla \phi) + \phi (\nabla \psi)$

より，

$\displaystyle \int_{C}\phi(\nabla \psi)\cdot d\boldsymbol{r} = -\int_{C}\psi(\nabla \phi)\cdot d\boldsymbol{r}$

2.

(1) 線積分を $\int_{C}{\bf A}\cdot d\boldsymbol{r}$の形に直す．そこで，任意の定ベクトル${\bf C}$とスカラー３重積を用いると，

$\displaystyle \int_{C}{\bf C} \cdot\boldsymbol{r} \times d\boldsymbol{r} = \int_{C}d\boldsymbol{r} \cdot{\bf C} \times \boldsymbol{r}$

と表せる．ここで，Stokesの定理を用いると，
$\displaystyle \int_{C}d\boldsymbol{r} \cdot{\bf C} \times \boldsymbol{r}$ $\displaystyle =$ $\displaystyle \int_{S} (\nabla \times ({\bf C} \times \boldsymbol{r}))\cdot\boldsymbol{n}dS$  

さて，
$\displaystyle \nabla \times ({\bf C} \times \boldsymbol{r})$ $\displaystyle =$ $\displaystyle (\boldsymbol{i}\frac{\partial}{\partial x} + \boldsymbol{j}\frac{... ...oldsymbol{k}\frac{\partial}{\partial z}) \times ({\bf C} \times \boldsymbol{r})$  
  $\displaystyle =$ $\displaystyle \boldsymbol{i} \times \frac{\partial ({\bf C} \times \boldsymbol{... ...oldsymbol{k} \times \frac{\partial ({\bf C} \times \boldsymbol{r})}{\partial z}$  
  $\displaystyle =$ $\displaystyle \boldsymbol{i} \times ({\bf C} \times \boldsymbol{r}_{x}) + \bold... ...\boldsymbol{r}_{y}) + \boldsymbol{k} \times ({\bf C} \times \boldsymbol{r}_{z})$  
  $\displaystyle =$ $\displaystyle (\boldsymbol{i} \cdot\boldsymbol{r}_{x}) {\bf C} - (\boldsymbol{i... ...dot\boldsymbol{r}_{z}){\bf C} - (\boldsymbol{k} \cdot{\bf C})\boldsymbol{r}_{z}$  
  $\displaystyle =$ $\displaystyle (\boldsymbol{i} \cdot\boldsymbol{r}_{x} + \boldsymbol{j} \cdot\bo... ...ot{\bf C})\boldsymbol{r}_{y} + (\boldsymbol{k} \cdot{\bf C})\boldsymbol{r}_{z})$  
  $\displaystyle =$ $\displaystyle (\nabla \cdot\boldsymbol{r}){\bf C} - ({\bf C} \cdot\nabla)\boldsymbol{r} +$  

ここで，

$\displaystyle \{({\bf C} \cdot\nabla)\boldsymbol{r}\} \cdot\boldsymbol{n} = {\bf C} \cdot\nabla (\boldsymbol{r} \cdot\boldsymbol{n})$

が成り立つので，
$\displaystyle \nabla \times ({\bf C} \times \boldsymbol{r}) \cdot\boldsymbol{n}$ $\displaystyle =$ $\displaystyle (\nabla \cdot\boldsymbol{r}){\bf C} \cdot\boldsymbol{n} - ({\bf C} \cdot\nabla){\bf A} \cdot\boldsymbol{n}$  
  $\displaystyle =$ $\displaystyle {\bf C} \cdot\{(\nabla \cdot\boldsymbol{r})\boldsymbol{n} - \nabla(\boldsymbol{r} \cdot\boldsymbol{n})\}$  

これより，

$\displaystyle \int_{C}{\bf C} \cdot\boldsymbol{r} \times d\boldsymbol{r} = \int... ...t\boldsymbol{r})\boldsymbol{n} - \nabla(\boldsymbol{r} \cdot\boldsymbol{n})\}dS$

ここで，$C$は任意の定ベクトルであるから，

$\displaystyle \int_{C} \boldsymbol{r} \times d\boldsymbol{r} = \int_{S} \{(\nab... ...t\boldsymbol{r})\boldsymbol{n} - \nabla(\boldsymbol{r} \cdot\boldsymbol{n})\}dS$

最後に，

$\displaystyle \nabla (\boldsymbol{r} \cdot\boldsymbol{n}) = (\boldsymbol{n} \cd... ...bol{r}) + \boldsymbol{r} \times (\nabla \times \boldsymbol{n}) = \boldsymbol{n}$

したがって，

$\displaystyle \int_{C}\boldsymbol{r} \times d\boldsymbol{r} = \int_{S}(3\boldsymbol{n} - \boldsymbol{n})dS = 2\int_{S}\boldsymbol{n}dS$

(2) Stokesの定理を用いると，

$\displaystyle \int_{C}r^{k}\boldsymbol{r}\cdot d\boldsymbol{r}$ $\displaystyle =$ $\displaystyle \int_{S} (\nabla \times (r^{k}\boldsymbol{r}))\cdot\boldsymbol{n}dS$  

ここで，

$\displaystyle (\nabla r^{k})\times \boldsymbol{r} + r^{k} \nabla \times \boldsy... ...bol{r} + r^{k}(0) = kr^{k-2} \frac{\boldsymbol{r}}{r} \times \boldsymbol{r} = 0$

に注意すると，

$\displaystyle \int_{C}r^{k}\boldsymbol{r}\cdot d\boldsymbol{r} = 0$

(3) $C$を任意の定数ベルトルとし，Stokesの定理を用いると，

$\displaystyle \int_{C}{\bf C} \cdot\boldsymbol{r}(\nabla \phi) \cdot d\boldsymbol{r}$ $\displaystyle =$ $\displaystyle \int_{S} (\nabla \times ({\bf C} \cdot\boldsymbol{r}(\nabla \phi)))\cdot\boldsymbol{n}dS$  

ここで，
$\displaystyle \nabla \times ({\bf C} \cdot\boldsymbol{r} (\nabla \phi))$ $\displaystyle =$ $\displaystyle \nabla({\bf C} \cdot\boldsymbol{r}) \times (\nabla \phi) + ({\bf ... ...a \times \nabla \phi = \nabla({\bf C} \cdot\boldsymbol{r}) \times (\nabla \phi)$  


$\displaystyle \nabla({\bf C} \cdot\boldsymbol{r})$ $\displaystyle =$ $\displaystyle (\boldsymbol{i}\frac{\partial}{\partial x} + \boldsymbol{j}\frac{... ...al y} + \boldsymbol{k}\frac{\partial ({\bf C} \cdot\boldsymbol{r})}{\partial z}$  
  $\displaystyle =$ $\displaystyle \boldsymbol{i} ({\bf C} \cdot\frac{\partial \boldsymbol{r}}{\part... ... y}) + \boldsymbol{k} ({\bf C} \cdot\frac{\partial \boldsymbol{r}}{\partial z})$  
  $\displaystyle =$ $\displaystyle \boldsymbol{i} ({\bf C} \cdot\boldsymbol{i}) + \boldsymbol{j} ({\... ...} \cdot\boldsymbol{j}) + \boldsymbol{k} ({\bf C} \cdot\boldsymbol{k}) = {\bf C}$  

より，
$\displaystyle \int_{C}{\bf C} \cdot(\nabla (\nabla \phi))\cdot d\boldsymbol{r}$ $\displaystyle =$ $\displaystyle \int_{S} {\bf C} \times ({\bf C} \cdot\boldsymbol{r}(\nabla \phi)... ...oldsymbol{n}dS = \int_{S}({\vert bf C} \times \nabla \phi)\cdot\boldsymbol{n}dS$  
  $\displaystyle =$ $\displaystyle \int_{S}{\bf C} \cdot(\nabla \phi \times \boldsymbol{n})dS$  

したがって，

$\displaystyle \int_{C}\boldsymbol{r}(\nabla \phi)\cdot d\boldsymbol{r} = \int_{S}(\nabla \phi \times \boldsymbol{n})dS$

3. Stokesの定理より，

$\displaystyle \int_{C}\phi{\bf A}\cdot d\boldsymbol{r} = \int_{S}(\nabla \times \phi {\bf A})\cdot\boldsymbol{n}dS$

ここで，

$\displaystyle \nabla \times (\phi {\bf A}) = (\nabla \phi)\times {\bf A} + \phi \nabla \times {\bf A}$

であることに注意すると，

$\displaystyle \int_{C}\phi{\bf A}\cdot d\boldsymbol{r} = \int_{S}(\nabla \times... ...t\boldsymbol{n}dS + \int_{S}\{\phi \nabla \times {\bf A}\}\cdot\boldsymbol{n}dS$

4. 線積分を $\int_{C}{\bf A}\cdot d\boldsymbol{r}$の形に直す．そこで，任意の定ベクトル${\bf C}$とスカラー３重積，Stokesの定理を用いると，

$\displaystyle \int_{C}{\bf C} \cdot\frac{\boldsymbol{r} \times d\boldsymbol{r}}... ...}(\nabla \times \frac{{\bf C} \times \boldsymbol{r}}{r^3})\cdot\boldsymbol{n}dS$

と表せる．

ここで，ベクトル３重積を用いると

$\displaystyle \nabla \times \frac{{\bf C} \times \boldsymbol{r}}{r^3}$ $\displaystyle =$ $\displaystyle (\boldsymbol{i}\frac{\partial}{\partial x} + \boldsymbol{j}\frac{... ...k}\frac{\partial}{\partial z}) \times {\bf C} \times \frac{\boldsymbol{r}}{r^3}$  
  $\displaystyle =$ $\displaystyle \boldsymbol{i} \times {\bf C} \times \frac{\partial}{\partial x}(... ...} \times {\bf C} \times \frac{\partial}{\partial z}(\frac{\boldsymbol{r}}{r^3})$  
  $\displaystyle =$ $\displaystyle \left(\boldsymbol{i}\cdot\frac{\partial}{\partial x}(\frac{\bolds... ...dsymbol{j} \cdot{\bf C})\frac{\partial}{\partial y}(\frac{\boldsymbol{r}}{r^3})$  
  $\displaystyle +$ $\displaystyle \left(\boldsymbol{k}\cdot\frac{\partial}{\partial z}(\frac{\bolds... ...dsymbol{k} \cdot{\bf C})\frac{\partial}{\partial z}(\frac{\boldsymbol{r}}{r^3})$  
  $\displaystyle =$ $\displaystyle {\bf C}(\nabla \cdot(\frac{\boldsymbol{r}}{r^3})) - {\bf C}\cdot\nabla (\frac{\boldsymbol{r}}{r^3})$  

さらに，

$\displaystyle \nabla \cdot(\frac{\boldsymbol{r}}{r^3}) = \nabla \cdot(r^{-3}\bo... ...}\frac{\boldsymbol{r}}{r}\cdot\boldsymbol{r} + 3r^{-3} = -3r^{-3} + 3r^{-3} = 0$

より，

$\displaystyle \int_{C}{\bf C} \cdot\frac{\boldsymbol{r} \times d\boldsymbol{r}}... ...int_{S}\{{\bf C} \cdot\nabla(\frac{\boldsymbol{r}}{r^3})\}\cdot\boldsymbol{n}dS$

ここで，${\bf C}$は任意の定ベクトルであるから，

$\displaystyle \int_{C}\frac{\boldsymbol{r} \times d\boldsymbol{r}}{r^3} = -\int_{S}\nabla(\frac{\boldsymbol{r}}{r^3})\cdot\boldsymbol{n}dS$

最後に，
$\displaystyle \boldsymbol{n} \cdot\nabla (\frac{\boldsymbol{r}}{r^3})$ $\displaystyle =$ $\displaystyle (n_{1}\frac{\partial}{\partial x} + n_{2}\frac{\partial}{\partial y} + n_{3}\frac{\partial}{\partial z})(r^{-3}\boldsymbol{r})$  
  $\displaystyle =$ $\displaystyle n_{1}(\frac{\partial}{\partial x}(\frac{\boldsymbol{r}}{r^3}) + \... ...boldsymbol{r}}{r^3}) + \frac{1}{r^3}\frac{\partial \boldsymbol{r}}{\partial z})$  
  $\displaystyle =$ $\displaystyle \boldsymbol{n} \cdot\nabla(\frac{1}{r^3})\boldsymbol{r} + \frac{\... ...oldsymbol{r}}{r}\cdot\boldsymbol{n} \boldsymbol{r} + \frac{\boldsymbol{n}}{r^3}$  
  $\displaystyle =$ $\displaystyle -3\frac{\boldsymbol{n} \cdot\boldsymbol{r}}{t^5}\boldsymbol{r} + \frac{\boldsymbol{n}}{r^3}$  

5. Stokesの定理より，

$\displaystyle 0 = \int_{C}{\bf A}\cdot d\boldsymbol{r} = \int_{S}(\nabla \times {\bf A})\cdot\boldsymbol{n}dS$

よって， $\nabla \times {\bf A} = 0$．したがって，${\bf A}$はスカラー・ポテンシャルをもつ．

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