벡터 미적분 전체 문제

(a) In the sketch below, the point $(x,y,z)$ deviates from the centre $(2,2,4)$ by $\sin\theta$ units in the $\hat{\mathbf{k}}$ direction, and by $\cos\theta$ units in the $\sqrt \frac1{\sqrt 2}(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})$ direction. So, $(x,y,z)=(2+\frac1{\sqrt 2}\cos\theta~,~2+\frac1{\sqrt 2}\cos\theta~,~4+\sin\theta)$.

So, we can parametrize the circle as $(x,y,z)=(2+\frac1{\sqrt 2}\cos\theta~,~2+\frac1{\sqrt 2}\cos\theta~,~4+\sin\theta)$, with $0 \le \theta \le 2\pi$.

Remark: it's easy to check that this equation satisfies the two properties we desire. Since the $x$- and $y$ coordinates match, it's in the plane $x=y$. To check that it's a circle centred at $(2,2,4)$, we note the distance from $(x,y,z)$ to $(2,2,4)$ is:

$$\begin{aligned} d&=\sqrt{(x-2)^2+(y-2)^2+(z-4)^2}=\sqrt{\left(\frac1{\sqrt 2}\cos\theta\right)^2+\left(\frac1{\sqrt 2}\cos\theta\right)^2+\left(\sin\theta\right)^2}\\ &=\sqrt{\frac12\cos^2\theta+\frac12\cos^2\theta+\sin^2\theta}=\sqrt{\cos^2\theta+\sin^2\theta}=1 \end{aligned}$$

So, our points all have distance one from the same point — that is, they lie on a circle of radius 1.

(b) Consider a point $(x,y,z)=(2+\frac1{\sqrt 2}\cos\theta~,~2+\frac1{\sqrt 2}\cos\theta~,~4+\sin\theta)$, rotating $\phi$ radians about the line $x=y=4$.

The new position of the point has the same height, $z=4+\sin\theta$. Its distance from the line $x=y=4$ is also preserved: $R=\sqrt{(x-4)^2+(y-4)^2+(z-z)^2}=\sqrt{(\frac1{\sqrt 2}\cos\theta-2)^2+(\frac1{\sqrt 2}\cos\theta-2)^2+0)}=\cos\theta-2\sqrt2$.

The circle traced out by a point $(x,y,z)=(2+\frac1{\sqrt2}\cos\theta,2+\frac{1}{\sqrt 2} \cos \theta,4+\sin\theta)$ on the circle is centred at $(4,4,z)$ with radius $\sqrt2(4-x)$, so it has equation $x=4+\sqrt2(2-\sqrt2\cos\theta)\cos\phi$, $y=4+\sqrt2(2-\sqrt2\cos\theta)\sin\phi$, $z=4\sin\theta$.

We shall show that $\frac{\partial G_3}{\partial y}-\frac{\partial G_2}{\partial z}=F_1$. The other components are similar. First we have

$$\begin{aligned} t\,\mathbf{F}\big(\mathbf{r}(t)\big)\times\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}(t) &=t\,\mathbf{F}\big(tx,ty,tz\big)\times\big(x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}\big) =t\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ F_1 & F_2 & F_3\\ x & y & z\end{matrix}\right] \end{aligned}$$

Reading off the $\hat{\mathbf{k}}$ and $\hat{\boldsymbol{\jmath}}$ components of the determinant gives

$$\begin{aligned} G_3(x,y,z)&=\int_0^1 t\big[F_1\big(tx,ty,tz\big)\,y-F_2\big(tx,ty,tz\big)\,x\big]\,dt\\ G_2(x,y,z)&=\int_0^1 t\big[F_3\big(tx,ty,tz\big)\,x-F_1\big(tx,ty,tz\big)\,z\big]\,dt \end{aligned}$$

So

$$\begin{aligned} \frac{\partial G_3}{\partial y} &=\int_0^1 t\Big[F_1\big(tx,ty,tz\big)+ \frac{\partial F_1}{\partial y}\big(tx,ty,tz\big)\,ty -\frac{\partial F_2}{\partial y}\big(tx,ty,tz\big)\,tx\Big]\,dt\\ \frac{\partial G_2}{\partial z}&=\int_0^1 t\Big[ \frac{\partial F_3}{\partial z}\big(tx,ty,tz\big)\,tx -\frac{\partial F_1}{\partial z}\big(tx,ty,tz\big)\,tz -F_1\big(tx,ty,tz\big)\Big]\,dt\\ \Rightarrow \frac{\partial G_3}{\partial y}-\frac{\partial G_2}{\partial z} &=\int_0^1 \Big[2t\,F_1\big(tx,ty,tz\big)+ t^2y\,\frac{\partial F_1}{\partial y}\big(tx,ty,tz\big) +t^2z\frac{\partial F_1}{\partial z}\big(tx,ty,tz\big)\\ & -t^2x\,\frac{\partial F_2}{\partial y}\big(tx,ty,tz\big) -t^2x\,\frac{\partial F_3}{\partial z}\big(tx,ty,tz\big)\Big]\,dt \end{aligned}$$

Since, by hypothesis, ${ \boldsymbol{\nabla} }\cdot\mathbf{F}=\frac{\partial F_1}{\partial x} +\frac{\partial F_2}{\partial y} +\frac{\partial F_3}{\partial z}=0$, the last two terms

$$\begin{aligned} -t^2x\Big\{\frac{\partial F_2}{\partial y}\big(tx,ty,tz\big) +\frac{\partial F_3}{\partial z}\big(tx,ty,tz\big)\Big\} =-t^2x\Big\{-\frac{\partial F_1}{\partial x}\big(tx,ty,tz\big)\Big\} \end{aligned}$$

so that

$$\begin{aligned} &\frac{\partial G_3}{\partial y}-\frac{\partial G_2}{\partial z} \\ &=\int_0^1 \Big[2tF_1\big(tx,ty,tz\big)+ t^2x\frac{\partial F_1}{\partial x}\big(tx,ty,tz\big) +t^2y\frac{\partial F_1}{\partial y}\big(tx,ty,tz\big) +t^2z\frac{\partial F_1}{\partial z}\big(tx,ty,tz\big)\Big]\,dt\\ &=\int_0^1 \frac{d}{dt}\Big[ t^2 F_1(tx,ty,tz)\Big]\,dt =\Big[ t^2 F_1(tx,ty,tz)\Big]^{t=1}_{t=0} = F_1(x,y,z) \end{aligned}$$

By the divergence theorem

$$\iint _{\partial V}\mathbf{r}\cdot\hat{\mathbf{n}}\, \mathrm{d}S =\iiint _V{ \boldsymbol{\nabla} }\cdot\mathbf{r}\,\mathrm{d}V =\iiint _V{ \boldsymbol{\nabla} }\cdot(x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}})\,\mathrm{d}V =\iiint _V 3\,\mathrm{d}V =3\,\text{Volume}(V)$$

Our gemetric explanation starts with the observation that the volume of the cone with vertex $(0,0,0)$ and base a tiny piece of surface $\mathrm{d}S$ is $\frac{1}{3}$ times the area of the base times the height of the cone. The height of the cone is $|\hat{\mathbf{n}}\cdot\mathbf{r}|$, where $\mathbf{r}$ is a point in $\mathrm{d}S$. So the volume of the cone is $\frac{1}{3}|\hat{\mathbf{n}}\cdot\mathbf{r}|\,\mathrm{d}S$.

First assume that $(0,0,0)$ is in $V$ and $V$ is convex. Then

• $\hat{\mathbf{n}}\cdot\mathbf{r}>0$, and the volume is $\frac{1}{3} \hat{\mathbf{n}}\cdot\mathbf{r} \,\mathrm{d}S$.
• the cone is contained in $V$ and
• $V$ is the union of all the tiny conical pieces with $\mathrm{d}S$ running over $\partial V$.

So

$$\text{Volume}(V)=\frac{1}{3}\iint _{\partial V}\mathbf{r}\cdot\hat{\mathbf{n}}\, \mathrm{d}S$$

To generalise to the case that $V$ is not convex or $(0,0,0)$ is not in $V$, write $V$ as the difference between a large convex solid and one or more smaller convex solids.

Pressure is force per unit surface area acting normally into a surface. So the force per unit surface area is $-p\hat{\mathbf{n}}$. The total force acting on $S$ is

$$-\iint _S p\hat{\mathbf{n}}\,\mathrm{d}S=-\iiint _E \nabla p\ \mathrm{d}V$$

We are assuming that $p$ is a constant, so that $\nabla p=0$ and the total force is zero.