벡터 미적분 전체 문제

(a) Since $m=1$, Newton's law of motion gives

$$\begin{aligned} \mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{F}(t) = \hat{\boldsymbol{\jmath}} - \sin t\, \hat{\mathbf{k}} \end{aligned}$$

Integating gives

$$\mathbf{v}(t) = t\,\hat{\boldsymbol{\jmath}} +\cos t\,\hat{\mathbf{k}} +\mathbf{c}$$

for some constant vector $\mathbf{c}$. Since $\mathbf{v}(0)=\hat{\boldsymbol{\imath}}+\hat{\mathbf{k}}$, we have $\mathbf{c} = \hat{\boldsymbol{\imath}}$ so that

$$\begin{aligned} \mathbf{r}'(t) = \mathbf{v}(t) = \hat{\boldsymbol{\imath}} + t\,\hat{\boldsymbol{\jmath}} +\cos t\,\hat{\mathbf{k}} \end{aligned}$$

Integating again gives

$$\mathbf{r}(t) = t\,\hat{\boldsymbol{\imath}}+\frac{t^2}{2}\,\hat{\boldsymbol{\jmath}} +\sin t\,\hat{\mathbf{k}} +\mathbf{c}$$

for some (new) constant vector $\mathbf{c}$. Since $\mathbf{r}(0)=\hat{\boldsymbol{\jmath}}$, we have $\mathbf{c} = \hat{\boldsymbol{\jmath}}$ so that

$$\begin{aligned} \mathbf{r}(t) = t\,\hat{\boldsymbol{\imath}}+\left(1+\frac{t^2}{2}\right)\,\hat{\boldsymbol{\jmath}} +\sin t\,\hat{\mathbf{k}} \end{aligned}$$

(b) The particle has $x=\pi/2$ when $t=\pi/2$ and then

$$\begin{aligned} \mathbf{r}_1=\mathbf{r}(\pi/2) = \frac{\pi}{2}\,\hat{\boldsymbol{\imath}}+\left(1+\frac{\pi^2}{8}\right)\,\hat{\boldsymbol{\jmath}} + \hat{\mathbf{k}} \end{aligned}$$

(c) The work done between time $t=0$ and time $t=\pi/2$ is

$$\begin{aligned} \int_0^{\pi/2} \mathbf{F}(t)\cdot \mathrm{d}\mathbf{r} &=\int_0^{\pi/2} \mathbf{F}(t)\cdot \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}(t)\,\mathrm{d}t =\int_0^{\pi/2} [\hat{\boldsymbol{\jmath}}-\sin t\,\hat{\mathbf{k}}]\cdot [\hat{\boldsymbol{\imath}}+t\,\hat{\boldsymbol{\jmath}}+\cos t\,\hat{\mathbf{k}}]\,\mathrm{d}t \\ &= \int_0^{\pi/2} [t -\sin t\cos t]\,\mathrm{d}t =\Big[\frac{t^2}{2}+\frac{1}{2}\cos^2 t\Big]_0^{\pi/2} =\frac{\pi^2}{8}-\frac{1}{2} \end{aligned}$$

The circle $x^2+y^2=4y$, or equivalently, $x^2+(y-2)^2=4$, has radius $2$ and centre $(0,2)$. On the bottom surface, $z=0$ and the outward normal is $-\hat{\mathbf{k}}$, so that $$\iint _{D}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S =-\iint _{D}\mathbf{F}\cdot\hat{\mathbf{k}}\,\mathrm{d}x\mathrm{d}y =-\iint _D(2x+3y)\,\mathrm{d}x\mathrm{d}y$$ By symmetry, the centre of mass, $(\bar x,\bar y)$, of the circle is $(0,2)$. Here $\bar x$ and $\bar y$ are the average values

$$\begin{aligned} \bar x = \frac{\iint _D x\,\mathrm{d}x\mathrm{d}y}{\iint _D \mathrm{d}x\mathrm{d}y}\qquad \bar y = \frac{\iint _D y\,\mathrm{d}x\mathrm{d}y}{\iint _D \mathrm{d}x\mathrm{d}y} \end{aligned}$$

of $x$ and $y$ over $D$. As the disk $D$ has area $4\pi$,

$$\iint _D x\,\mathrm{d}x\mathrm{d}y =4\pi\bar x =0\qquad \iint _D y\,\mathrm{d}x\mathrm{d}y =4\pi\bar y =8\pi$$

and $$\iint _{D}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S =-4\pi(2\bar x+3\bar y) =-4\pi(2\times 0+3\times 2) =-24\pi$$ As

$$\begin{aligned} { \boldsymbol{\nabla} }\cdot\mathbf{F} &=\frac{\partial }{\partial x}(x+x^2y) +\frac{\partial }{\partial y}(y-xy^2) +\frac{\partial }{\partial z}(z+2x+3y) =(1+2xy)+(1-2xy)+(1)\\ &=3 \end{aligned}$$

the divergence theorem gives

$$\begin{aligned} \iint _{\mathcal{S}}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &=\iiint _R{ \boldsymbol{\nabla} }\cdot\mathbf{F}\ \mathrm{d}V -\iint _{D}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S\\ &=\iiint _R3\ \mathrm{d}V -(-24\pi) =3\text{Vol}(R)+24\pi =3\times10+24\pi \\ &=30+24\pi \end{aligned}$$

(a) We parametrize $S$ in cylindrical coordinates:

$$\mathbf{r}(r, \theta) = r\cos \theta\, \hat{\boldsymbol{\imath}} +r\sin \theta\, \hat{\boldsymbol{\jmath}} + r\,\hat{\mathbf{k}} \qquad \text{with $0\le r\le 1$, $0\le \theta \le \pi$}$$

(b) We compute

$$\begin{aligned} \frac{\partial \mathbf{r}}{\partial r} &= \cos \theta \,\hat{\boldsymbol{\imath}} +\sin \theta \,\hat{\boldsymbol{\jmath}} +\,\hat{\mathbf{k}} \\ \frac{\partial \mathbf{r}}{\partial \theta} &= -r \sin \theta \,\hat{\boldsymbol{\imath}} + r\cos \theta \,\hat{\boldsymbol{\jmath}} \\ \hat{\mathbf{n}}\, \mathrm{d}S &= \pm \frac{\partial \mathbf{r}}{\partial r}\times \frac{\partial \mathbf{r}}{\partial \theta} \ \mathrm{d}r \mathrm{d}\theta = \pm \big(-r\cos \theta \,\hat{\boldsymbol{\imath}} - r \sin \theta \,\hat{\boldsymbol{\jmath}} + r \,\hat{\mathbf{k}}\big)\, \mathrm{d}r \mathrm{d}\theta \end{aligned}$$

To calculate the downward flux, we use the minus sign. We find

$$\begin{aligned} \iint _S \mathbf{v} \cdot \hat{\mathbf{n}}\, \mathrm{d}S & = \int_0^\pi \mathrm{d}\theta \int_0^1 \mathrm{d}r\ (r \cos \theta, r\sin \theta, -2r)\cdot (r\cos \theta, r\sin \theta, -r) \\ &= \int_0^\pi \mathrm{d}\theta \int_0^1 \mathrm{d}r\ 3 r^2 = \pi r^3\Big|_{r=0}^1 = \pi \end{aligned}$$

(c) Solution 1: Let $\mathcal{P}$ be the path along line segments from $(1,0,1)$ to $(0,0,0)$ and from $(0,0,0)$ to $(-1,0,1)$. Here is a sketch. $\mathcal{P}$ is in blue.

Then

$$\int_\mathcal{C} \mathbf{F} \cdot \mathrm{d}\mathbf{r} + \int_\mathcal{P} \mathbf{F} \cdot \mathrm{d}\mathbf{r} = \iint _S { \boldsymbol{\nabla} } \times \mathbf{F} \cdot \hat{\mathbf{n}}\, \mathrm{d}S$$

by Stokes' Theorem. Along $\mathcal{P}$, the vector field $\mathbf{F}$ is orthogonal to the curve so that $\int_\mathcal{P} \mathbf{F}\cdot \mathrm{d}\mathbf{r}=0$. Note that $\nabla\times \mathbf{F}$ is the vector field $\mathbf{v}$ from part (b). Thus

$$\int_\mathcal{C} \mathbf{F}\cdot \mathrm{d}\mathbf{r} = \iint _S \mathbf{v} \cdot \hat{\mathbf{n}}\, \mathrm{d}S = \pi$$

(c) Solution 2: Let $\mathcal{L}$ be the line segment from $(1,0,1)$ to $(-1,0,1)$ and let

$$\begin{aligned} \mathcal{R}=\left\{(x,y,z)\right\}{x^2+y^2\le1, y\ge 0, z=1} \end{aligned}$$

Here is a sketch. $\mathcal{L}$ is in blue and $\mathcal{R}$ is shaded.

Then

$$\int_\mathcal{C} \mathbf{F} \cdot \mathrm{d}\mathbf{r} + \int_\mathcal{L} \mathbf{F} \cdot \mathrm{d}\mathbf{r} = \iint _{\mathcal{R}} { \boldsymbol{\nabla} } \times \mathbf{F} \cdot (-\hat{\mathbf{k}})\, \mathrm{d}S$$

by Stokes' Theorem. Along $\mathcal{L}$, the vector field $\mathbf{F}=\hat{\boldsymbol{\jmath}}$ is orthogonal to the curve (which has direction $-\hat{\boldsymbol{\imath}}$ so that $\int_\mathcal{L} \mathbf{F}\cdot \mathrm{d}\mathbf{r}=0$. Note that $\nabla\times \mathbf{F}$ is the vector field $\mathbf{v}$ from part (b). Thus

$$\int_\mathcal{C} \mathbf{F}\cdot \mathrm{d}\mathbf{r} = -\iint _\mathcal{R} \mathbf{v} \cdot \hat{\mathbf{k}}\, \mathrm{d}S = \iint _\mathcal{R} 2z \, \mathrm{d}S = 2\iint _\mathcal{R} \, \mathrm{d}S =2\,\text{Area}(\mathcal{R}) = \pi$$