벡터 미적분 전체 문제

(a) The curve is $\mathbf{r}(\theta) = x(\theta)\,\hat{\boldsymbol{\imath}}+ y(\theta)\,\hat{\boldsymbol{\jmath}}$ with $x(\theta)=r(\theta)\cos\theta$, $y(\theta)=r(\theta)\sin\theta$ and $\theta_1\le \theta\le \theta_2$. On this curve

$$\begin{aligned} \mathbf{v}(\theta) =\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}\theta}(\theta)&= x'(\theta)\hat{\boldsymbol{\imath}}+ y'(\theta)\hat{\boldsymbol{\jmath}} =\big[r'(\theta)\cos\theta-r(\theta)\sin\theta\big]\hat{\boldsymbol{\imath}}+ \big[r'(\theta)\sin\theta+r(\theta)\cos\theta\big]\hat{\boldsymbol{\jmath}}\\ \implies \frac{\mathrm{d}s}{\mathrm{d}\theta}(\theta) &=\sqrt{\big[r'(\theta)\cos\theta-r(\theta)\sin\theta\big]^2 + \big[r'(\theta)\sin\theta+r(\theta)\cos\theta\big]^2}\\ &=\sqrt{r'(\theta)^2+r(\theta)^2} \end{aligned}$$

Hence

$$\begin{aligned} \int_\mathcal{C} f(x,y)\,ds &= \int_{\theta_1}^{\theta_2}\!\!\!f\big(x(\theta), y(\theta)\big) \frac{\mathrm{d}s}{\mathrm{d}\theta}\,\mathrm{d}\theta \\ &=\int_{\theta_1}^{\theta_2}\!\!\!f\big(r(\theta)\cos\theta, r(\theta)\sin\theta\big) \sqrt{r(\theta)^2 +\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}(\theta)\right)^2}\, \mathrm{d}\theta \end{aligned}$$

(b) In this case $f(x,y)=1$, $r(\theta)=1+\cos\theta$, $\theta_1=0$ and $\theta_2=2\pi$,

$$\begin{aligned} \int_C \,ds &=\int_0^{2\pi} \sqrt{[1+\cos\theta]^2+[-\sin\theta]^2}\,\mathrm{d}\theta =\int_0^{2\pi} \sqrt{2(1+\cos\theta)}\,\mathrm{d}\theta\\ &=\int_0^{2\pi} \sqrt{4\cos^2\frac{\theta}{2}}\,\mathrm{d}\theta =2\int_0^{2\pi} \left|\cos\frac{\theta}{2}\right|\,\mathrm{d}\theta =4\int_0^{\pi} \cos\frac{\theta}{2}\ \mathrm{d}\theta =8 \sin\frac{\theta}{2}\bigg|_0^{\pi} =8 \end{aligned}$$

a. If $\mathbf{F}+\mathbf{G}$ is conservative for any particular $\mathbf{F}$ and $\mathbf{G}$, then by definition, there exists a potential $\varphi$ with $\mathbf{F}+\mathbf{G} = { \boldsymbol{\nabla} } \varphi$.

Since $\mathbf{F}$ is conservative, there also exists a potential $\psi$ with $\mathbf{F} = { \boldsymbol{\nabla} } \psi$.

But now $\mathbf{G} = (\mathbf{F}+\mathbf{G})-\mathbf{F}={ \boldsymbol{\nabla} } \varphi - { \boldsymbol{\nabla} } \psi = { \boldsymbol{\nabla} }(\varphi-\psi)$. That means the function $(\varphi-\psi)$ is a potential for $G$. However, this is impossible: since $\mathbf{G}$ is non-conservative, no function with this property exists.

So it is not possible that $\mathbf{F}+\mathbf{G}$ is conservative. It must be non-conservative. b. Counterexample: if $\mathbf{F} = -\mathbf{G}$, then $\mathbf{F}+\mathbf{G} = \mathbf 0 = { \boldsymbol{\nabla} } c$ for any constant $c$. c. Since both fields are conservative, they both have potentials, say $\mathbf{F}={ \boldsymbol{\nabla} } \varphi$ and $\mathbf{G} = { \boldsymbol{\nabla} } \psi$. Then $\mathbf{F}+\mathbf{G} = { \boldsymbol{\nabla} }\varphi+{ \boldsymbol{\nabla} }\psi={ \boldsymbol{\nabla} }(\varphi+\psi)$. That is, $(\varphi+\psi)$ is a potential for $\mathbf{F}+\mathbf{G}$, so $\mathbf{F}+\mathbf{G}$ is conservative.

Denote by $S$ the surface $G(x,y,z)=0$ and by $C$ the parametrized curve $\mathbf{r}(t)=\big(x(t),y(t),z(t)\big)$. To start, we'll find the tangent plane to $S$ at $\mathbf{r}_0$ and the tangent line to $C$ at $\mathbf{r}_0$.

• The tangent vector to $C$ at $\mathbf{r}_0$ is $\left( x'(t_0)\,,\,y'(t_0)\,,\,z'(t_0) \right)$, so the parametric equations for the tangent line to $C$ at $\mathbf{r}_0$ are

$$x-x_0 = t x'(t_0)\qquad y-y_0 = t x'(t_0)\qquad z-z_0 = t x'(t_0) \tag{$E_1$}$$

• The gradient $\left(\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\right)$ is a normal vector to the surface $S$ at $(x_0,y_0,z_0)$. So the tangent plane to the surface $S$ at $(x_0,y_0,z_0)$ is

$$\left(\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\right) \cdot \left( x-x_0\,,\, y-y_0\,,\,z-z_0\right) = 0$$

or

$$\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ (x-x_0) +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ (y-y_0) +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ (z-z_0) = 0 \tag{$E_2$}$$

Next, we'll show that the tangent vector $\left( x'(t_0)\,,\,y'(t_0)\,,\,z'(t_0) \right)$ to $C$ at $\mathbf{r}_0$ and the normal vector $\left(\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\right)$ to $S$ at $\mathbf{r}_0$ are perpendicular to each other. To do so, we observe that, for every $t$, the point $\big(x(t),y(t),z(t)\big)$ lies on the surface $G(x,y,z)=0$ and so obeys

$$\begin{aligned} G\big(x(t),y(t),z(t)\big) =0 \end{aligned}$$

Differentiating this equation with respect to $t$ gives, by the chain rule,

$$\begin{aligned} 0&= \frac{\mathrm{d}}{\mathrm{d}t}G\big(x(t),y(t),z(t)\big) \\ &=\frac{\partial G}{\partial x}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ x'(t) +\frac{\partial G}{\partial y}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ y'(t) +\frac{\partial G}{\partial z}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ z'(t) \end{aligned}$$

Then setting $t=t_0$ gives

$$\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ x'(t_0) +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ y'(t_0) +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ z'(t_0) = 0 \tag{$E_3$}$$

Finally, we are in a position to show that if $(x,y,z)$ is any point on the tangent line to $C$ at $\mathbf{r}_0$, then $(x,y,z)$ is also on the tangent plane to $S$ at $\mathbf{r}_0$. As $(x,y,z)$ is on the tangent line to $C$ at $\mathbf{r}_0$ then there is a $t$ such that, by $(E_1)$,

$$\begin{aligned} &\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ \{x-x_0\} +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ \{y-y_0\} +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ \{z-z_0\} \\ &=\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ \big\{ t\,x'(t_0)\big\} +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ \big\{ t\,y'(t_0)\big\} +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ \big\{ t\,z'(t_0)\big\}\\ &= t\left[\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ x'(t_0) +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ y'(t_0) +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ z'(t_0) \right] =0 \end{aligned}$$

by $(E_3)$. That is, $(x,y,z)$ obeys the equation, $(E_2)$, of the tangent plane to $S$ at $\mathbf{r}_0$ and so is on that tangent plane. So the tangent line to $C$ at $\mathbf{r}_0$ is contained in the tangent plane to $S$ at $\mathbf{r}_0$.

(a) Expressing the left hand side as an iterated integral, with $z$ as the innermost integration variable, we have

$$\begin{aligned} \iiint _V\frac{\partial f}{\partial z}(x,y,z)\ \mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z &=\int_0^1\mathrm{d}x\int_0^1\mathrm{d}y\left[\int_0^1\mathrm{d}z\ \frac{\partial f}{\partial z}(x,y,z)\right] \\ &=\int_0^1\mathrm{d}x\int_0^1\mathrm{d}y\ \big[f(x,y,1)-f(x,y,0)\big] \\ &\text{by the fundamental theorem of calculus} \\ &=\iint _R\big[f(x,y,1)-f(x,y,0)\big]\ \mathrm{d}x\,\mathrm{d}y \\ &=\iint _R f(x,y,1)\ \mathrm{d}x\,\mathrm{d}y - \iint _R f(x,y,0)\ \mathrm{d}x\,\mathrm{d}y \end{aligned}$$

(b) Define the vector field $\mathbf{F}(x,y,z) = f(x,y,z)\,\hat{\mathbf{k}}$. Then the divergence of $\mathbf{F}$ is ${ \boldsymbol{\nabla} }\cdot\mathbf{F}(x,y,z) = \frac{\partial f}{\partial z}(x,y,z)$. The boundary of the cube $V$ is the union of six faces

$$\begin{aligned} S_1&=\big\{ \ (x,y,z)\ \big|\ 0\le x\le 1,\ 0\le y\le 1,\ z=1\ \big\}\quad \text{with outward normal $\hat{\mathbf{n}}=\hat{\mathbf{k}}$} \\ S_2&=\big\{ \ (x,y,z)\ \big|\ 0\le x\le 1,\ 0\le y\le 1,\ z=0\ \big\}\quad \text{with outward normal $\hat{\mathbf{n}}=-\hat{\mathbf{k}}$} \\ S_3&=\big\{ \ (x,y,z)\ \big|\ 0\le x\le 1,\ 0\le z\le 1,\ y=1\ \big\}\quad \text{with outward normal $\hat{\mathbf{n}}=\hat{\boldsymbol{\jmath}}$} \\ S_4&=\big\{ \ (x,y,z)\ \big|\ 0\le x\le 1,\ 0\le z\le 1,\ y=0\ \big\}\quad \text{with outward normal $\hat{\mathbf{n}}=-\hat{\boldsymbol{\jmath}}$} \\ S_5&=\big\{ \ (x,y,z)\ \big|\ 0\le y\le 1,\ 0\le z\le 1,\ x=1\ \big\}\quad \text{with outward normal $\hat{\mathbf{n}}=\hat{\boldsymbol{\imath}}$} \\ S_6&=\big\{ \ (x,y,z)\ \big|\ 0\le y\le 1,\ 0\le z\le 1,\ x=0\ \big\}\quad \text{with outward normal $\hat{\mathbf{n}}=-\hat{\boldsymbol{\imath}}$} \end{aligned}$$