벡터 미적분 전체 문제

Let $\varphi$ be a potential for $\mathbf{F}$. Define $\phi=\varphi+ax+by+cz$. Then ${ \boldsymbol{\nabla} } \phi = { \boldsymbol{\nabla} }\varphi+(a,b,c)=\mathbf{F}+(a,b,c)$. So, $\mathbf{F}+(a,b,c)$ is also conservative.

The square has four sides, each of which is a line segment.

• On the first side, $y=0$ and $\mathrm{d}y=0$. That is, we may parametrize the first side by $\mathbf{r}(x)=x\,\hat{\boldsymbol{\imath}}$ with $0\le x\le 1$.
• On the second side, $x=1$ and $\mathrm{d}x=0$. We may parametrize the second side by $\mathbf{r}(y)=\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}$ with $0\le y\le 1$.
• On the third side, $y=1$ and $\mathrm{d}y=0$. We may parametrize the third side by $\mathbf{r}(x)=x\,\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}$ with $x$ running from $1$ to $0$.
• On the final side, $x=0$ and $\mathrm{d}x=0$. We may parametrize the fourth side by $\mathbf{r}(y)=y\,\hat{\boldsymbol{\jmath}}$ with $y$ running from $1$ to $0$.

So

$$\begin{aligned} \int_\mathcal{C} x^2y^2\,\mathrm{d}x+x^3y\,\mathrm{d}y &=\int_0^1 x^2\times 0^2\,\mathrm{d}x +\int_0^1 1^3\times y\,\mathrm{d}y +\int_1^0 x^2\times 1^2\,\mathrm{d}x +\int_1^0 0^3\times y\,\mathrm{d}y \\ &=\frac{1}{2}-\frac{1}{3} =\frac{1}{6} \end{aligned}$$

Since $\mathbf{F}$ is conservative, $\int_C\mathbf{F}\cdot\mathrm{d}\mathbf{r}=0$ over any closed curve $C$. The given curve is closed, so the integral is simply zero.

The point here is that $\mathbf{F}$ is conservative, as $\mathbf{F}=\nabla\phi$ with

$$\phi=\frac{x^2}{2}+yx-yz+\frac{z^2}{2}$$

So, for all paths from $\mathbf{r}(t_0)=(1,0,-1)$ to $\mathbf{r}(t_1)=(0,-2,3)$,

$$\begin{aligned} \int_\mathcal{C}\mathbf{F}\cdot \mathrm{d}\mathbf{r} =\phi\big(\mathbf{r}(t_1)\big)-\phi\big(\mathbf{r}(t_0)\big) &=\phi(0, -2, 3)-\phi(1,0,-1) \\ &=\left[0+0+6+\frac{9}{2}\right]-\left[\frac{1}{2}+0-0+\frac{1}{2}\right] \\ &=9\frac{1}{2} \end{aligned}$$

.

This parametrization is almost trivial. We know it will have the form $\mathbf{r}(x,y)= \psi_1(x,y)\hat{\boldsymbol{\imath}}+ \psi_2(x,y)\hat{\boldsymbol{\jmath}}+ \psi_3(x,y)\hat{\mathbf{k}}$ where $\psi_1$ gives the $x$-component (i.e. $x$), $\psi_2$ gives the $y$-component (i.e. $y$), and $\psi_3$ gives the $z$-component (i.e. $e^{x+1}+xy$). So, $\mathbf{r}(x,y)= x\hat{\boldsymbol{\imath}}+ y\hat{\boldsymbol{\jmath}}+ (e^{x+1}+xy)\hat{\mathbf{k}}$

Our parametrization is

$$\begin{aligned} x(u,v) &= u+v \\ y(u,v) &= u^2+v^2 \\ z(u,v) &= u-v \end{aligned}$$

• Adding $x(u,v)$ and $z(u,v)$ gives $x(u,v)+z(u,v) = 2u$.
• Subtracting $z(u,v)$ from $x(u,v)$ gives $x(u,v)-z(u,v) = 2v$.

So $u=\frac{1}{2}\big(x(u,v)+z(u,v)\big)$ and $v=\frac{1}{2}\big(x(u,v)-z(u,v)\big)$. So on our surface

$$\begin{aligned} y(u,v) &= u^2+v^2 = \frac{1}{4}\big(x(u,v)+z(u,v)\big)^2 +\frac{1}{4}\big(x(u,v)-z(u,v)\big)^2 \\ &= \frac{1}{2} x(u,v)^2 + \frac{1}{2} z(u,v)^2 \end{aligned}$$

All points of our surface lie on $2y= x^2+z^2$. This is a parabolic bowl:

• no points have $y<0$ and
• the $y=Y$ (with $Y> 0$) cross-section is the circle $x^2+z^2=2Y$, $y=Y$
• the $x=0$ cross-section is the parabola $2y=z^2$, $x=0$
• the $z=0$ cross-section is the parabola $2y=x^2$, $z=0$

Write $F(x,y,z) = x^2+y^2+(z-1)^2-1$ and $G(x,y,z) = x^2+y^2+(z+1)^2-1$. Let $S_1$ denote the surface $F(x,y,z)=0$ and $S_2$ denote the surface $G(x,y,z)=0$. First note that $F(0,0,0)=G(0,0,0)=0$ so that the point $(0,0,0)$ lies on both $S_1$ and $S_2$. The gradients of $F$ and $G$ are

$$\begin{aligned} { \boldsymbol{\nabla} } F(x,y,z) &=\left(\frac{\partial F}{\partial x}(x,y,z)\,,\, \frac{\partial F}{\partial y}(x,y,z)\,,\, \frac{\partial F}{\partial z}(x,y,z)\right) =\left( 2x\,,\,2y\,,\,2(z-1)\right) \\ { \boldsymbol{\nabla} } G(x,y,z) &=\left(\frac{\partial G}{\partial x}(x,y,z)\,,\, \frac{\partial G}{\partial y}(x,y,z)\,,\, \frac{\partial G}{\partial z}(x,y,z)\right) =\left( 2x\,,\,2y\,,\,2(z+1)\right) \end{aligned}$$

In particular,

$${ \boldsymbol{\nabla} } F(0,0,0)=\left( 0,0,-2\right)\qquad { \boldsymbol{\nabla} } G(0,0,0)=\left( 0,0,2\right)$$

so that the vector $\hat{\mathbf{k}}=-\frac{1}{2}{ \boldsymbol{\nabla} } F(0,0,0) =\frac{1}{2}{ \boldsymbol{\nabla} } G(0,0,0)$ is normal to both surfaces at $(0,0,0)$. So the tangent plane to both $S_1$ and $S_2$ at $(0,0,0)$ is

$$\hat{\mathbf{k}}\cdot\left(x-0,y-0,z-0\right)=0\qquad\text{or}\qquad z=0$$

Denote by $P$ the plane $z=0$. Thus $S_1$ is tangent to $P$ at $(0,0,0)$ and $P$ is tangent to $S_2$ at $(0,0,0)$. So it is reasonable to say that $S_1$ and $S_2$ are tangent at $(0,0,0)$.

No. The vector field $\mathbf{F}(x,y,z) = \hat{\boldsymbol{\imath}} +y\,\hat{\mathbf{k}}$ has

$$\begin{aligned} { \boldsymbol{\nabla} }\times \mathbf{F} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ 1 & 0 & y \end{matrix}\right] \\ &=\hat{\boldsymbol{\imath}} \end{aligned}$$

has dot product $1$ with $\mathbf{F}(x,y,z)$ (for all $x$, $y$, $z$) and so is not perpendicular to it.

(a) By the product rule

$$\begin{aligned} { \boldsymbol{\nabla} }\cdot(f\mathbf{F}) &=\frac{\partial }{\partial x}(fF_1) +\frac{\partial }{\partial y}(fF_2) +\frac{\partial }{\partial z}(fF_3) \\ &=\phantom{+}f\frac{\partial F_1}{\partial x} +f\frac{\partial F_2}{\partial y} +f\frac{\partial F_3}{\partial z} \\ &\phantom{=}+F_1\frac{\partial f}{\partial x} +F_2\frac{\partial f}{\partial y} +F_3\frac{\partial f}{\partial z} \\ &=f\,{ \boldsymbol{\nabla} }\cdot\mathbf{F}+\mathbf{F}\cdot{ \boldsymbol{\nabla} } f \end{aligned}$$

(b) Again by the product rule

$$\begin{aligned} { \boldsymbol{\nabla} }\cdot(\mathbf{F}\times\mathbf{G}) &=\frac{\partial }{\partial x}(F_2G_3-F_3G_2) +\frac{\partial }{\partial y}(F_3G_1-F_1G_3) +\frac{\partial }{\partial z}(F_1G_2-F_2G_1) \\ &=\phantom{+}\frac{\partial F_2}{\partial x}G_3 -\frac{\partial F_3}{\partial x}G_2 +\frac{\partial F_3}{\partial y}G_1 - \frac{\partial F_1}{\partial y}G_3 +\frac{\partial F_1}{\partial z}G_2 - \frac{\partial F_2}{\partial z}G_1 \\ &\phantom{=}+F_2\frac{\partial G_3}{\partial x} -F_3\frac{\partial G_2}{\partial x} +F_3\frac{\partial G_1}{\partial y} -F_1\frac{\partial G_3}{\partial y} +F_1\frac{\partial G_2}{\partial z} - F_2\frac{\partial G_1}{\partial z} \\ &=\phantom{+} \left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)G_1 +\left(\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x}\right)G_2 +\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)G_3 \\ &\phantom{=} -F_1\left(\frac{\partial G_3}{\partial y}-\frac{\partial G_2}{\partial z}\right) -F_2\left(\frac{\partial G_1}{\partial z}-\frac{\partial G_3}{\partial x}\right) -F_3\left(\frac{\partial G_2}{\partial x}-\frac{\partial G_1}{\partial y}\right) \\ &=\mathbf{G}\cdot({ \boldsymbol{\nabla} }\times\mathbf{F})- \mathbf{F}\cdot({ \boldsymbol{\nabla} }\times\mathbf{G}) \end{aligned}$$

(c) Recall that ${ \boldsymbol{\nabla} }^2(fg) = { \boldsymbol{\nabla} }\cdot\big[{ \boldsymbol{\nabla} }(fg)\big]$. First

$$\begin{aligned} { \boldsymbol{\nabla} }(fg) &=\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x}(fg) +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y}(fg) +\hat{\mathbf{k}}\frac{\partial }{\partial z}(fg) \\ &=\phantom{+}\hat{\boldsymbol{\imath}} g\frac{\partial f}{\partial x} +\hat{\boldsymbol{\jmath}} g\frac{\partial f}{\partial y} +\hat{\mathbf{k}} g\frac{\partial f}{\partial z} \\ &\phantom{=}+\hat{\boldsymbol{\imath}} f\frac{\partial g}{\partial x} +\hat{\boldsymbol{\jmath}} f\frac{\partial g}{\partial y} +\hat{\mathbf{k}} f\frac{\partial g}{\partial z} \\ &=g{ \boldsymbol{\nabla} } f+f{ \boldsymbol{\nabla} } g \end{aligned}$$

So by part (a), twice,

$$\begin{aligned} { \boldsymbol{\nabla} }^2(fg) & = { \boldsymbol{\nabla} }\cdot\big(g{ \boldsymbol{\nabla} } f\big) +{ \boldsymbol{\nabla} }\cdot\big(f{ \boldsymbol{\nabla} } g\big) \\ &= g\big({ \boldsymbol{\nabla} }\cdot { \boldsymbol{\nabla} } f\big) + \big({ \boldsymbol{\nabla} } g\big)\cdot\big({ \boldsymbol{\nabla} } f\big) +f\big({ \boldsymbol{\nabla} }\cdot{ \boldsymbol{\nabla} } g\big) + \big({ \boldsymbol{\nabla} } f\big)\cdot\big({ \boldsymbol{\nabla} } g\big) \\ &=f\,{ \boldsymbol{\nabla} }^2 g+2{ \boldsymbol{\nabla} } f\cdot{ \boldsymbol{\nabla} } g+g\,{ \boldsymbol{\nabla} }^2 f \end{aligned}$$