벡터 미적분 전체 문제

The surface

$$\begin{aligned} S&=\big\{ \ (x,y,z)\ \big|\ -1\le x\le 1,\ -1\le y\le 1,\ z\ge 0,\ z = (1-x^2)(1-y^2)\ \big\} \\ &=\big\{ \ (x,y,z)\ \big|\ -1\le x\le 1,\ -1\le y\le 1,\ \ z = (1-x^2)(1-y^2)\ \big\} \end{aligned}$$

Note that when $x=1$ or $x=-1$ or $y=1$ or $y=-1$, we have $z= (1-x^2)(1-y^2)=0$. So the boundary of $S$, call it $C$, is the boundary of the square $-1\le x,y\le 1$, $z=0$, oriented counterclockwise. Here is a sketch of $C$.

Apply Stokes' theorem. Observing that $z=0$ on $C$ so that $\mathbf{F} = -y\,\hat{\boldsymbol{\imath}} +x^3\,\hat{\boldsymbol{\jmath}}$,

$$\begin{aligned} \iint _S{ \boldsymbol{\nabla} }\times\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &=\oint_C\mathbf{F}\cdot \mathrm{d}\mathbf{r} =\oint_C[-y\,\hat{\boldsymbol{\imath}}+x^3\hat{\boldsymbol{\jmath}}]\cdot \mathrm{d}\mathbf{r}\\ &=\underbrace{\int_{-1}^1 -(-1)\,\mathrm{d}x}_{y=-1{\rm\ side}} +\underbrace{\int_{-1}^1 (1)^3\,\mathrm{d}y}_{x=1{\rm\ side}} + \underbrace{\int^{-1}_1 -(1)\,\mathrm{d}x}_{y=1{\rm\ side}} + \underbrace{\int^{-1}_1 (-1)^3\,\mathrm{d}y}_{x=-1{\rm\ side}} \\ &=8 \end{aligned}$$

(a) Note that all three vertices, $(2,0,0)$, $(0,2,0)$ and $0,0,2)$, lie in the plane $x+y+z=2$. So the entire path lies in that plane too.

In part (b) we will need to evaluate a line integral that clearly cannot be computed directly — we will need to use Stokes' theorem. So let's use Stokes's theorem in part (a) too. First, we find

$$\begin{aligned} { \boldsymbol{\nabla} }\times\mathbf{F} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} &\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}} \\ \tfrac{\partial}{\partial x} & \tfrac{\partial}{\partial y} & \tfrac{\partial}{\partial z} \\ z^2 & x^2 & y^2 \end{matrix} \right] =2y\,\hat{\boldsymbol{\imath}} + 2z\,\hat{\boldsymbol{\jmath}} + 2x\,\hat{\boldsymbol{\imath}} \end{aligned}$$

Let $S$ be the triangular surface that is contained in the plane $x+y+z=2$ and is bounded by $L_1$, $L_2$ and $L_3$. Orient $S$ by the normal vector $\hat{\mathbf{n}} = \frac{1}{\sqrt{3}}(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+\hat{\mathbf{k}})$. Then,

$${ \boldsymbol{\nabla} }\times\mathbf{F}\cdot\hat{\mathbf{n}} =\frac{1}{\sqrt{3}}\big(2y\,\hat{\boldsymbol{\imath}} + 2z\,\hat{\boldsymbol{\jmath}} + 2x\,\hat{\boldsymbol{\imath}}\big)\cdot(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+\hat{\mathbf{k}}) =\frac{2}{\sqrt{3}}(x+y+z)$$

and, by Stokes' theorem,

$$\begin{aligned} \oint_C\mathbf{F}\cdot \mathrm{d}\mathbf{r} &=\iint _S{ \boldsymbol{\nabla} }\times\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S =\frac{2}{\sqrt{3}}\iint _S(x+y+z)\,\mathrm{d}S =\frac{4}{\sqrt{3}}\iint _S\mathrm{d}S =\frac{4}{\sqrt{3}}\text{Area}(S) \end{aligned}$$

The triangle $S$ is half of the prallelogram with sides $(0,2,0)-(2,0,0) =(-2,2,0)$ and $(0,0,2)-(2,0,0)=(-2,0,2)$. The area of the parallelogram is

$$\begin{aligned} \big|(-2,2,0)\times (-2,0,2)\big| =\big|(4,4,4)\big| =4\sqrt{3} \end{aligned}$$

So

$$\begin{aligned} \oint_C\mathbf{F}\cdot \mathrm{d}\mathbf{r} = \frac{4}{\sqrt{3}} 2\sqrt{3} =8 \end{aligned}$$

(b) Let $\tilde S$ be the specified surface. Then, as in part (a),

$$\begin{aligned} \oint_C\mathbf{F}\cdot \mathrm{d}\mathbf{r} &=\iint _{\tilde S}{ \boldsymbol{\nabla} }\times\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S =\frac{2}{\sqrt{3}}\iint _{\tilde S}(x+y+z)\,\mathrm{d}S =\frac{4}{\sqrt{3}}\iint _{\tilde S}\mathrm{d}S =\frac{4}{\sqrt{3}}\text{Area}(\tilde S) \\ &=4\sqrt{3} \end{aligned}$$

We are to evaluate the line integral of a complicated vector field around a relatively complicated closed curve. That certainly suggests that we should not try to evaluate the integral directly. To see if Stokes' theorem looks promising, let's compute the curl

$$\begin{aligned} { \boldsymbol{\nabla} }\times\mathbf{F} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} &\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}} \\ \tfrac{\partial}{\partial x} & \tfrac{\partial}{\partial y} & \tfrac{\partial}{\partial z} \\[0.05in] -y+e^x\sin x & y^4 & \sqrt{z}\tan z \end{matrix} \right] =\hat{\mathbf{k}} \end{aligned}$$

That's suggestive. Next we need to find a surface whose boundary is $C$. First, here is a sketch of $C$.

We can choose the surface $S$ to be the union of two flat parts:

• the quadralateral $Q$ in the $yz$-plane with vertices $(0,0,0)$, $(0,1,1)$, $(0,1,2)$ and $(0,2,0)$ and
• the triangle $T$ in the $xy$-plane with vertices $(0,0,0)$, $(0,2,0)$ and $(2,2,0)$.

The normal to $Q$ is $-\hat{\boldsymbol{\imath}}$ and the normal to $T$ is $-\hat{\mathbf{k}}$. Then Stokes' theorem gives

$$\begin{aligned} \int_C\mathbf{F}\cdot\mathrm{d}\mathbf{r} &=\iint _S { \boldsymbol{\nabla} }\times\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S \\ &=\iint _Q \hat{\mathbf{k}}\cdot(-\hat{\boldsymbol{\imath}})\,\mathrm{d}S + \iint _T \hat{\mathbf{k}}\cdot(-\hat{\mathbf{k}})\,\mathrm{d}S \\ &= - \iint _T \mathrm{d}S \\ &=-\text{Area}(T) \\ &= -\tfrac{1}{2}\overbrace{2}^{\text{base}}\overbrace{2}^{\text{height}} \\ &=-2 \end{aligned}$$

We are to evaluate the line integral of a complicated vector field around a relatively complicated closed curve. That certainly suggests that we should not try to evaluate the integral directly. As we are to use Stokes' theorem, let's compute the curl

$$\begin{aligned} { \boldsymbol{\nabla} }\times\mathbf{F} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} &\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}} \\ \tfrac{\partial}{\partial x} & \tfrac{\partial}{\partial y} & \tfrac{\partial}{\partial z} \\ x \sin y & - y \sin x & (x - y)z^2 \end{matrix} \right] =-z^2\,\hat{\boldsymbol{\imath}}-z^2\,\hat{\boldsymbol{\jmath}} -(y\cos x +x\cos y)\hat{\mathbf{k}} \end{aligned}$$

Next we need to find a surface whose boundary is $C$. First, here is a sketch of $C$.

We can choose the surface $S$ to be the union of two flat parts:

• the rectangle $S_x$ in the $xz$-plane with vertices $(0,0,0)$, $({\pi}/{2},0,0)$, $({\pi}/{2},0,1)$ and $(0,0,1)$ and
• the rectangle $S_y$ in the $yz$-plane with vertices $(0,0,0)$, $(0,0,1)$, $(0,{\pi}/{2},1)$ and $(0,{\pi}/{2},0)$

The normal to $S_x$ is $-\hat{\boldsymbol{\jmath}}$ and the normal to $S_y$ is $-\hat{\boldsymbol{\imath}}$. Then Stokes' theorem gives

$$\begin{aligned} \int_C\mathbf{F}\cdot\mathrm{d}\mathbf{r} &=\iint _S { \boldsymbol{\nabla} }\times\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S \\ &=\iint _{S_x} { \boldsymbol{\nabla} }\times\mathbf{F}\cdot(-\hat{\boldsymbol{\jmath}})\,\mathrm{d}S + \iint _{S_y} { \boldsymbol{\nabla} }\times\mathbf{F}\cdot(-\hat{\boldsymbol{\imath}})\,\mathrm{d}S \\ &=\int_0^{{\pi}/{2}} \mathrm{d}x\int_0^1\mathrm{d}z\ z^2 + \int_0^{{\pi}/{2}} \mathrm{d}y\int_0^1\mathrm{d}z\ z^2 \\ &=\int_0^{{\pi}/{2}} \mathrm{d}x\ \frac{1}{3} + \int_0^{{\pi}/{2}} \mathrm{d}y\ \frac{1}{3} \\ &=\frac{\pi}{3} \end{aligned}$$

(a) Here is a sketch.

(b) We are to evaluate the line integral of a complicated vector field around a relatively complicated closed curve. That certainly suggests that we should not try to evaluate the integral directly. Let's try Stokes' theorem. First, we compute the curl

$$\begin{aligned} { \boldsymbol{\nabla} }\times\mathbf{F} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} &\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}} \\ \tfrac{\partial}{\partial x} & \tfrac{\partial}{\partial y} & \tfrac{\partial}{\partial z} \\ \frac{2z}{1+y}+\sin(x^2) & \frac{3z}{1+x}+\sin(y^2) & 5(x+1)(y+2) \end{matrix} \right] \\ &= \Big(5(x+1)-\frac{3}{1+x}\Big)\,\hat{\boldsymbol{\imath}} -\Big(5(y+2)-\frac{2}{1+y}\Big)\,\hat{\boldsymbol{\jmath}} +\Big(-\frac{3z}{(1+x)^2}+\frac{2z}{(1+y)^2}\Big)\,\hat{\mathbf{k}} \end{aligned}$$

Next we need to find a surface $S$ whose boundary is $C$. We can choose the surface $S$ to be the union of two flat parts:

• the triangle $S_x$ in the $xz$-plane with vertices $(0,0,0)$, $(2,0,0)$, and $(0,0,2)$ and
• the triangle $S_y$ in the $yz$-plane with vertices $(0,0,0)$, $(0,0,2)$, and $(0,3,0)$

Note that

• The normal to $S_x$ specified by Stokes' theorem is $-\hat{\boldsymbol{\jmath}}$. On $S_x$ we have $y=0$, so that ${ \boldsymbol{\nabla} }\times\mathbf{F}\cdot\hat{\boldsymbol{\jmath}}$ simplifies to $-\big(5(0+2)-\frac{2}{1+0}\big)=-8$.
• The normal to $S_y$ specified by Stokes' theorem is $-\hat{\boldsymbol{\imath}}$. On $S_y$ we have $x=0$, so that ${ \boldsymbol{\nabla} }\times\mathbf{F}\cdot\hat{\boldsymbol{\imath}}$ simplifies to $\big(5(0+1)-\frac{3}{1+0}\big)=2$.

So Stokes' theorem gives

$$\begin{aligned} \int_C\mathbf{F}\cdot\mathrm{d}\mathbf{r} &=\iint _S { \boldsymbol{\nabla} }\times\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S =\iint _{S_x} \overbrace{{ \boldsymbol{\nabla} }\times\mathbf{F}\cdot(-\hat{\boldsymbol{\jmath}})}^{8}\,\mathrm{d}S + \iint _{S_y} \overbrace{{ \boldsymbol{\nabla} }\times\mathbf{F}\cdot(-\hat{\boldsymbol{\imath}})}^{-2}\,\mathrm{d}S \\ &=8\,\text{Area}(S_x) -2\,\text{Area}(S_y) =8\frac{1}{2}(2)(2) -2\frac{1}{2}(3)(2) \\ &=10 \end{aligned}$$