벡터 미적분 전체 문제

(a) Parametrize $C$ by $x$. When the first component of a point on the curve is $x$, then the second component, $y$, must be $x^2$ and the third component, $z$, must be $x^3$. So

$$\begin{aligned} \mathbf{r}(x)&=x\,\hat{\boldsymbol{\imath}}+x^2\,\hat{\boldsymbol{\jmath}}+x^3\,\hat{\mathbf{k}}\qquad 0\le x\le 1\\ \mathbf{r}'(x)&=\hat{\boldsymbol{\imath}}+2x\,\hat{\boldsymbol{\jmath}}+3x^2\,\hat{\mathbf{k}}\\ \frac{\mathrm{d}s}{\mathrm{d}x}(x)&=\sqrt{1+4x^2+9x^4} \end{aligned}$$

and

$$\begin{aligned} \rho(x)\ \frac{\mathrm{d}s}{\mathrm{d}x}(x) &=\big(8x+36x^3\big)\sqrt{1+4x^2+9x^4}\\ \int_C\rho\ \mathrm{d}s &=\int_0^1 \big(8x+36x^3\big)\sqrt{1+4x^2+9x^4}\,\mathrm{d}x \end{aligned}$$

Substituting $u=1+4x^2+9x^4$, $\mathrm{d}u = \big(8x+36x^3\big)\,\mathrm{d}x$, $u(0)=1$, $u(1)=14$,

$$\begin{aligned} \int_C\rho\ \mathrm{d}s &=\int_1^{14} \sqrt{u}\ \mathrm{d}u =\frac{2}{3}u^{3/2}\bigg|_1^{14}\\ &=\frac{2}{3}\big[14^{3/2}-1\big]\approx 34.26 \end{aligned}$$

(b) Since $\mathbf{F}(x,y,z)=\nabla f(x,y,z)$ with $f(x,y,z)=x\sin y+yz+\tfrac{1}{2} z^2$,

$$\int_C \mathbf{F}\cdot \mathrm{d}\mathbf{r}=f(1,1,1)-f(0,0,0) =\sin 1+\frac{3}{2} \approx 2.3415$$

The potential $f$ was just guessed. Alternatively, it can be found by solving

$$\begin{aligned} \frac{\partial f}{\partial x}(x,y,z) &= \sin y \\ \frac{\partial f}{\partial y}(x,y,z) &= x\cos y+z \\ \frac{\partial f}{\partial z}(x,y,z) &= y+z \end{aligned}$$

Integrating the first of these equations gives

$$f(x,y,z) = x\sin y +g(y,z)$$

Substituting this into the second equation gives

$$x\cos y+\frac{\partial g}{\partial y}(y,z) = x\cos y+z\qquad\text{or}\qquad \frac{\partial g}{\partial y}(x,z)=z$$

which forces

$$g(y,z) = yz + h(z)$$

Finally, substituting $f(x,y,z) = x\sin y +yz + h(z)$ into the last equation gives

$$y + h'(z) =y+z \quad\text{or}\quad h'(z) = z$$

So $h(x) = \frac{z^2}{2}+C$ and hence $f(x,y,z) = x\sin y + yz + \frac{z^2}{2}+C$ for any constant $C$.

(a) Let's evaluate the integral directly using the parametrization

$$\mathbf{r}(x) = x\,\hat{\boldsymbol{\imath}} +(9-x^2)\,\hat{\boldsymbol{\jmath}}$$

with $-3\le x \le 3$.

Since $\mathbf{r}'(x) = \hat{\boldsymbol{\imath}} -2x\,\hat{\boldsymbol{\jmath}}$,

$$\begin{aligned} \int_C (x^2 + y)\,\mathrm{d}x + x\,\mathrm{d}y &=\int_{-3}^3 \big(x^2\ +\ \overbrace{9-x^2}^{y}\ +x\overbrace{(-2x)}^{\frac{\mathrm{d}y}{\mathrm{d}x}}\big)\,\mathrm{d}x =\int_{-3}^3 \big(9-2x^2\big)\,\mathrm{d}x \\ &=2\int_0^3 \big(9-2x^2\big)\,\mathrm{d}x =2\left(27-2\frac{3^3}{3}\right) =18 \end{aligned}$$

(b) In this solution, we'll evaluate the integral directly. Label the four sides of the square $L_1$, $L_2$, $L_3$ and $L_4$ as in the figure

The parametrization of $L_1$ by arc length is $\mathbf{r}(s) = s\,\hat{\boldsymbol{\imath}}$, $0\le s\le 1$. As the outward pointing normal to $L_1$ is $-\hat{\boldsymbol{\jmath}}$,

$$\begin{aligned} \int_{L_1} \mathbf{F} \cdot\hat{\mathbf{n}}\, \mathrm{d}s =\int_0^1 \mathbf{F}(s,0) \cdot(-\hat{\boldsymbol{\jmath}})\, \mathrm{d}s =\int_0^1 (-0)\, \mathrm{d}s =0 \end{aligned}$$

The parametrization of $L_2$ by arc length is $\mathbf{r}(s) = \hat{\boldsymbol{\imath}}+s\,\hat{\boldsymbol{\jmath}}$, $0\le s\le 1$. As the outward pointing normal to $L_2$ is $\hat{\boldsymbol{\imath}}$,

$$\begin{aligned} \int_{L_2} \mathbf{F} \cdot\hat{\mathbf{n}}\, \mathrm{d}s =\int_0^1 \mathbf{F}(1,s) \cdot \hat{\boldsymbol{\imath}}\, \mathrm{d}s =\int_0^1 2\, \mathrm{d}s =2 \end{aligned}$$

The parametrization of $L_3$ by arc length (starting at $(1,1)$) is $\mathbf{r}(s) = (1-s)\,\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}$, $0\le s\le 1$. As the outward pointing normal to $L_3$ is $\hat{\boldsymbol{\jmath}}$,

$$\begin{aligned} \int_{L_3} \mathbf{F} \cdot\hat{\mathbf{n}}\, \mathrm{d}s =\int_0^1 \mathbf{F}(1-s,1) \cdot\hat{\boldsymbol{\jmath}}\, \mathrm{d}s =\int_0^1 e^{1-s}\, \mathrm{d}s =\Big[-e^{1-s}\Big]_0^1 =e-1 \end{aligned}$$

The parametrization of $L_4$ by arc length (starting at $(0,1)$) is $\mathbf{r}(s) = (1-s)\,\hat{\boldsymbol{\jmath}}$, $0\le s\le 1$. As the outward pointing normal to $L_4$ is $-\hat{\boldsymbol{\imath}}$,

$$\begin{aligned} \int_{L_2} \mathbf{F} \cdot\hat{\mathbf{n}}\, \mathrm{d}s =\int_0^1 \mathbf{F}(0,1-s) \cdot(- \hat{\boldsymbol{\imath}})\, \mathrm{d}s =\int_0^1 (-0)\, \mathrm{d}s =0 \end{aligned}$$

All together

$$\begin{aligned} \int_C \mathbf{F} \cdot\hat{\mathbf{n}}\, \mathrm{d}s &= \int_{L_1} \mathbf{F} \cdot\hat{\mathbf{n}}\, \mathrm{d}s +\int_{L_2} \mathbf{F} \cdot\hat{\mathbf{n}}\, \mathrm{d}s +\int_{L_3} \mathbf{F} \cdot\hat{\mathbf{n}}\, \mathrm{d}s +\int_{L_4} \mathbf{F} \cdot\hat{\mathbf{n}}\, \mathrm{d}s \\ &= 0+2+(e-1) +0 =e+1 \end{aligned}$$

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 $2y=x^2$, $z=0$

Note that, since $x^2+y^2=1+2z^2$ on $S$, the condition $z\ge 1$ is equivalent to $x^2+y^2\ge 3$, $z\ge 0$. So the hyperboloid is $\big\{ \ (x,y,z)\ \big|\ x^2+y^2=1+2z^2,\ 3\le x^2+y^2\le 9, z\ge 0\ \big\}$.

(a) No. Under this parametrization, the condition $3\le x^2+y^2\le 9$ is $3\le u^2+v^2\le 9$, not $2\le u^2+v^2\le 9$.

(b) Yes. Under this parametrization, $x=u\sin v$, $y=-u\cos v$ and $z=\sqrt{\frac{u^2}{2}-\frac{1}{2}}$. So

• $x^2+y^2-2z^2 = u^2-2\left(\frac{u^2}{2}-\frac{1}{2}\right) = 1$, as desired.
• The condition $x^2+y^2\le 9$ is equivalent to $u\le 3$, since $u\ge 0$.
• The condition $x^2+y^2\ge 3$ is equivalent to $u\ge\sqrt{3}$, since $u\ge 0$.
• $z=\sqrt{\frac{u^2}{2}-\frac{1}{2}}\ge 0$

(c) Yes. Under this parametrization, $x=\sqrt{1+2v^2}\cos u$, $y=\sqrt{1+2v^2}\sin u$ and $z=v$. So

• $x^2+y^2-2z^2 = 1+2v^2 -2v^2 = 1$, as desired.
• The condition $x^2+y^2\le 9$ is equivalent to $1+2v^2\le 9$, which is equivalent to $v\le 2$, since $v\ge 0$.
• The condition $x^2+y^2\ge 3$ is equivalent to $1+2v^2\ge 3$, which is equivalent to $v\ge 1$, since $v\ge 0$.
• $z=v\ge 0$