선적분

(a) For $\mathbf{F}$ to be conservative, it must pass the screening test

$$\begin{aligned} \mathbf{0} = { \boldsymbol{\nabla} }\times\mathbf{F} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} &\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ (x - a)ye^x & xe^x + z^3 & byz^2 \end{matrix} \right] \\ &= \big(bz^2-3z^2)\,\hat{\boldsymbol{\imath}} -\big(0-0\big)\,\hat{\boldsymbol{\jmath}} +\big(e^x+xe^x-(x-a)e^x\big)\,\hat{\mathbf{k}} \end{aligned}$$

This is the case if and only if $b=3$ and $a=-1$

(b) Set $a=-1$ and $b=3$. For $f$ to be a potential for $\mathbf{F}$, it must obey

$$\begin{aligned} \frac{\partial f}{\partial x}(x,y,z) &= (x + 1)ye^x \\ \frac{\partial f}{\partial y}(x,y,z) &= xe^x + z^3 \\ \frac{\partial f}{\partial z}(x,y,z) &= 3yz^2 \end{aligned}$$

Integrating the second of these equations gives

$$f(x,y,z) = xye^x + yz^3 +g(x,z)$$

Substituting this into the last equation gives

$$3yz^2+\frac{\partial g}{\partial z}(x,z) = 3yz^2\qquad\text{or}\qquad \frac{\partial g}{\partial z}(x,z)=0$$

which forces

$$g(x,z) = h(x)$$

Finally, substituting $f(x,y,z) = xye^x + yz^3 + h(x)$ into the first equation gives

$$xye^x + ye^x + h'(x) =(x + 1)ye^x \quad\text{or}\quad h'(x) = 0$$

So $h(x) = C$ and hence $f(x,y,z) = xye^x + yz^3 + C$ works for any constant $C$.

(c) Since $\mathbf{F}={ \boldsymbol{\nabla} } f$,

$$\begin{aligned} \int_C\mathbf{F}\cdot\mathrm{d}\mathbf{r} &=\int_C{ \boldsymbol{\nabla} } f\cdot\mathrm{d}\mathbf{r} =f\big(\mathbf{r}(\pi)) - f(\big(\mathbf{r}(0)\big) =f(\pi,1,-1)-f(0,1,1) \\ &=\big[\pi e^\pi -1 \big]-\big[ 1\big] =\pi e^\pi-2 \end{aligned}$$

(d) Since

$$\int_C\mathbf{F}\cdot\mathrm{d}\mathbf{r} =\int_C (x + 1)ye^x \,\mathrm{d}x + (xe^x + z^3 ) \,\mathrm{d}y + 3yz^2 \,\mathrm{d}z$$

we have

$$\begin{aligned} I &= \int_C\mathbf{F}\cdot\mathrm{d}\mathbf{r} + \int_C yz^2\mathrm{d}z \\ &= \pi e^\pi-2 +\int_0^\pi \overbrace{(\cos 2t)}^{y} \overbrace{\cos^2 t}^{z^2} \overbrace{(-\sin t)\,\mathrm{d}t}^{\mathrm{d}z} \\ &= \pi e^\pi-2 +\int_0^\pi (2\cos^2t-1) \cos^2 t (-\sin t)\,\mathrm{d}t \\ &= \pi e^\pi-2 +\int_1^{-1} (2u^2-1)u^2\,\mathrm{d}u \qquad\text{with } u=\cos t,\ \mathrm{d}u=-\sin t\,\mathrm{d}t \\ &= \pi e^\pi-2 +\Big[\frac{2u^5}{5} -\frac{u^3}{3}\Big]_{1}^{-1}\\ &= \pi e^\pi-2 +\Big[-\frac{4}{5} +\frac{2}{3}\Big] \\ &= \pi e^\pi -\frac{32}{15} \end{aligned}$$

(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

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

(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

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