11. 벡터장과 선적분Vector Fields and Line Integrals

Solution. Parametrize the segment $C$ as $\mathbf{r}(t)=t\,\mathbf{i}+t\,\mathbf{j}+t\,\mathbf{k}$ $(0\le t\le 1)$; then $\mathbf{r}'(t)=(1,1,1)$, so $|\mathbf{r}'(t)|=\sqrt{3}$.

Since $f(\mathbf{r}(t))=t-3t^2+t=2t-3t^2$,

$$\int_C f\,ds=\int_0^1 (2t-3t^2)\sqrt{3}\,dt=\sqrt{3}\left[t^2-t^3\right]_0^1=\sqrt{3}(1-1)=0.$$

Solution. Parametrize $C_1$ as $\mathbf{r}_1(t)=t\,\mathbf{i}+t\,\mathbf{j}$ $(0\le t\le 1)$; then $\mathbf{r}_1'(t)=(1,1,0)$, $|\mathbf{r}_1'(t)|=\sqrt2$, and $f(\mathbf{r}_1(t))=t-3t^2$, so

$$\int_{C_1} f\,ds=\int_0^1 (t-3t^2)\sqrt2\,dt=\sqrt2\left[\frac{t^2}{2}-t^3\right]_0^1=\sqrt2\left(\frac12-1\right)=-\frac{\sqrt2}{2}.$$

Parametrize $C_2$ as $\mathbf{r}_2(t)=\mathbf{i}+\mathbf{j}+t\,\mathbf{k}$ $(0\le t\le 1)$; then $\mathbf{r}_2'(t)=(0,0,1)$, $|\mathbf{r}_2'(t)|=1$, and $f(\mathbf{r}_2(t))=1-3+t=t-2$, so

$$\int_{C_2} f\,ds=\int_0^1(t-2)\,dt=\left[\frac{t^2}{2}-2t\right]_0^1=\frac12-2=-\frac32.$$

Hence

$$\int_{C_1\cup C_2} f\,ds=\int_{C_1}f\,ds+\int_{C_2}f\,ds=-\frac{\sqrt2}{2}-\frac32.$$

Solution. Since $\mathbf{r}'(t)=(-\sin t,\cos t,1)$, we get $|\mathbf{r}'(t)|=\sqrt{\sin^2t+\cos^2t+1}=\sqrt2$. Since $f(\mathbf{r}(t))=2\cos t\sin t+\sqrt t=\sin 2t+\sqrt t$,

$$\int_C f\,ds=\sqrt2\int_0^\pi(\sin 2t+\sqrt t)\,dt=\sqrt2\left(\left[-\frac{\cos 2t}{2}\right]_0^\pi+\left[\frac23 t^{3/2}\right]_0^\pi\right).$$

Here $\left[-\dfrac{\cos 2t}{2}\right]_0^\pi=-\dfrac{\cos2\pi}{2}+\dfrac{\cos0}{2}=0$ and $\left[\dfrac23t^{3/2}\right]_0^\pi=\dfrac23\pi^{3/2}$, so

$$\int_C f\,ds=\sqrt2\cdot\frac23\pi^{3/2}=\frac{2\sqrt2}{3}\pi^{3/2}.$$

Solution.

(1) Since $x=t^2,\ y=t^3$, we have $dx/dt=2t$, $dy/dt=3t^2$, and $ds=\sqrt{(2t)^2+(3t^2)^2}\,dt=\sqrt{4t^2+9t^4}\,dt=t\sqrt{4+9t^2}\,dt$ (as $t\gt0$ on $[1,2]$). Also $\dfrac{x^2}{y^{4/3}}=\dfrac{t^4}{(t^3)^{4/3}}=\dfrac{t^4}{t^4}=1$, so

$$\int_C \frac{x^2}{y^{4/3}}\,ds=\int_1^2 t\sqrt{4+9t^2}\,dt.$$

Substituting $u=4+9t^2$, $du=18t\,dt$, and noting $u=13$ at $t=1$, $u=40$ at $t=2$,

$$\int_1^2 t\sqrt{4+9t^2}\,dt=\frac{1}{18}\int_{13}^{40}\sqrt u\,du=\frac{1}{18}\cdot\frac23\left[u^{3/2}\right]_{13}^{40}=\frac{1}{27}\left(40^{3/2}-13^{3/2}\right)=\frac{80\sqrt{10}-13\sqrt{13}}{27}$$

(using $40^{3/2}=40\sqrt{40}=80\sqrt{10}$).

(2) Since $x=0,\ y=t,\ z=2-2t$, we have $dx/dt=0$, $dy/dt=1$, $dz/dt=-2$, so $ds=\sqrt{0+1+4}\,dt=\sqrt5\,dt$. Also $xy+y+z=0+t+(2-2t)=2-t$, so

$$\int_C(xy+y+z)\,ds=\sqrt5\int_0^1(2-t)\,dt=\sqrt5\left[2t-\frac{t^2}{2}\right]_0^1=\sqrt5\left(2-\frac12\right)=\frac{3\sqrt5}{2}.$$

Solution. The gradient field of $T=100-x^2-y^2-z^2$ is

$$\nabla T=(-2x,-2y,-2z)=-2(x,y,z).$$

At the point $(x,y,z)$, $\nabla T$ is opposite in direction to the position vector $(x,y,z)$, so it always points toward the origin, and its magnitude $|\nabla T|=2\sqrt{x^2+y^2+z^2}$ grows in proportion to the distance from the origin. The level surfaces of $T$ are spheres $x^2+y^2+z^2=100-T$ centered at the origin, so $\nabla T$ is perpendicular to these spheres and points in the direction in which $T$ increases, i.e., toward the center (the origin). The figure below depicts this vector field restricted to the plane $z=0$: the farther a point is from the origin, the longer its arrow.

Solution. For a point $(x,y)\neq(0,0)$ in the plane, let $r=\sqrt{x^2+y^2}$ be its distance from the origin. The unit vector pointing toward the origin is $-\dfrac{(x,y)}{r}$, and since the magnitude of $\mathbf F$ is inversely proportional to $r^2$, we have $|\mathbf F|=\dfrac{c}{r^2}$ for some constant $c\gt0$. Hence

$$\mathbf F=|\mathbf F|\cdot\left(-\frac{(x,y)}{r}\right)=-\frac{c}{r^2}\cdot\frac{(x,y)}{r}=-\frac{c\,(x,y)}{r^3},$$

that is,

$$\mathbf F=-\frac{c\,x}{(x^2+y^2)^{3/2}}\,\mathbf i-\frac{c\,y}{(x^2+y^2)^{3/2}}\,\mathbf j\qquad(c\gt0\text{ an arbitrary constant})$$

is the complete family of vector fields satisfying the condition.

Solution.

(1) $\mathbf F=-\dfrac{y}{\sqrt{x^2+y^2}}\mathbf i+\dfrac{x}{\sqrt{x^2+y^2}}\mathbf j=\dfrac1r(-y,x)$ (where $r=\sqrt{x^2+y^2}$) has constant magnitude $|\mathbf F|=\dfrac{\sqrt{y^2+x^2}}{r}=1$ at every point, and points in the direction obtained by rotating the position vector $(x,y)$ by $90^\circ$ counterclockwise — the tangent direction to the circle centered at the origin through that point. That is, $\mathbf F$ is a vector field of constant magnitude $1$ that circulates counterclockwise about the origin.

(2) $\mathbf F=x\,\mathbf i+y\,\mathbf j$ is simply the position vector, so at every point it points outward, away from the origin, with magnitude $|\mathbf F|=\sqrt{x^2+y^2}=r$ equal to the distance from the origin. That is, $\mathbf F$ radiates outward from the origin, and its arrows grow longer with distance.

Solution.

(1) Since $\mathbf r(t)=t^2\,\mathbf i+t\,\mathbf j+\sqrt t\,\mathbf k$ $(0\le t\le1)$, we have $\mathbf r'(t)=\left(2t,\,1,\,\dfrac{1}{2\sqrt t}\right)$. Since $x=t^2,\ y=t,\ z=\sqrt t$,

$$\mathbf F(\mathbf r(t))=z\,\mathbf i+xy\,\mathbf j-y^2\,\mathbf k=\sqrt t\,\mathbf i+t^3\,\mathbf j-t^2\,\mathbf k,$$

and

$$\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)=\sqrt t\cdot 2t+t^3\cdot1+(-t^2)\cdot\frac{1}{2\sqrt t}=2t^{3/2}+t^3-\frac12t^{3/2}=\frac32t^{3/2}+t^3.$$

Hence

$$\int_C\mathbf F\cdot d\mathbf r=\int_0^1\left(\frac32t^{3/2}+t^3\right)dt=\left[\frac32\cdot\frac25t^{5/2}+\frac{t^4}{4}\right]_0^1=\frac35+\frac14=\frac{17}{20}.$$

(2) Since $\mathbf r(t)=\cos t\,\mathbf i+\sin t\,\mathbf j+t\,\mathbf k$ $(0\le t\le2\pi)$, we have $x=\cos t,\ y=\sin t,\ z=t$, and $dx=-\sin t\,dt,\ dy=\cos t\,dt,\ dz=dt$. Hence

$$-y\,dx+z\,dy+2x\,dz=\sin^2t\,dt+t\cos t\,dt+2\cos t\,dt,$$

so

$$\int_C-y\,dx+z\,dy+2x\,dz=\int_0^{2\pi}\sin^2t\,dt+\int_0^{2\pi}t\cos t\,dt+\int_0^{2\pi}2\cos t\,dt.$$

Now $\displaystyle\int_0^{2\pi}\sin^2t\,dt=\pi$, $\displaystyle\int_0^{2\pi}t\cos t\,dt=\big[t\sin t+\cos t\big]_0^{2\pi}=(0+1)-(0+1)=0$, and $\displaystyle\int_0^{2\pi}2\cos t\,dt=\big[2\sin t\big]_0^{2\pi}=0$, so

$$\int_C-y\,dx+z\,dy+2x\,dz=\pi+0+0=\pi.$$

Solution.

(1) Parametrize $C_1$ as $\mathbf r(t)=t\,\mathbf i+t\,\mathbf j+t\,\mathbf k$ $(0\le t\le1)$; then $\mathbf r'(t)=(1,1,1)$ and $\mathbf F(\mathbf r(t))=3t\,\mathbf i+2t\,\mathbf j+4t\,\mathbf k$. Hence $\mathbf F\cdot\mathbf r'(t)=3t+2t+4t=9t$, so

$$\int_{C_1}\mathbf F\cdot d\mathbf r=\int_0^1 9t\,dt=\frac92.$$

(2) Parametrize $C_2$ as $\mathbf r(t)=t\,\mathbf i+t^2\,\mathbf j+t^4\,\mathbf k$ $(0\le t\le1)$; then $\mathbf r'(t)=(1,2t,4t^3)$ and $\mathbf F(\mathbf r(t))=3t^2\,\mathbf i+2t\,\mathbf j+4t^4\,\mathbf k$. Hence

$$\mathbf F\cdot\mathbf r'(t)=3t^2\cdot1+2t\cdot2t+4t^4\cdot4t^3=3t^2+4t^2+16t^7=7t^2+16t^7,$$

so

$$\int_{C_2}\mathbf F\cdot d\mathbf r=\int_0^1\left(7t^2+16t^7\right)dt=\left[\frac{7}{3}t^3+2t^8\right]_0^1=\frac73+2=\frac{13}{3}.$$

Solution. Since $x=\cos t,\ y=\sin t,\ z=-\cos t$ $(0\le t\le2\pi)$, we have $dx=-\sin t\,dt$, $dy=\cos t\,dt$, $dz=\sin t\,dt$.

(1) $xz\,dx=\cos t\cdot(-\cos t)\cdot(-\sin t)\,dt=\cos^2t\sin t\,dt$, so

$$\int_C xz\,dx=\int_0^{2\pi}\cos^2t\sin t\,dt=\left[-\frac{\cos^3t}{3}\right]_0^{2\pi}=-\frac13+\frac13=0.$$

(2) $xz\,dy=\cos t\cdot(-\cos t)\cdot\cos t\,dt=-\cos^3t\,dt$, so

$$\int_C xz\,dy=-\int_0^{2\pi}\cos^3t\,dt=0$$

($\cos^3t$ integrates to $0$ over a full period).

(3) $xyz\,dz=\cos t\cdot\sin t\cdot(-\cos t)\cdot\sin t\,dt=-\cos^2t\sin^2t\,dt=-\dfrac14\sin^2(2t)\,dt$, so

$$\int_C xyz\,dz=-\frac14\int_0^{2\pi}\sin^2(2t)\,dt=-\frac14\cdot\pi=-\frac{\pi}{4}$$

(the integral of $\sin^2(2t)$ over one period equals $\pi$).

Solution.

(1) Parametrize the curve as $\mathbf r(t)=t\,\mathbf i+t^2\,\mathbf j+t^3\,\mathbf k$ $(0\le t\le1)$; then $\mathbf r'(t)=(1,2t,3t^2)$. Since $x=t,\ y=t^2,\ z=t^3$,

$$\mathbf F(\mathbf r(t))=(y-x^2)\,\mathbf i+(z-y^2)\,\mathbf j+(x-z^2)\,\mathbf k=(t^2-t^2)\,\mathbf i+(t^3-t^4)\,\mathbf j+(t-t^6)\,\mathbf k=(t^3-t^4)\,\mathbf j+(t-t^6)\,\mathbf k,$$

and

$$\mathbf F\cdot\mathbf r'(t)=0\cdot1+(t^3-t^4)\cdot2t+(t-t^6)\cdot3t^2=2t^4-2t^5+3t^3-3t^8.$$

Hence the work is

$$W=\int_0^1\left(3t^3+2t^4-2t^5-3t^8\right)dt=\left[\frac34t^4+\frac25t^5-\frac13t^6-\frac13t^9\right]_0^1=\frac34+\frac25-\frac13-\frac13.$$

With common denominator $60$: $\dfrac34=\dfrac{45}{60}$, $\dfrac25=\dfrac{24}{60}$, $\dfrac13=\dfrac{20}{60}$, so

$$W=\frac{45+24-20-20}{60}=\frac{29}{60}.$$

(2) Parametrize the curve as $\mathbf r(t)=\cos t\,\mathbf i+\sin t\,\mathbf j+t\,\mathbf k$ $\left(0\le t\le\dfrac\pi2\right)$; then $\mathbf r'(t)=(-\sin t,\cos t,1)$. Since $x=\cos t,\ z=t,\ y=\sin t$, $\mathbf F(\mathbf r(t))=x\,\mathbf i+z\,\mathbf j+y\,\mathbf k=\cos t\,\mathbf i+t\,\mathbf j+\sin t\,\mathbf k$, so

$$\mathbf F\cdot\mathbf r'(t)=\cos t\cdot(-\sin t)+t\cos t+\sin t\cdot1=-\sin t\cos t+t\cos t+\sin t.$$

The flow is

$$\int_C\mathbf F\cdot d\mathbf r=\int_0^{\pi/2}(-\sin t\cos t+t\cos t+\sin t)\,dt,$$

and $\displaystyle\int_0^{\pi/2}-\sin t\cos t\,dt=\left[-\frac{\sin^2t}{2}\right]_0^{\pi/2}=-\frac12$, $\displaystyle\int_0^{\pi/2}t\cos t\,dt=\big[t\sin t+\cos t\big]_0^{\pi/2}=\frac\pi2-1$, $\displaystyle\int_0^{\pi/2}\sin t\,dt=\big[-\cos t\big]_0^{\pi/2}=1$, so

$$\int_C\mathbf F\cdot d\mathbf r=-\frac12+\left(\frac\pi2-1\right)+1=\frac\pi2-\frac12.$$

(3) Parametrize the circle $C:x^2+y^2=1$ counterclockwise as $\mathbf r(t)=\cos t\,\mathbf i+\sin t\,\mathbf j$ $(0\le t\le2\pi)$. Rotating the tangent direction $90^\circ$ clockwise gives the outward unit normal, so $\mathbf n\,ds=(dy,-dx)$, and for $\mathbf F=M\,\mathbf i+N\,\mathbf j$ the flux is $\displaystyle\oint_C \mathbf F\cdot\mathbf n\,ds=\oint_C M\,dy-N\,dx$. Here $M=x-y,\ N=x$, and $dx=-\sin t\,dt,\ dy=\cos t\,dt$, so

$$M\,dy-N\,dx=(\cos t-\sin t)\cos t\,dt-\cos t\cdot(-\sin t)\,dt=(\cos^2t-\sin t\cos t+\sin t\cos t)\,dt=\cos^2t\,dt.$$

Hence the flux is

$$\oint_C M\,dy-N\,dx=\int_0^{2\pi}\cos^2t\,dt=\pi.$$

Solution.

(1) Parametrize the curve as $\mathbf r(t)=\cos(\pi t)\,\mathbf i+t^2\,\mathbf j+\sin(\pi t)\,\mathbf k$ $(0\le t\le1)$; then $\mathbf r'(t)=(-\pi\sin(\pi t),\,2t,\,\pi\cos(\pi t))$. Since $\mathbf F=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k$, $\mathbf F(\mathbf r(t))=\cos(\pi t)\,\mathbf i+t^2\,\mathbf j+\sin(\pi t)\,\mathbf k$, so

$$\mathbf F\cdot\mathbf r'(t)=-\pi\sin(\pi t)\cos(\pi t)+2t^3+\pi\sin(\pi t)\cos(\pi t)=2t^3$$

(the first and last terms cancel). Hence

$$W=\int_0^1 2t^3\,dt=\left[\frac{t^4}{2}\right]_0^1=\frac12.$$

(2) Parametrize the curve as $\mathbf r(t)=\cos t\,\mathbf i+\sin t\,\mathbf j$ $(0\le t\le2\pi)$; then $\mathbf r'(t)=(-\sin t,\cos t)$. Since $\mathbf F=(x-y)\,\mathbf i+x\,\mathbf j$, $\mathbf F(\mathbf r(t))=(\cos t-\sin t)\,\mathbf i+\cos t\,\mathbf j$, so

$$\mathbf F\cdot\mathbf r'(t)=(\cos t-\sin t)(-\sin t)+\cos t\cdot\cos t=-\sin t\cos t+\sin^2t+\cos^2t=1-\sin t\cos t.$$

Hence the circulation is

$$\oint_C\mathbf F\cdot d\mathbf r=\int_0^{2\pi}(1-\sin t\cos t)\,dt=2\pi-0=2\pi$$

(the integral of $\sin t\cos t$ over one period is $0$).

(3) Let the vertices of the triangle be $A=(1,0),\ B=(0,1),\ C_0=(-1,0)$; then $A\to B\to C_0\to A$ is counterclockwise (the shoelace formula gives area $=1\gt0$). By 예제 04(3), for $\mathbf F=M\,\mathbf i+N\,\mathbf j$ ($M=x+y,\ N=-(x^2+y^2)$) the flux is

$$\oint_{\partial} M\,dy-N\,dx=\oint_\partial (x+y)\,dy+(x^2+y^2)\,dx.$$

We parametrize each side.

$\overline{AB}$: $x=1-t,\ y=t$ $(0\le t\le1)$, $dx=-dt,\ dy=dt$. Then $(x+y)dy+(x^2+y^2)dx=dt-[(1-t)^2+t^2]\,dt=(2t-2t^2)\,dt$, so $\displaystyle\int_{\overline{AB}}=\int_0^1(2t-2t^2)\,dt=1-\frac23=\frac13$.

$\overline{BC_0}$: $x=-t,\ y=1-t$ $(0\le t\le1)$, $dx=-dt,\ dy=-dt$. Then $(x+y)dy=(1-2t)(-dt)$ and $(x^2+y^2)dx=-[t^2+(1-t)^2]\,dt=-(1-2t+2t^2)\,dt$, so the sum is $[-(1-2t)-(1-2t+2t^2)]\,dt=(-2+4t-2t^2)\,dt$, giving $\displaystyle\int_{\overline{BC_0}}=\int_0^1(-2+4t-2t^2)\,dt=-2+2-\frac23=-\frac23$.

$\overline{C_0A}$: $x=-1+2t,\ y=0$ $(0\le t\le1)$, $dx=2dt,\ dy=0$. Then $(x^2+y^2)dx=2(-1+2t)^2\,dt$, so substituting $u=-1+2t$, $\displaystyle\int_{\overline{C_0A}}=\int_0^1 2(-1+2t)^2\,dt=\int_{-1}^1u^2\,du=\frac23$.

Hence the flux is $\dfrac13-\dfrac23+\dfrac23=\dfrac13$.

Solution. When the object is at $(x,y)$, the unit vector pointing toward the origin is $-\dfrac{(x,y)}{\sqrt{x^2+y^2}}$, so the force of constant magnitude $k$ always directed toward the origin is

$$\mathbf F(x,y)=-\frac{k}{\sqrt{x^2+y^2}}(x,y).$$

Parametrize the curve $y=f(x)$ as $\mathbf r(t)=t\,\mathbf i+f(t)\,\mathbf j$ $(a\le t\le b)$; then $\mathbf r'(t)=(1,f'(t))$ and

$$\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)=-\frac{k\big(t+f(t)f'(t)\big)}{\sqrt{t^2+f(t)^2}}.$$

Since $\dfrac{d}{dt}\sqrt{t^2+f(t)^2}=\dfrac{2t+2f(t)f'(t)}{2\sqrt{t^2+f(t)^2}}=\dfrac{t+f(t)f'(t)}{\sqrt{t^2+f(t)^2}}$,

$$\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)=-k\cdot\frac{d}{dt}\sqrt{t^2+f(t)^2}.$$

Hence the work done by the force is

$$W=\int_a^b\mathbf F\cdot d\mathbf r=-k\int_a^b\frac{d}{dt}\sqrt{t^2+f(t)^2}\,dt=-k\Big[\sqrt{t^2+f(t)^2}\Big]_a^b=k\left(\sqrt{a^2+f(a)^2}-\sqrt{b^2+f(b)^2}\right).$$

Solution. Parametrize the curve of intersection of $y=x^2$ and $z=x$ by $x=t$, so $y=t^2,\ z=t$; the interval from $(0,0,0)$ to $(1,1,1)$ corresponds to $t:0\to1$. With $\mathbf r(t)=t\,\mathbf i+t^2\,\mathbf j+t\,\mathbf k$, $\mathbf r'(t)=(1,2t,1)$. Since $\mathbf F=xy\,\mathbf i+y\,\mathbf j-yz\,\mathbf k$,

$$\mathbf F(\mathbf r(t))=t\cdot t^2\,\mathbf i+t^2\,\mathbf j-t^2\cdot t\,\mathbf k=t^3\,\mathbf i+t^2\,\mathbf j-t^3\,\mathbf k,$$

and

$$\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)=t^3\cdot1+t^2\cdot2t+(-t^3)\cdot1=t^3+2t^3-t^3=2t^3.$$

Hence the flow is

$$\int_C\mathbf F\cdot d\mathbf r=\int_0^1 2t^3\,dt=\left[\frac{t^4}{2}\right]_0^1=\frac12.$$