Tutorial 12: Line Integrals

Tutorial Question

Tutorial Solution

1. If $A=\left(3 x^{2}+6 y\right) \hat{i}-14 y z \hat{j}+20 x z^{2} \hat{k}$, evaluate $\int_{C} A \bullet d r$ from $(0,0,0)$ to $(1,1,1)$ along the following paths $\mathrm{C}$ :

   a. $x=t, y=t^{2}, z=t^{3}$.

   Solution

   $$\begin{aligned} r(t)&=x(t)\hat{i}+y(t)\hat{j}+z(t)\hat{k}\\ &=t\hat{i}+t^2\hat{j}+t^3\hat{k} \end{aligned}$$$$\begin{aligned} A(r(t))&=(3t^2+6t^2)\hat{i}-14(t^2)(t^3)\hat{j}+20(t)(t^3)^2\hat{k}\\ &=9t^2\hat{i}-14t^5\hat{j}+20t^7\hat{k} \end{aligned}$$$$\frac{dr}{dt}=\hat{i}+2t\hat{j}+3t^2\hat{k}$$$$\begin{aligned} \int_C F\cdot dr &=\int_0^1 F(r(t))\cdot \frac{dr}{dt}\cdot dt\\ &=\int_0^1(9t^2\hat{i}-14t^5\hat{j}+20t^7\hat{k})(\hat{i}+2t\hat{j}+3t^2\hat{k})dt\\ &=\int_0^1 9t^2-28t^6+60t^9 dt\\ &=\left. 3t^3-4t^7+6t^10 \right|_0^1\\ &=3-4+6\\ &=5 \end{aligned}$$

   b. The straight line from $(0,0,0)$ to $(1,0,0)$ then to $(1,1,0)$, and then to $(1,1,1)$.

   Solution

   From $(0,0,0)$ to $(1,0,0)$,

   $$r=(0,0,0)+t(1,0,0)=t\hat{i}+0\hat{j}+0\hat{k}\\\Rightarrow x=t,y=0,z=0$$$$\frac{dr}{dt}=1\hat{i}$$$$\begin{aligned} \int_0^1 F(r(t))\cdot\frac{dr}{dt}\cdot dt &=\int_0^1 (3t^2 \hat{i})\cdot(\hat{i})dt\\ &=\left. t^3\right|_0^1=1 \end{aligned}$$

   From $(1,0,0)$ to $(1,1,0)$,

   $$r=(1,0,0)+t(0,1,0)=\hat{i}+t\hat{j}+0\hat{k}\\\Rightarrow x=1,y=t,z=0$$$$\frac{dr}{dt}=1\hat{j}$$$$F(r,t)=(3(1)+6t)\hat{i}-14(t)(0)\hat{j}+20(1)(0)\hat{k}=(3+6t)\hat{i}$$$$\begin{aligned} \int_0^1 F(r(t))\cdot\frac{dr}{dt}\cdot dt &=\int_0^1 (3+6t \hat{i})\cdot(\hat{j})dt\\ &=0 \end{aligned}$$

   From $(1,1,0)$ to $(1,1,1)$,

   $$r=(1,1,0)+t(0,0,1)=\hat{i}+\hat{j}+t\hat{k}\\\Rightarrow x=1,y=1,z=t$$$$\frac{dr}{dt}=1\hat{k}$$$$\begin{aligned} F(r,t)&=(3(1)+6(1))\hat{i}-14(1)(t)\hat{j}+20(1)(t)^2\hat{k}\\ &=9\hat{i}-14t\hat{j}+20t^2\hat{k} \end{aligned}$$$$\begin{aligned} \int_0^1 F(r(t))\cdot\frac{dr}{dt}\cdot dt &=\int_0^1 (9\hat{i}-14t\hat{j}+20t^2\hat{k})\cdot(\hat{k})dt\\ &=\int_0^1 20t^2 dt\\ &=\left.\frac{20}{3}t^3\right|_0^1=\frac{20}{3} \end{aligned}$$

   $\therefore \text{Total Line Integral}=1+0+\frac{20}{3}=\frac{23}{3}$

   c. The straight line joining $(0,0,0)$ and $(1,1,1)$.

   Solution

   $$r=t\hat{i}+t\hat{j}+t\hat{k}\\\Rightarrow x=t,y=t,z=t$$$$\frac{dr}{dt}=\hat{i}+\hat{j}+\hat{k}$$$$\begin{aligned} &\int_0^1\left[(3t^2+6t)\hat{i}-14(t)(t)\hat{j}+20(t)(t)^2\hat{k}\right]\cdot\left[\hat{i}+\hat{j}+\hat{k}\right]\\ &=\int_0^1(3t^2+6t-14t^2+20t^3)dt\\&=\int_0^1(6t-11t^2+20t^3)dt\\ &=\left. 3t^2-\frac{11}{3}t^3+5t^4\right|_0^1\\ &=\frac{13}{3} \end{aligned}$$

2. Find the work done in moving a particle in a force field given by $F=3 x y \hat{i}-5 z \hat{j}+10 x \hat{k}$ along the curve $x=t^{2}+1, y=2 t^{2}, z=t^{3}$ from $t=1$ to $t=2$.

   Solution

   $$r(t)=x(t)\hat{i}+y(t)\hat{j}+z(t)\hat{k}=(t^2+1)\hat{i}+2t^2\hat{j}+t^3\hat{k}$$$$\frac{dr}{dt}=2t\hat{i}+4t\hat{j}+3t^2\hat{k}$$$$F(r(t))=3(t^2+1)(2t^2)\hat{i}-5(t^3)\hat{j}+10(t^2+1)\hat{k}$$$$\begin{aligned} \int_C F\cdot dr&=\int_1^2 (6t^4+6t^2)\hat{i}-5t^3\hat{j}+(10t^2+10)\hat{k} \cdot (2t\hat{i}+4t\hat{j}+3t^2\hat{k}) dt\\ &=\int_1^2 (12t^5+12t^3-20t^4+30t^4+30t^2)dt\\ &=\left. 2t^6+3t^4-4t^5+6t^5+10t^3\right|_1^2\\ &=\left.2t^6+2t^5+3t^4+10t^3\right|_1^2\\ &=(128+64+48+80)-(2+2+3+10)\\ &=320-17=303 \end{aligned}$$

3. Determine the work done in moving a particle counterclockwise once around a circle $\mathrm{C}$ in the $x y$-plane, if the circle has center at the origin and radius of 3 and if those field is given by $F=(2 x-y+z) \hat{i}+\left(x+y-z^{2}\right) \hat{j}+(3 x-2 y+4 z) \hat{k}$

   Solution

   $$r=x\hat{i}+y\hat{j}=3\cos{t}\hat{i}+3\sin{t}\hat{j}$$

   In the plane $z=0$,

   $$\mathbf{F}=(2x-y)\hat{i}+(x+y)\hat{j}+(3x-2y)\hat{k}$$$$dr=dx\hat{i}+dy\hat{j}$$

   The work done is,

   $$\begin{aligned} \int_C \mathbf{F}\cdot d\mathbf{r}&=\int_C \left[(2x-y)\hat{i}+(x+y)\hat{j}+(3x-2y)\hat{k}\right]\cdot [dx\hat{i}+dy\hat{j}]\\ &=\int_C (2x-y)dx+(x+y)dy \end{aligned}$$

   Parametric equation of the circle: $x=3\cos{t}$, $y=3\sin{t}$, where $t$ varies from $0$ to $2\pi$. $\Rightarrow$ $dx=-3\sin t dt, dy=3\cos t dt$

   The line integral,

   $$\begin{aligned} &\int_{t=0}^{2\pi} \left[2(3\cos{t}-3\sin{t})(-3\sin{t})dt\right]+[3\cos{t}+3\sin{t}][3\cos{t}dt]\\ &=\int_0^{2\pi} -18\sin t\cos t+9\sin^2 t+9\cos^2 t+9\sin t\cos t dt\\ &=\int_0^{2\pi} (9-9\sin{t}\cos{t})dt \\ &=\left. 9t-\frac{9}{2}\sin^2{t}\right|_0^{2\pi}\\ &=18\pi \end{aligned}$$

   Note: We choose counterclockwise direction, which is known as positive direction. If traversing from counterclockwise (negative) direction, the value of integral would be $-18\pi$.

4. Find the line integral for $\oint_{C}\left(x y+y^{2}\right) d x+x^{2} d y$ where $C$ is the closed curve of the region bounded by $y=x$ and $y=x^{2}$ in the positive direction in traversing C. $y=x$ and $y=x^{2}$ intersect at $(0,0)$ and $(1,1)$.

   Solution

   

   Along $y=x^2$, where $dy=2xdx$ the line integral equals

   $$\int_0^1\left((x)(x^2)+(x^2)^2\right)dx+(x^2)(2x)dx=\int_0^1(3x^3+x^4)dx=\frac{19}{20}$$

   Along $y=x$ from $(1,1)$ to $(0,0)$, the line integral equals

   $$\int_0^1\left((x)(x)+(x)^2\right)dx+x^2dx=\int_0^13x^2dx=1$$

   L.H.S = $\displaystyle\text{The required line integral}=\frac{19}{20}-1=-\frac{1}{20}$

   $$\begin{aligned} \iint_R\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dxdy&=\iint_R\left[\frac{\partial}{\partial x}(x^2)-\frac{\partial}{\partial y}(xy+y^2)\right]dxdy\\ &=\iint_R(x-2y)dxdy\\ &=\int_{x=0}^1\int_{y=x^2}^{x}(x-2y)dydx\\ &=\int_0^1\left[\int_{x^2}^x(x-2y)dy\right]dx\\ &=\int_0^1\left.(xy-y^2)\right|_{x^2}^xdx\\ &=\int_{0}^1\left[(x^2-x^2)-(x^3-x^4)\right]dx\\ &=\int_{0}^1\left(x^4-x^3\right)dx\\ &=\left.\frac{x^5}{5}-\frac{x^4}{4}\right|_0^1\\ &=\frac{1}{5}-\frac{1}{4}=-\frac{1}{20}\\ \end{aligned}$$

   L.H.S = R.H.S = $\displaystyle-\frac{1}{20}$.

5. Evaluate $\oint_{C}(y-\sin x) d x+\cos x d y$ where $C$ is the triangle of the figure below.

   
   Solution

   $M=y-\sin x$, $N=\cos x$, $\displaystyle\frac{\partial N}{\partial x}=-\sin x$, $\displaystyle\frac{\partial M}{\partial y}=1$,

   $$\begin{aligned}\oint_C Mdx+Ndy &=\iint_R \left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dxdy\\ &=\iint_R(-\sin x-1)dydx\\ &=\int_{x=0}^{\frac{\pi}{2}}\left[\int_{y=0}^{\frac{2x}{\pi}} (-\sin x-1) dy \right] dx\\ &=\int_{x=0}^{\frac{\pi}{2}}\left.(-y\sin x-y)\right|_0^{\frac{2x}{\pi}}dx\\ &=\int_0^{\frac{\pi}{2}}\left(-\frac{2x}{\pi}\sin x-\frac{2x}{\pi}\right)dx\\ &=\left.-\frac{2}{\pi}(-x\cos x+\sin x)-\frac{x^2}{\pi}\right|_0^{\frac{\pi}{2}}\\ &=-\frac{2}{\pi}-\frac{\pi}{4} \end{aligned}$$

   Note: Although there exist lines parallel to the coordinate axes (coincident with the coordinate axes in this case) which meet $C$ in an infinite number of points, Green’s theorem in the plane still holds. In general the theorem is valid when $C$ is composed of a finite number of straight line segments.

6. Calculate $\oint_{C}-x^{2} y d x+x y^{2} d y$ where $C$ is the circle of radius 2 centered on the origin.

   Solution

   $$\oint_CMdx+Ndy=\iint_RN_x-M_ydA$$

   We let $M=-x^2y$ and $N=xy^2$ to get

   $$\begin{aligned}\oint_C -x^2ydx+xy^2dy&=\iint_Ry^2-(-x^2)dA\\ &=\iint_Rx^2+y^2dA\\&=\int_0^{2\pi}\int_0^2 r^2r dr d\theta\\ &=\int_0^{2\pi}\left. \frac{r^4}{4}\right|_0^2d\theta\\ &=\int_0^{2\pi}4d\theta\\ &=\left.4\theta\right|_0^{2\pi}=8\pi \end{aligned}$$

7. Compute the line integral of $\boldsymbol{F}(x, y)=\left\langle x^{3}, 4 x\right\rangle$ along the path $C$ shown below against a grid of unit-sized squares.

   
   Solution

   Let $L$ be the line segment going from $B$ to $A$. Then, we can now apply Green’s theorem to combination of $C$ and $L$. Let $D$ be the region bounded by these two paths. Then, by Green’s theorem, since we are oriented correctly,

   $$\begin{aligned} \int_{\partial D}\mathbf{F}\cdot d\mathbf{r}&=\iint_D \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}dA\\ &=\iint_D\frac{\partial}{\partial x}(4x)-\frac{\partial}{\partial y}(x^3)dA\\ &=\iint_D4dA\\ &=4\cdot\text{Area}(D)\\ &=16 \end{aligned}$$

   because the area of the region is made of exactly 4 unit squares. The boundary of $D$ is $C$ and $L$:

   $$\int_{\partial D}\mathbf{F}\cdot d\mathbf{r}=\int_C\mathbf{F}\cdot d\mathbf{r}+\int_L\mathbf{F}\cdot d\mathbf{r}=16$$

   The line integral along $L$ is easier: parametrizing $L$ by $\mathbf{r}(t)=(-1,t)$ for $-1\leq t\leq 0$, we get

   $$\int_L\mathbf{F}\cdot d\mathbf{r}=\int_{-1}^0(-1,-4)\cdot (0,1)dt=\int_{-1}^0-4dt=-4$$

   Putting it together,

   $$\int_C \mathbf{F}\cdot d\mathbf{r}=16-\int_L\mathbf{F}\cdot d\mathbf{r}=16-(-4)=20$$

8. A particle starts at $(-2,0)$ and moves along the $x$-axis to $(2,0)$. Then it moves along the upper part of the circle $x^{2}+y^{2}=4$ and back to $(-2,0)$. Compute the work done on this particle by the force field $\boldsymbol{F}(x, y)=\left\langle x, x^{3}+3 x y^{2}\right\rangle$.

   Solution

   Let $\mathbf{F}(x,y)=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j}$ with $P(x,y)=x$ and $Q(x,y)=x^3+3xy^2$.

   The path is the boundary of the upper half disk $D=\{x^2+y^2\le 4,\ y\ge 0\}$, traversed counter-clockwise (the region lies to the left of the direction of travel), so Green's theorem applies:

   $$\begin{aligned} \int_C Pdx+Qdy&=\iint_D \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy\\ &=\iint_D \left(3x^2+3y^2-0\right)dxdy\\ &=3\iint_D (x^2+y^2)\,dA \end{aligned}$$

   Switching to polar coordinates over the upper half disk ($0\le r\le 2$, $0\le\theta\le\pi$),

   $$\begin{aligned} W&=3\int_0^{\pi}\int_0^2 r^{2}\cdot r\,drd\theta =3\int_0^{\pi}\left[\frac{r^{4}}{4}\right]_0^{2}d\theta =3\int_0^{\pi}4\,d\theta =12\pi\\ &\approx 37.70 \end{aligned}$$

9. Let $\mathbf{F}(x, y)=\langle\sin x, \cos y\rangle$ and let $C$ be the curve that is the top half of the circle $x^{2}+y^{2}=1$, traversed counterclockwise from $(1,0)$ to $(-1,0)$, and the line segment from $(-1,0)$ to $(-2,3)$. Evaluate the line integral $\int \mathbf{F} \cdot \mathbf{T} d s=\int_{C} \sin x d x+\cos y d y$

   Solution

   We consider the integrals over the semicircle, denoted by $C_1$, and the line segment, denoted by $C_2$, seperately. We then have,

   $$\int_C\sin{x}dx+\cos{y}dy=\int_{C_1}\sin{x}dx+\cos{y}dy+\int_{C_2}\sin{x}dx+\cos{y}dy$$

   For the semicircle, we use the parametric equations

   $$x=\cos{t},\quad y=\sin{t},\quad 0\leq t\leq\pi$$

   This yields

   $$\begin{aligned} \int_{C_1}\sin x\,dx+\cos y\,dy &=\int_0^\pi\sin(\cos t)(-\sin t)\,dt+\int_0^\pi\cos(\sin t)(\cos t)\,dt\\ &=\left.-\cos(\cos t)\right|_0^\pi+\left.\sin(\sin t)\right|_0^\pi\\ &=(-\cos(-1)+\cos(1))+(\sin(0)-\sin(0))\\ &=(-\cos(1)+\cos(1))+0\\ &=0 \end{aligned}$$

   For the line segment, we use the parametric equations

   $$x=-1-t,\quad y=3t,\quad 0\leq t \leq 1$$

   This yields

   $$\begin{aligned}\int_{C_2}\sin xdx+\cos ydy&=\int_0^1 \sin (-1-t)(-1)dt+\cos(3t)(3)dt\\ &=\left. -\cos(-1-t)\right|_0^1+\left.\sin(3t)\right|_0^1\\&=-\cos(-2)+\cos(-1)+\sin(3)-\sin(0)\\ &=-\cos(2)+\cos(1)+\sin(3) \end{aligned}$$

   We conclude

   $$\int_C \sin xdx+\cos ydy=\cos(1)-\cos(2)+\sin(3)$$

   In evaluating these integrals, we have taken advantage of the rule,

   $$\int_a^b f'(g(t))g'(t)dt=f(g(b))-f(g(a))$$