Line Integrals

Definition If $f$ is a function defined on a curve $C$ that is parametrized by $\vec r(t)=g(t)\vec i+h(t)\vec j+k(t)\vec k$, $a\leq t\leq b$, then the line integral of $f$ over $C$ is $$\int_Cf(x,y,z)ds=\lim_{n\rightarrow\infty}\sum_{k=1}^nf(x_k,y_k,z_k)\Delta s_k$$ provided this limit exists.

Line integrals in the 2D plane

Let $f(x,y)\geq 0$ for $(x,y)\in C$ (the function value is nonnegative over the line). The line integral $\int_Cfds$ is the area of the portion of the cylindrical surface or ''wall'' beneath the function $f(x,y)$.

How to evaluate a line integral with curve $\vec{r}(t)$

Recall that we can define the variable $s(t)$ as $$s(t)=\int_a^t|\vec v(\tau)|d\tau, \qquad \vec v(t)={d\vec r\over dt}(t).$$

Then we have $ds=|\vec v(t)|dt$ and the change of variable gives $$\int_Cf(x,y,z)ds=\int_a^bf(g(t),h(t),k(t)){\color{red}|\vec v(t)|}dt.$$

Steps to integrate a continuous function $f(x, y, z)$ over a curve $C$:

1. Find a smooth parametrization of $C$, $$\vec r(t) = g(t)\vec i + h(t)\vec j + k(t)\vec k,\qquad a \leq t \leq b.$$

2. Evaluate the integral as $$\int_Cf(x,y,z)ds=\int_a^bf(g(t),h(t),k(t))|\vec v(t)|dt.$$

Example: Integrate $f(x, y, z) = x - 3y^2 + z$ over the line segment $C$ joining the origin to the point $(1, 1, 1)$.

Solution: Let $\vec r(t)=t\vec i+t\vec j+t\vec k$ for $t\in[0,1]$ be the line segment $C$, and we have $$\int_Cf(x,y,z)ds=\int_0^1 (t-3t^2+t)\sqrt{3}dt=0.$$

Additivity

For a piecewise smooth curve $C$ made by joining $C_1,C_2,\cdots,C_n$, we have $$\int_C ƒ ds = \int_{C_1}f ds + \int_{C_2}fds + \cdots+\int_{C_n}fds.$$ That is, we sum the line integrals over all disjoint subcurves to get the line integral over the curve $C$.

Example: Integrate $f(x, y, z) = x - 3y^2 + z$ over $C$ which goes from the origin directly to the point $(1, 1, 0)$, and from $(1,1,0)$ to $(1,1,1)$.

Solution:

For $C_1:(0,0,0)\rightarrow(1,1,0)$: let $\vec r(t)=t\vec i+t\vec j+0\vec k$, we have $$\int_Cf(x,y,z)ds=\int_0^1(t-3t^2)\sqrt{2}dt=-{\sqrt{2}\over 2}.$$ For $C_2:(1,1,0)\rightarrow (1,1,1)$: let $\vec r(t)=\vec i+\vec j+t\vec k$, we have $$\int_Cf(x,y,z)ds=\int_0^1(1-3+t)\sqrt{1}dt=-{3\over 2}.$$ The line integral is the sum of the two values: $$-{\sqrt{2}\over2}-{3\over2}.$$

Example: Find the line integral of $f(x,y,z)=2xy+\sqrt{z}$ over the helix $\vec r(t)=\cos t\vec i+\sin t\vec j +t\vec k$, $0\leq t\leq \pi$.

Solution: $$\int_0^\pi (2\cos t\sin t +\sqrt{t} )\sqrt{(-\sin t)^2+(\cos t)^2+1}dt={2\sqrt{2}\over 3}{\pi}^{3/2}.$$

Mass and moment formulas for coil springs, wires, and thin rods lying along a smooth curve $C$ in space

• Mass: $$M=\int_C\delta ds,\quad \delta=\delta(x,y,z)\text{ is the density at }(x,y,z)$$

• First moments about the coordinate planes: $$M_{yz}=\int_Cx\delta ds,\quad M_{xz}=\int_Cy\delta ds,\quad M_{xy}=\int_Cz\delta ds$$

• Coordinates of the center of mass: $$\bar x=M_{yz}/M,\quad \bar y=M_{xz}/M,\quad \bar z=M_{xy}/M$$

• Moments of inertia about axes and other straight lines $$I_x=\int_C(y^2+z^2)\delta ds,\quad I_y=\int_C(x^2+z^2)\delta ds,\quad I_z=\int_C(x^2+y^2)\delta ds$$ $$I_L=\int_Cr^2\delta ds,\quad r(x,y,z)=\text{distance from the point }(x,y,z)\text{ to line }L$$

Example: A slender metal arch, denser at the bottom than at the top, lies along the semicircle $y^2 + z^2 = 1$, $z \geq0$, in the $yz$-plane. Find the center of the arch's mass if the density on the arch is $\delta(x, y, z) = 2 - z$.

Solution:

Let $\vec r(t)=0\vec i+ \cos t\vec j+ \sin t\vec j$ for $0\leq t\leq \pi$. Then the total mass is $$M=\int_0^\pi (2-\sin t)dt=2\pi-2$$ The first moment is $$\begin{align*}M_{xy}=\int_0^\pi \sin t(2-\sin t)dt=&{8-\pi\over 2}\\ M_{xz}=\int_0^\pi \cos t(2-\sin t)dt=&{0} \end{align*}$$ So, the center is $$\bar x= M_{yz}/M=0,\quad \bar y= M_{xz}/M=0,\quad \bar z= M_{xy}/M={8-\pi\over 4\pi -4}.$$

Vector Fields

A vector field is a function that assigns a vector to each point in its domain. A vector field on a 3D domain might have a formula like $$\vec F(x, y, z) = M(x, y, z)\vec i + N(x, y, z)\vec j + P(x, y, z)\vec k.$$

• A field is continuous if its component functions $M$, $N$, and $P$ are continuous;

• A field is differentiable if its component functions are differentiable.

• Example: tangent vector $\vec T$ and normal vector $\vec{N}$ of a curve.

Special vector fields: gradient fields

We define the gradient field of a differentiable function $f(x, y, z)$ to be the field of gradient vectors $$\nabla f={\partial f\over\partial x}\vec i+{\partial f\over\partial y}\vec j+{\partial f\over\partial z}\vec k.$$

• At $(x, y, z)$, the gradient field gives a vector pointing toward the greatest increase of $f$.

• Its magnitude is the directional derivative in the direction of the greatest increase of $f$.

Find the vector field $\vec F$

Suppose that the temperature $T$ at $(x, y, z)$ in a region of space is given by $$T=100-x^2-y^2-z^2,$$ and that $\vec F(x,y,z)$ is the gradient of $T$.

Line integrals of vector fields

Work done by a force over a curve

Let the tangent vector $\vec T=d\vec r/ds$, a unit vector tangent to the path. The work done along the subarc from $P_{k-1}$ to $P_k$ shown below is approximately $$\underbrace{\vec{F}(x_k,y_k,z_k)\cdot \vec{T}(x_k,y_k,z_k)}_{\text{the portion contributes}}\Delta s_k.$$

Definition Let $C$ be a smooth curve parametrized by $\vec r(t)$, $a \leq t \leq b$, and $\vec F$ be a continuous force field over a region containing $C$. Then, the work done in moving an object from the point $A = \vec r(a)$ to the point $B = \vec r(b)$ along $C$ is $$W=\int_C\vec F\cdot\vec T ds=\int_C\left(\vec F\cdot{d\vec r\over ds}\right)ds=\int_a^b\vec F(\vec r(t))\cdot{d\vec r\over dt}dt.$$

Definition Let $\vec F$ be a vector field with continuous components defined along a smooth curve $C$ parametrized by $\vec r(t)$, $a \leq t \leq b$. The line integral of $F$ along C is $$\int_C\vec F\cdot \vec T ds= \int_C\left(\vec F\cdot{d\vec r\over ds}\right)ds=\int_C\vec F\cdot d\vec r.$$

Line integrals of vector fields

Evaluate the integral of $\vec F=M\vec i+N\vec j+P\vec k$ along $C: \vec r(t ) = g (t )\vec i + h(t )\vec j + k(t )\vec k$ in the following steps

• Express the vector field $\vec F$ as $\vec F(\vec r(t))$

• Find the derivative vector $d\vec r/dt$.

• Evaluate the line integral $$\int_C\vec F\cdot d\vec r=\int_a^b\vec F(\vec r(t))\cdot {d\vec r\over dt}dt.$$

Example: Let $\vec F(x,y,z)=z\vec i+xy\vec j-y^2\vec k$ and $C:\vec r(t)=t^2\vec i+t\vec j+\sqrt{t}\vec k$, $0\leq t\leq 1$.

Solution: $$\int_C\vec F\cdot d\vec r=\int_0^1\langle \sqrt{t},t^3,-t^2\rangle\cdot\langle 2t,1,{1\over2\sqrt{t}}\rangle dt=\int_0^12t^{3/2}+t^3-{1\over2}t^{3/2}dt={17\over 20}.$$

Scalar differential form

line integral with respect to $dx$, $dy$, or $dz$

Let $\vec F=M\vec i+N\vec j+P\vec k$ and $C: \vec r(t ) = g (t )\vec i + h(t )\vec j + k(t )\vec k$ $$\begin{align*} \int_C M\vec{i}\cdot d\vec{r} &=\int_CM(x,y,z)dx=\int_a^bM(g(t),h(t),k(t))g'(t)dt\\ \int_C N\vec{j}\cdot d\vec{r} &=\int_CN(x,y,z)dy=\int_a^bN(g(t),h(t),k(t))h'(t)dt\\ \int_C P\vec{k}\cdot d\vec{r} &=\int_CP(x,y,z)dz=\int_a^bP(g(t),h(t),k(t))k'(t)dt \end{align*}$$ $$\int_CMdx+Ndy+Pdz=\int_C\vec{F}\cdot d\vec{r}$$

Different ways to write the work integral

Different ways to write the work integral for $\vec{F}-M\vec{i}+N\vec{j}+P\vec{k}$ over the curve $C:\vec{t}=g(t)\vec{i}+h(t)\vec{j}+k(t)\vec{k},~a\leq t\leq b.$

Example: Evaluate $\int_C-ydx+zdy+2xdz$, where $C$ is the helix $\vec r(t)=\cos t\vec i+\sin t \vec j+t\vec k$, $0\leq t\leq 2\pi$

Solution:

Example: Find the work done by the force field $\vec F = (y - x^2)\vec i + (z - y^2)\vec j + (x - z^2)\vec k$ along the curve $\vec r(t) = t \vec i + t^2\vec j + t^3\vec k$, $0\leq t \leq 1$, from $(0, 0, 0)$ to $(1, 1, 1)$.

Solution: $$\begin{align*} &\int_0^1 (t^2-t^2)+(t^3-t^4)2t+(t-t^6)3t^2dt\\ =&\int_0^1 2t^4-2t^5+3t^3-3t^8dt\\ =& {2\over 5}-{2\over 6}+{3\over 4}-{3\over 9}\\ =&{29\over 60} \end{align*}$$

Flow integrals and circulation for velocity fields

Definition If $\vec r(t)$ parametrizes a smooth curve $C$ in the domain of a continuous velocity field $\vec F$, the flow along the curve from $A = \vec r(a)$ to $B = \vec r(b)$ is $$\mbox{Flow} = \int_C \vec F\cdot \vec T ds=\int_C \vec F\cdot d\vec{r}.$$ The integral is called a flow integral. If the curve starts and ends at the same point, so that $A = B$, the flow is called the circulation around the curve.

Example: A fluid's velocity field is $\vec F=x\vec i+z\vec j+y\vec k$. Find the flow along the helix $\vec r(t)=\cos t\vec i+\sin t \vec j+t\vec k$, $0\leq t\leq \pi/2$.

Solution:

Example: Find the circulation of the field $\vec F = (x - y)\vec i + x \vec j$ around the circle $\vec r(t) = (\cos t)\vec i + (\sin t)\vec j,~0 \leq t \leq 2\pi$.

Solution: $$\begin{align*} & \int_0^{2\pi} (\cos t-\sin t)(-\sin t)+ \cos t(\cos t)dt\\=&{1\over 4}\cos(2t)+t\Big|_0^{2\pi}\\=&2\pi \end{align*}$$

Flux across a simple closed-plane curve

Definition A curve in the $xy$-plane is simple if it does not cross itself. When a curve starts and ends at the same point, it is a closed curve or loop.

Definition If $C$ is a smooth, simple closed curve in the domain of a continuous vector field $\vec F = M(x, y)\vec i + N(x, y)\vec j$, and if $\vec n$ is the outward-pointing unit normal vector on $C$, the flux of $\vec F$ across $C$ is $$\mbox{Flux of }\vec F \mbox{ across }C=\int_C\vec F\cdot\vec n ds.$$

How to evaluate the flux

The $\vec n$ is orthogonal to $\vec T$. Note that there are two unit vectors orthogonal to the vector $\vec T$, and they are in the opposite direction.

• If the curve is clockwise, then the vector $\vec n$ is to the left of the curve while a particle is moving on the curve, so the angle of the vector $\vec n$ is obtained by adding $\pi/2$ to the angle of the vector $\vec T$. That is $$\cos(\theta+\pi/2)=-\sin\theta, ~\sin(\theta+\pi/2)=\cos\theta$$

• If the curve is counterclockwise, then the vector $\vec n$ is to the right of the curve while a particle is moving on the curve, so the angle of the vector $\vec n$ is obtained by subtractinging $\pi/2$ from the angle of the vector $\vec T$. That is $$\cos(\theta-\pi/2)={\sin\theta}, ~\sin(\theta-\pi/2)={-\cos\theta}$$

Remind that we have $$\vec T = {dx\over ds}\vec i + {dy\over ds}\vec j=\cos \theta \vec i+\sin\theta\vec j.$$ Therefore for the counterclockwise case, we have $$\vec n={dy\over ds}\vec i-{dx\over ds}\vec j.$$

So we have $$\begin{align*} \oint_C\vec F\cdot\vec n ds=&\oint_C\left(M{dy\over ds}-N{dx\over ds}\right)ds\\ =&\oint_CMdy-Ndx. \end{align*}$$

Example: Find the flux of $\vec F=(x-y)\vec i+x\vec j$ across the circle $x^2+y^2=1$.