Line integral

Given $C$ the curve connecting two points R，Q. However, this curve should be a smooth curve. Let $s$ be the arc length measured along $C$．Then, as we learned in Chapter 2, the points on the curve $(x,y,z)$ can be expressed by $s$ as a parameter．So the equation for the curve $C$ is

$$\boldsymbol{r} = \boldsymbol{r}(s) = x(s)\:\boldsymbol{i} + y(s)\:\boldsymbol{j} + z(s)\:\boldsymbol{k}$$

However, it is assumed that points R and Q correspond to $s = a$   and$b (a < b)$, respectively. At this time, the scalar field defined for any point P $(x,y,z)$ on the curve $C:\boldsymbol{r} = \boldsymbol{r}(s)$ is $f({\ rm P}) = f(x,y,z)$.

The curve $C$ is divided into $n$ arcs $s_{1},s_{2}, \ldots,s_{n}$, and this division is represented by $\Delta$. Let the arc length of each curve $s_{i}$ be $\Delta s_{i}$, take an arbitrary point ${\ rm P}_{i}$ in $s_{i}$,

$$J(\Delta) = \sum_{i=1}^{n} f({\rm P}_{i})\Delta s_{i}$$

Here, if $J(\Delta)$ approaches the limit value $J$ when this division is made finer and $\Delta$ is made as small as possible, then this limit value $J$ is called line integral of the scalar field $f$ along C and denoted by

$$\int_{C}f({\rm P}) ds$$

When the curve $C$ is closed, it is expressed as

$$\oint_{C} f({\rm P}) ds$$

Since the definition of line integral is based on the same Riemann sum as the previous integral, it is clear that the following formula holds for line integral.

$$\begin{array}{l} 1. \ \int_{C} k f({\rm P}) ds = k \int_{C}f(... ... = \int_{C} f({\rm P})ds + \int_{C}g({\rm P})ds \\ \end{array}$$

Also, when the curve $C$ is not smooth but is made up of a finite number of smooth curves $C_{1},C_{2},\ldots,C_{n}$, this curve is Piecewise smooth curve , and the line integral along such a curve is

$$\int_{C} = \int_{C_{1}} + \int_{C_{2}} + \cdots + \int_{C_{n}}$$

Example 3..7  

Evaluate the line integral $${\int_{C}(x^3 + y^4)ds}$$, where $C$ is a line connecting the points $(0,0,0)$ and $(1,1,1)$.

Answer When the straight line connecting the point $(0,0,0)$ and the point $(1,1,1)$ is displayed as a parameter

$$\left\{\begin{array}{l} x = t\\ y = t\\ z = t \end{array}\right. \ \ 0 \leq t \leq 1$$

Then the curve $C$ is expressed as $\boldsymbol{r}(t) = t\:\boldsymbol{i} + t\:\boldsymbol{j} + t\:\boldsymbol{k}$ and the length $s$ of $C$ is

$$s(t) = \int_{0}^{t}\vert\frac{d \boldsymbol{r}}{dt}\vert dt = \in... ...}^{t} \vert\boldsymbol{i} + \boldsymbol{j} + \boldsymbol{k}\vert dt = \sqrt{3}t$$

Thus, $ds = \sqrt{3}dt$ and the line integral is

$$\int_{C}(x^3 + y^4)ds = \int_{0}^{1} (t^3 + t^4) \sqrt{3}dt = \sqrt{3}(\frac{1}{4} + \frac{1}{5}) = \frac{9\sqrt{3}}{20}$$

Line integral of vector field

Oriented curve $C : \boldsymbol{r} = \boldsymbol{r}(t)$ and the vector field defined on $C$, $\boldsymbol{F} = F_{1}\:\boldsymbol{i} + F_{2}\:\boldsymbol{j} + F_{3}\:\boldsymbol{k}$ are given．Here, the unit tangent vector ${\bf t}$ of $C$ is positive direction．Then $\boldsymbol{F}\cdot{\bf t}$ is a scalar field defined on $C$ and the line integral of this scalar field on $C$ is expressed by

$$\int_{C} \boldsymbol{F} \cdot{\bf t} ds$$

This is called a line integral along the $C$ oriented curve $\boldsymbol{F}$．EEspecially, note that

$${\bf t} ds = \frac{d\boldsymbol{r}}{ds} ds = d \boldsymbol{r}$$

Then

$$\int_{C} \boldsymbol{F} \cdot{\bf t} ds = \int_{C} \boldsymbol{F} \cdot d \boldsymbol{r}$$

Figure 3.2: Line integral of vector field

Considering the case where the vector field $\boldsymbol{F}$ is an electric field. $\int_{{\rm P}}^{{\rm S}}\boldsymbol{F}\cdot d \boldsymbol{r}$ can be thought of as the work per unit charge performed by the electric field $\boldsymbol{F}$. when the positive charge moves from point P to point S along the curve $C$, which is called the potential difference or voltage between two points.

Example 3..8  

Find the amount of word done by the mass point to go around $${\frac{x^2}{16} + \frac{y^2}{9} = 1}$$. Note that vector field is given by

$$\boldsymbol{F} = (3x-4y+2z)\:\boldsymbol{i} + (4x+2y-3z^2)\:\boldsymbol{j} + (2xz-4y^2 + z^3)\:\boldsymbol{k}$$

Answer Let $x = 4\cos{t}, y = 3\sin{t}$. Then

$$\boldsymbol{r} = \boldsymbol{r}(t) = 4\cos{t}\:\boldsymbol{i} + 3... ...\ d\boldsymbol{r}(t) = (-4\sin{t}\:\boldsymbol{i} + 3\cos{t}\:\boldsymbol{j})dt$$

Also,

$$\boldsymbol{F} = (12\cos{t} - 12\sin{t})\:\boldsymbol{i} + (16\cos{t} + 6\sin{t})\:\boldsymbol{j} -36\sin^{2}{t}\:\boldsymbol{k}$$

Thus

$$\int_{C} \boldsymbol{F} \cdot d \boldsymbol{r} = \int_{0}^{2\pi} [48 - 30\sin{t}\cos{t}]dt = 96\pi$$

Example 3..9  

Find the line integral $${\int_{C}(x^2 y\boldsymbol{i} + (x+y)\:\boldsymbol{j}) \cdot{\bf t}ds}$$．Here， $C$ is the curve $${y = x^2}$$ from $(0,0)$ to $(2,4)$.

Answer

$$\int_{C} \boldsymbol{F} \cdot{\bf t} ds = \int_{C}\boldsymbol{F}\cdot d\boldsymbol{r}$$

$$= \int_{C} (x^2 y\boldsymbol{i} + (x+y)\:\boldsymbol{j})\cdot(dx\:\boldsymbol{i} + dy\:\boldsymbol{j}) = \int_{C} (x^2 ydx + (x+y)dy)$$

$$= \int_{0}^{2} x^2 x^2 dx + \int_{0}^{2}(x + x^2)(2xdx) = \left[\fr... ...} + \left[\frac{2}{3}x^{3} + \frac{2}{4}x^{4} \right ]_{0}^{2} = \frac{296}{15}$$

Alternate answer Express the curve $C$ using a parameter.

$$C : \boldsymbol{r}(t) = x\:\boldsymbol{i} + y\:\boldsymbol{j} = t\:\boldsymbol{i} + t^2\:\boldsymbol{j}, \ 0 \leq t \leq 2$$

Then

$$d\boldsymbol{r} = (\boldsymbol{i} + 2t\:\boldsymbol{j})dt, \ \bol... ...mbol{i} + (x+y)\:\boldsymbol{j} = t^4\:\boldsymbol{i} + (t+t^2)\:\boldsymbol{j}$$

Thus

$$\int_{C} \boldsymbol{F} \cdot{\bf t} ds = \int_{C}\boldsymbol{F}\cdot d\boldsymbol{r}$$

$$= \int_{0}^{2}(t^4\:\boldsymbol{i} + (t+t^2)\:\boldsymbol{j}) \cdot... ...5}t^{5} + \frac{2}{3}t^{3} + \frac{2}{4}t^{4} \right ]_{0}^{2} = \frac{296}{15}$$

In this example, $-C$ is a reversed curve of $C$ so that the direction is from $(2,4)$ to $(0,0)$. Then the parameteric expression of $-C$ is

$$-C : \boldsymbol{r}(t) = \boldsymbol{r}(2 - t) = (2-t)\:\boldsymbol{i} + (2-t)^2\:\boldsymbol{j}, \ 0 \leq t \leq 2$$

Thus,

$$d\boldsymbol{r} = (-\boldsymbol{i} -2(2-t)\:\boldsymbol{j})dt, \ ... ...x+y)\:\boldsymbol{j} = (2-t)^4\:\boldsymbol{i} + (2-t)+(2-t)^2\:\boldsymbol{j}$$

Therefore，

$$\int_{-C}\boldsymbol{F}\cdot d\boldsymbol{r} = \int_{0}^{2} ((2-t... ...2-t)+(2-t)^2)\:\boldsymbol{j})\cdot(-\boldsymbol{i} -2(2-t)\:\boldsymbol{j})dt$$

Note that let $2-t = u$. Then $du = -dt$ and

$$\int_{-C}\boldsymbol{F}\cdot d\boldsymbol{r} = - \int_{2}^{0}(u^4\:\boldsymbol{i} + (u+u^2)\:\boldsymbol{j}) \cdot(-\boldsymbol{i} -2u\:\boldsymbol{j})du$$

$$= - \int_{0}^{2}(u^4\:\boldsymbol{i} + (u+u^2)\:\boldsymbol{j}) \cd... ...bol{i} -2u\:\boldsymbol{j})du = - \int_{C} \boldsymbol{F} \cdot d\boldsymbol{r}$$

Therefore，

$$\int_{-C}\boldsymbol{F}\cdot d\boldsymbol{r} = - \int_{C}\boldsymbol{F}\cdot d\boldsymbol{r}$$

or

$$\int_{-C}\boldsymbol{F}\cdot {\bf t}ds = - \int_{C}\boldsymbol{F}\cdot {\bf t}ds$$

So far, in the conservation field, we have already learned that the vector field is equal in magnitude to the gradient of the scalar field. Now let's investigate what holds true in relation to line integrals.

Theorem 3..1  

In the vector field $\boldsymbol{F}(x,y,z)$, the following conditions are equivalent．

(1) there exists a scalar function $\phi(x,y,z)$ so that $\boldsymbol{F} = -\nabla \phi$

(2) For any closed curve $C$, $\oint_{C} \boldsymbol{F}\cdot d \boldsymbol{r} = 0$ holds. (independent of path)．

Potential energy

Suppose the force field $\boldsymbol{F}$ has the potential $U$. That is, suppose $\boldsymbol{F} =-\nabla U$ holds. Consider the curve $C$ in the field of this force, and assume that the curve $C$ goes from the point P to Q.．When a mass moves from point P to Q along this curve $C$ while being affected by this force field, the work $W$ that this mass receives from $\boldsymbol{F}$ is

$$W = \int_{C}\boldsymbol{F}\cdot d\boldsymbol{r} = U(P) - U(Q)$$

It doesn't matter how you choose the curve $C$ that connects the two points P and Q．Therefore, the value $U({\rm P})$ at the point P of the potential $U$ is called potential energy at the point P of this force field．

Example 3..10  

Let $\boldsymbol{r}= x\:\boldsymbol{i} + y\:\boldsymbol{j} + z\:\boldsymbol{k}$， $r= \vert\boldsymbol{r}\vert$. Then show that the vector field $\boldsymbol{F} = \frac{\boldsymbol{r}}{r^3}$ has the scalar potential $\phi = \frac{1}{r}$ and find the potential energy at each point in space.

Answer $\nabla \frac{1}{r} = -\frac{\boldsymbol{r}}{r^3}$ implies $\boldsymbol{F} = -\nabla \phi$. or，

$$\nabla \frac{1}{r} = \frac{\partial (r^{-1})}{\partial x}\:\boldsymbol{i} + \frac{\par... ...} + \frac{\partial (r^{-1})}{\partial r}\frac{d(r)}{\partial z}\:\boldsymbol{k}$$

$$= -r^{-2}(\frac{x\:\boldsymbol{i} + y\:\boldsymbol{j} + z\:\boldsymbol{k}}{r}) = - \frac{\boldsymbol{r}}{r^3}$$

Thus， $\boldsymbol{F} = - \nabla \frac{1}{r}$ and the vector field $\boldsymbol{F}$ has a potential $\phi = \frac{1}{r}$. Therefore, the potential energy at P is $\frac{1}{r}$．

Question 3..1  

Suppose the force field $\boldsymbol{F}$ has the potential $U$. Prove that the following equation holds when a mass point with mass $m$ moves in this force field and moves from point A to point B.

$$\frac{1}{2}mv_{A}^2 + U(A) = \frac{1}{2}mv_{B}^2 + U(B)$$

Here， $v_{A},v_{B}$ are magnitude of the velocity vectors at $A,B$, respectively．

Exercise3.3

1.

Given the force field ${\mathbf F} = 3xy \boldsymbol{i} - 5{\bf z} + 10 x\boldsymbol{k}$. Find the work done by ${\mathbf F}$ which is $W = \int_{C}{\mathbf F}\cdot d\boldsymbol{r}$, moving along the curve

$$C : x = t^2 + 1,\ y = 2t^2,\ z = t^3$$

from $t = 1$ to $t = 2$.

2.

Given the scalar field $\phi = 2xyz^2$，the field ${\mathbf F} = xy\boldsymbol{i} - z\boldsymbol{j} + x^2 \boldsymbol{k}$. Let the curve $C$ be parametrized by $x = t^2, y = 2t, z = t^3\ (0 \leq t \leq 1)$．Find the following line integrals.

(1) $\int_{C}\phi d\boldsymbol{r}$(2) $\int_{C}{\mathbf F} \times d\boldsymbol{r}$

3.

Let $\boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z \boldsymbol{k}$．Then for any closed curve $C$, prove that $\int_{C}\boldsymbol{r}\cdot d\boldsymbol{r} = 0$．

4.

Suppose that the force field ${\mathbf F}$ has the potential $U$．Within the force field, the point mass with the mass $m$ moves from the point A to the point B, show that following equation holds．

$$\frac{1}{2}mv_{A}^2 + U(A) = \frac{1}{2}mv_{B}^2 + U(B)$$

Here， $v_{A},v_{B}$ are the magnitude of velocity vectors of A and B.

5.

$\Phi = \tan^{-1}{\frac{y}{x}}$ is defined in the domain $D$ excluding the $z$ axis from the whole space． Let $C$ be a circle with a radius of $a$ centered at the origin O on the $xy$ plane. Prove the following equation.

$$\int_{C}(\nabla \phi)\cdot d\boldsymbol{r} = 2\pi$$