Definition

Suppose that we can describe a curve $C$ by the vector function $\mathbf{r}=\mathbf{r}\left(s\right),$ $0\le s\le S,$ where the variable $s$ is the arc length of the curve (Figure $1\text{).}$

If a scalar function $F$ is defined over the curve $C,$ then the integral $\underset{0}{\overset{S}{\int }}F\left(\mathbf{r}\left(s\right)\right)ds$ is called a line integral of scalar function $F$ along the curve $C$ and denoted as

The line integral $\underset{C}{\int }Fds$ exists if the function $F$ is continuous on the curve $C.$

Properties of Line Integrals of Scalar Functions

The line integral of a scalar function has the following properties:

1. The line integral of a scalar function over the smooth curve $C$ does not depend on the orientation of the curve;
2. If $C_1$ is a curve that begins at $A$ and ends at $B,$ and if $C_2$ is a curve that begins at $B$ and ends at $D$ (Figure $2$), then their union is defined to be the curve $C_1\cup C_2$ that progresses along the curve $C_1$ from $A$ to $B,$ and then along $C_2$ from $B$ to $D,$ so that
   $$\underset{C_1\cup C_2}{\int }Fds=\underset{C_1}{\int }Fds+\underset{C_2}{\int }Fds;$$



3. If the smooth curve $C$ is parameterized by $\mathbf{r}=\mathbf{r}\left(t\right),$ $\alpha \le t\le \beta$ and the scalar function $F$ is continuous on the curve $C,$ then
   $$\underset{C}{\int }F\left(x,y,z\right)ds=\underset{\alpha }{\overset{\beta }{\int }}F\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)\sqrt{{\left(x^{\prime }\left(t\right)\right)}^2+{\left(y^{\prime }\left(t\right)\right)}^2+{\left(z^{\prime }\left(t\right)\right)}^2}dt.$$
4. If $C$ is a smooth curve in the $xy$-plane given by the equation $y=f\left(x\right),$ $a\le x\le b,$ then
   $$\underset{C}{\int }F\left(x,y\right)ds=\underset{a}{\overset{b}{\int }}F\left(x,f\left(x\right)\right)\sqrt{1+{\left(f^{\prime }\left(x\right)\right)}^2}dx.$$
5. Similarly, if a smooth curve $C$ in the $xy$-plane is defined by the equation $x=\phi \left(y\right),$ $c\le y\le d,$ then
   $$\underset{C}{\int }F\left(x,y\right)ds=\underset{c}{\overset{d}{\int }}F\left(\phi \left(y\right),y\right)\sqrt{1+{\left({\phi }^{\prime }\left(y\right)\right)}^2}dy;$$
6. In polar coordinates the line integral $\underset{C}{\int }F\left(x,y\right)ds$ becomes
   $$\underset{C}{\int }F\left(x,y\right)ds=\underset{\alpha }{\overset{\beta }{\int }}F\left(r\cos \theta ,r\sin \theta \right)\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}d\theta ,$$
   where the curve $C$ is defined by the polar function $r\left(\theta \right).$

Solved Problems

Click or tap a problem to see the solution.

Solution.

Solution.

The arc length differential is

Then applying the formula

in the $xy$-plane, we obtain