# Calculus

## Line Integrals # Line Integrals of Scalar Functions

## Definition

Suppose that we can describe a curve C by the vector function r = r (s), 0 ≤ sS, where the variable s is the arc length of the curve (Figure 1).

If a scalar function $$F$$ is defined over the curve $$C,$$ then the integral $$\int\limits_0^S {F\left( {\mathbf{r}\left( s \right)} \right)ds}$$ is called a line integral of scalar function $$F$$ along the curve $$C$$ and denoted as

$\int\limits_C {F\left( {x,y,z} \right)ds} \;\;\text{or}\;\;\int\limits_C {Fds} .$

The line integral $$\int\limits_C {Fds}$$ 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
$\int\limits_{{C_1} \cup {C_2}} {Fds} = \int\limits_{{C_1}} {Fds} + \int\limits_{{C_2}} {Fds} ;$
4. 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
$\int\limits_C {F\left( {x,y,z} \right)ds} = \int\limits_\alpha ^\beta {F\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) \sqrt {{{\left( {x'\left( t \right)} \right)}^2} + {{\left( {y'\left( t \right)} \right)}^2} + {{\left( {z'\left( t \right)} \right)}^2}} dt}.$
5. 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
$\int\limits_C {F\left( {x,y} \right)ds} = \int\limits_a^b {F\left( {x,f\left( x \right)} \right) \sqrt {1 + {{\left( {f'\left( x \right)} \right)}^2}} dx} .$
6. Similarly, if a smooth curve $$C$$ in the $$xy$$-plane is defined by the equation $$x = \varphi \left( y \right),$$ $$c \le y \le d,$$ then
$\int\limits_C {F\left( {x,y} \right)ds} = \int\limits_c^d {F\left( {\varphi \left( y \right),y} \right) \sqrt {1 + {{\left( {\varphi'\left( y \right)} \right)}^2}} dy} ;$
7. In polar coordinates the line integral $$\int\limits_C {F\left( {x,y} \right)ds}$$ becomes
$\int\limits_C {F\left( {x,y} \right)ds} = \int\limits_\alpha ^\beta {F\left( {r\cos \theta ,r\sin \theta } \right) \sqrt {{r^2} + {{\left( {\frac{{dr}}{{d\theta }}} \right)}^2}} d\theta } ,$
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.

### Example 1

Evaluate the line integral $\int\limits_C {{x^2}yds}$ along the segment of the line $$y = x$$ from the origin up to the point $$\left( {2,2} \right)$$ (see Figure $$3$$).

### Example 2

Calculate the line integral $\int\limits_C {{y^2}ds},$ where $$C$$ is a part of the circle

$x = a\cos t, y = a\sin t, 0 \le t \le {\frac{\pi }{2}}.$

### Example 1.

Evaluate the line integral $\int\limits_C {{x^2}yds}$ along the segment of the line $$y = x$$ from the origin up to the point $$\left( {2,2} \right)$$ (see Figure $$3$$).

Solution.

$\int\limits_C {{x^2}yds} = \int\limits_0^2 {{x^2} \cdot x\sqrt {1 + {1^2}} dx} = \sqrt 2 \int\limits_0^2 {{x^3}dx} = \sqrt 2 \left[ {\left. {\left( {\frac{{{x^4}}}{4}} \right)} \right|_0^2} \right] = 4\sqrt 2 .$

### Example 2.

Calculate the line integral $\int\limits_C {{y^2}ds},$ where $$C$$ is a part of the circle

$x = a\cos t, y = a\sin t, 0 \le t \le {\frac{\pi }{2}}.$

Solution.

The arc length differential is

$ds = \sqrt {{{\left( {x'\left( t \right)} \right)}^2} + {{\left( {y'\left( t \right)} \right)}^2}} dt = \sqrt {{a^2}{{\sin }^2}t + {a^2}{{\cos }^2}t} \,dt = adt.$

Then applying the formula

$\int\limits_C {F\left( {x,y,z} \right)ds} = \int\limits_\alpha ^\beta {F\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) \sqrt {{{\left( {x'\left( t \right)} \right)}^2} + {{\left( {y'\left( t \right)} \right)}^2} + {{\left( {z'\left( t \right)} \right)}^2}} dt}$

in the $$xy$$-plane, we obtain

$\int\limits_C {{y^2}ds} = \int\limits_0^{\frac{\pi }{2}} {{a^2}{{\sin }^2}t \cdot adt} = {a^3}\int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}tdt} = \frac{{{a^3}}}{2}\int\limits_0^{\frac{\pi }{2}} {\left( {1 - \cos 2t} \right)dt} = \frac{{{a^3}}}{2}\left[ {\left. {\left( {t - \frac{{\sin 2t}}{2}} \right)} \right|_0^{\frac{\pi }{2}}} \right] = \frac{{{a^3}}}{2} \cdot \frac{\pi }{2} = \frac{{{a^3}\pi }}{4}.$

See more problems on Page 2.