# Calculus

## Line Integrals # Line Integrals of Scalar Functions

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

${\int\limits_C {F\left( {x,y,z} \right)ds} \;\;\;}\kern-0.3pt{\text{or}\;\;\int\limits_C {Fds} .}$

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
${\int\limits_C {F\left( {x,y,z} \right)ds} } = {\int\limits_\alpha ^\beta {F\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) \cdot}}\kern0pt{{ \sqrt {{{\left( {x’\left( t \right)} \right)}^2} + {{\left( {y’\left( t \right)} \right)}^2} + {{\left( {z’\left( t \right)} \right)}^2}} dt} }$
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
${\int\limits_C {F\left( {x,y} \right)ds} } = {\int\limits_a^b {F\left( {x,f\left( x \right)} \right) \cdot}}\kern0pt{{ \sqrt {1 + {{\left( {f’\left( x \right)} \right)}^2}} dx} ;}$
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
${\int\limits_C {F\left( {x,y} \right)ds} } = {\int\limits_c^d {F\left( {\varphi \left( y \right),y} \right) \cdot}}\kern0pt{{ \sqrt {1 + {{\left( {\varphi’\left( y \right)} \right)}^2}} dy} ;}$
7. In polar coordinates the line integral $$\int\limits_C {F\left( {x,y} \right)ds}$$ becomes
${\int\limits_C {F\left( {x,y} \right)ds} } = {\int\limits_\alpha ^\beta {F\left( {r\cos \theta ,r\sin \theta } \right) \cdot}}\kern0pt{{ \sqrt {{r^2} + {{\left( {\frac{{dr}}{{d\theta }}} \right)}^2}} d\theta } ,}$
where the curve $$C$$ is defined by the polar function $$r\left( \theta \right).$$

• ## Solved Problems

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

### Example 3

Evaluate the line integral $$\int\limits_C {{x^2}ds},$$ where $$C$$ is a the curve given by the equation $$y = f\left( x \right) = \ln x,$$ $$1 \le x \le e.$$

### Example 4

Evaluate the integral $$\int\limits_C {ds}$$ over the plane curve, where $$C$$ is the line segment from $$O\left( {0,0} \right)$$ to $$A\left( {1,2} \right)$$ (see Figure $$4$$).

### Example 5

Calculate the integral $$\int\limits_C {\left( {{x^2} + {y^2}} \right)zds},$$ where the curve $$C$$ is parameterized by $$\mathbf{r}\left( t \right) = \left( {\sin 3t,\cos 3t,4t} \right)$$, $$0 \le t \le \pi.$$

### Example 6

Calculate the line integral $$\int\limits_C {\large\frac{{ds}}{{y – x}}\normalsize},$$ where the curve $$C$$ is the line segment from $$\left( {0, – 2} \right)$$ to $$\left( {4, 0} \right)$$ (Figure $$5$$).

### Example 7

Calculate the line integral $$\int\limits_C {xyds},$$ where the curve $$C$$ is the part of ellipse $${\large\frac{{{x^2}}}{{{a^2}}}\normalsize} + {\large\frac{{{y^2}}}{{{b^2}}}\normalsize} = 1$$ lying in the first quadrant (Figure $$6$$).

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

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

${\int\limits_C {F\left( {x,y,z} \right)ds} } = {\int\limits_\alpha ^\beta {F\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) \cdot}}\kern0pt{{ \sqrt {{{\left( {x’\left( t \right)} \right)}^2} + {{\left( {y’\left( t \right)} \right)}^2} + {{\left( {z’\left( t \right)} \right)}^2}} dt} }$

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

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

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Problems 1-2
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Problems 3-7