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Calculus

# Properties and Applications of Line Integrals

Scalar functions: $$F\left( {x,y,z} \right),$$ $$F\left( {x,y} \right),$$ $$f\left( x \right)$$
Scalar potential: $$u\left( {x,y,z} \right)$$
Curves: $$C,$$ $${C_1},$$ $${C_2}$$
Limits of integration: $$a,$$ $$b,$$ $$\alpha,$$ $$\beta$$
Parameters: $$t$$, $$s$$
Polar coordinates: $$r$$, $$\theta$$
Vector field: $$\mathbf{F}\left( {P,Q,R} \right)$$
Position vector: $$\mathbf{r}\left( s \right)$$
Unit vectors: $$\mathbf{i},$$ $$\mathbf{j},$$ $$\mathbf{k},$$ $$\vec{\tau}$$
Area of a region: $$S$$
Length of a curve: $$L$$

Mass of a wire: $$m$$
Density: $$\rho\left( {x,y,z} \right),$$ $$\rho\left( {x,y} \right)$$
Coordinates of the center of mass: $$\bar x,$$ $$\bar y,$$ $$\bar z$$
First moments: $${M_{xy}},$$ $${M_{yz}},$$ $${M_{xz}}$$
Moments of inertia: $${I_x},$$ $${I_y},$$ $${I_z}$$
Volume of a solid: $$V$$
Work: $$W$$
Magnetic field: $$\mathbf{B}$$
Current: $$I$$
Electromotive force: $$\varepsilon$$
Magnetic flux: $$\psi$$

1. Line integral of a scalar function
Let a curve $$C$$ be given by the vector function $$\mathbf{r} = \mathbf{r}\left( s \right)$$, $$0 \le s \le S,$$ and a scalar function $$F$$ is defined over the curve $$C$$.
The line integral of the scalar function $$F$$ over the curve $$C$$ is written in the form
$${\large\int\limits_0^S\normalsize} {F\left( {\mathbf{r}\left( s \right)} \right)ds} =$$ $${\large\int\limits_C\normalsize} {F\left( {x,y,z} \right)ds} =$$ $${\large\int\limits_C\normalsize} {Fds},$$
where $$ds$$ is the arc length differential.
2. The line integral of a scalar function over a union of two curves is equal to the sum of the line integrals over each curve:
$${\large\int\limits_{{C_1} \cup {C_2}}\normalsize} {Fds} =$$ $${\large\int\limits_{{C_1}}\normalsize} {Fds} + {\large\int\limits_{{C_2}}\normalsize} {Fds}$$
1. If a smooth curve $$C$$ is parametrized by the equation $$\mathbf{r} = \mathbf{r}\left( t \right),$$ $$\alpha \le t \le \beta,$$ then the line integral of a scalar function is expressed by the formula
$${\large\int\limits_C\normalsize} {F\left( {x,y,z} \right)ds} =$$ $${\large\int\limits_\alpha^\beta\normalsize} {F\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) }$$ $${ \sqrt {{{\left( {x’\left( t \right)} \right)}^2} + {{\left( {y’\left( t \right)} \right)}^2} + {{\left( {z’\left( t \right)} \right)}^2}} dt}$$
2. If $$C$$ is a smooth curve lying in the $$xy$$-plane and defined by the explicit equation $$y = f\left( x \right)$$, $$a \le x \le b$$, then the line integral is given by the expression
$${\large\int\limits_C\normalsize} {F\left( {x,y} \right)ds} =$$ $${\large\int\limits_a^b\normalsize} {F\left( {x,f\left( x \right)} \right) }$$ $${\sqrt {1 + {{\left( {f’\left( x \right)} \right)}^2}} dx}$$
3. Line integral of a scalar function in polar coordinates
$${\large\int\limits_C\normalsize} {F\left( {x,y} \right)ds} =$$ $${\large\int\limits_\alpha^\beta\normalsize} {F\left( {r\cos \theta ,r\sin \theta } \right) }$$ $${\sqrt {{r^2} + {{\left( {{\large\frac{{dr}}{{d\theta }}}\normalsize} \right)}^2}} d\theta } ,$$
where the curve $$C$$ is defined by the polar function $$r\left( \theta \right).$$
4. Line integral of a vector field
Let a curve $$C$$ be defined by the vector function $$\mathbf{r} = \mathbf{r}\left( s \right),$$ $$0 \le s \le S.$$ The vector
$${\large\frac{{d\mathbf{r}}}{{ds}}\normalsize} = \vec{\tau} =$$ $$\left( {\cos \alpha ,\cos \beta ,\cos \gamma } \right)$$
is the unit vector of the tangent line to this curve.

Let also a vector field $$\mathbf{F}\left( {P,Q,R} \right)$$ be defined over the curve $$C$$. Then the line integral of the vector function $$\mathbf{F}$$ along the curve $$C$$ is expressed in the form
$${\large\int\limits_C\normalsize} {Pdx + Qdy + Rdz} =$$ $${\large\int\limits_0^S\normalsize} {\big( {P\cos \alpha + Q\cos \beta }}$$ $$+\;{{ R\cos \gamma } \big)ds}$$

1. Properties of line integrals of vector fields
$${\large\int\limits_{ – C}\normalsize} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} =$$ $$– {\large\int\limits_C\normalsize} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)},$$
where $$-C$$ denotes the curve with the opposite orientation.
$${\large\int\limits_C\normalsize} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} =$$ $${\large\int\limits_{{C_1} \cup {C_2}}\normalsize} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} =$$ $${\large\int\limits_{C_1}\normalsize} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)}$$ $$+\;{\large\int\limits_{C_2}\normalsize} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} ,$$
where $$C$$ is the union of the curves $${C_1}$$ and $${C_2}$$.
2. If the curve $$C$$ is parametrized by $$\mathbf{r}\left( t \right) =$$ $$\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right),$$ $$\alpha \le t \le \beta,$$ then the line integral of the vector field is written as
$${\large\int\limits_C\normalsize} {Pdx + Qdy + Rdz} =$$ $${\large\int\limits_\alpha^\beta\normalsize} {\Big[ {P\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) {\large\frac{{dx}}{{dt}}\normalsize} }}$$
$${{+\; {Q\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) {\large\frac{{dy}}{{dt}}\normalsize}} }}$$ $$+\;{{R\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right){\large\frac{{dz}}{{dt}}\normalsize}} \Big]dt}$$
3. If the curve $$C$$ lies in the $$xy$$-plane and defined by the equation $$y = f\left( x \right),$$ $$a \le x \le b,$$ then the line integral of the vector field is given by
$${\large\int\limits_C\normalsize} {Pdx + Qdy} =$$ $${\large\int\limits_\alpha^\beta\normalsize} {\Big[ {P\left( {x,f\left( x \right)} \right) }}$$ $$+\;{{ Q\left( {x,f\left( x \right)} \right){\large\frac{{df}}{{dx}}\normalsize}} \Big] dx}$$
4. Green’s theorem
$${\large\iint\limits_C\normalsize} {\left( {{\large\frac{{\partial Q}}{{\partial x}}\normalsize} – {\large\frac{{\partial P}}{{\partial y}}\normalsize}} \right)dxdy} =$$ $${\large\oint\limits_C\normalsize} {Pdx + Qdy},$$
where $$\mathbf{F} =$$ $$P\left( {x,y} \right)\mathbf{i}$$ $$+\; Q\left( {x,y} \right)\mathbf{j}$$ is a continuous vector function with continuous first partial derivatives $$\partial P/\partial y,$$ $$\partial Q/\partial x$$ in a some domain $$R,$$ which is bounded by a closed, piecewise smooth curve $$C.$$
5. Area of a region $$R$$ bounded by a closed curve $$C$$
$$S = {\large\iint\limits_R\normalsize} {dxdy} =$$ $${\large\frac{1}{2}\oint\limits_C\normalsize} {xdy – ydx}$$
6. Path independence of line integrals
The line integral of a vector function $$\mathbf{F} = P\mathbf{i}$$ $$+\; Q\mathbf{j}$$ $$+\; R\mathbf{k}$$ is said to be path independent, if and only if $$P$$, $$Q$$ and $$R$$ are continuous in a domain $$D$$ and if there exists some scalar function $$u = u\left( {x,y,z} \right)$$ (a scalar potential) such that
$$\mathbf{F} = \text{grad }u$$ or $$\partial u/\partial x = P,$$ $$\partial u/\partial y = Q,$$ $$\partial u/\partial z = R.$$
Then the line integral is given by
$${\large\int\limits_C\normalsize} {\mathbf{F}\left( \mathbf{r} \right) \cdot d\mathbf{r}} =$$ $${\large\int\limits_C\normalsize} {Pdx + Qdy + Rdz} =$$ $$u\left( B \right) – u\left( A \right).$$
7. Test for a conservative field
A vector field of the form $$\mathbf{F} = \text{grad }u$$ is called a conservative field. The line integral of a vector function $$\mathbf{F} = P\mathbf{i} + Q\mathbf{j}$$ $$+\; R\mathbf{k}$$ is path independent if and only if
$$\text{rot }\mathbf{F} =$$ $$\left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {\large\frac{\partial }{{\partial x}}\normalsize} & {\large\frac{\partial }{{\partial y}}\normalsize} & {\large\frac{\partial }{{\partial z}}\normalsize}\\ P & Q & R \end{array}} \right|$$ $$= \mathbf{0}.$$
If the line integral is taken in the $$xy$$-plane, then the following formula for a conservative field is valid:
$${\large\int\limits_C\normalsize} {Pdx + Qdy} =$$ $$u\left( B \right) – u\left( A \right).$$
For the two-dimensional case, the test for a conservative field can be written in the form
$${\large\frac{{\partial P}}{{\partial y}}\normalsize} = {\large\frac{{\partial Q}}{{\partial x}}\normalsize}.$$
8. Length of a curve
$$L = {\large\int\limits_C\normalsize} {ds} =$$ $${\large\int\limits_\alpha^\beta\normalsize} {\left| {{\large\frac{{d\mathbf{r}}}{{dt}}\normalsize}\left( t \right)} \right|dt} =$$ $${\large\int\limits_\alpha^\beta\normalsize} {\sqrt {{{\left( {\large\frac{{dx}}{{dt}}\normalsize} \right)}^2} + {{\left( {\large\frac{{dy}}{{dt}}\normalsize} \right)}^2} + {{\left( {\large\frac{{dz}}{{dt}}\normalsize} \right)}^2}} dt}$$ where $$C$$ is a piecewise smooth curve defined by the position vector $$\mathbf{r}\left( t \right),$$ $$\alpha \le t \le \beta.$$
If the curve $$C$$ is two-dimensional, then
$$L = {\large\int\limits_C\normalsize} {ds} =$$ $${\large\int\limits_\alpha^\beta\normalsize} {\left| {{\large\frac{{d\mathbf{r}}}{{dt}}\normalsize}\left( t \right)} \right|dt} =$$ $${\large\int\limits_\alpha^\beta\normalsize} {\sqrt {{{\left( {\large\frac{{dx}}{{dt}}\normalsize} \right)}^2} + {{\left( {\large\frac{{dy}}{{dt}}\normalsize} \right)}^2}} dt} .$$
If the curve $$C$$ lies in the $$xy$$-plane and is described by the explicit function $$y = f\left( x \right)$$, $$a \le x \le b,$$ then its length is given by
$$L = {\large\int\limits_a^b\normalsize} {\sqrt {1 + {{\left( {\large\frac{{dy}}{{dx}}\normalsize} \right)}^2}} dx}.$$
9. Length of a curve in polar coordinates
$$L = {\large\int\limits_\alpha^\beta\normalsize} {\sqrt {{{\left( {\large\frac{{dr}}{{d\theta }}\normalsize} \right)}^2} + {r^2}} d\theta } ,$$
where the curve $$C$$ is determined by the equation $$r = r\left( \theta \right),$$ $$\alpha \le \theta \le \beta$$ in polar coordinates.
10. Mass of a wire
$$m = {\large\int\limits_C\normalsize} {\rho \left( {x,y,z} \right)ds} ,$$
where $${\rho \left( {x,y,z} \right)}$$ is the mass per unit length of the wire.
If the curve $$C$$ is parametrized by the vector function
$$\mathbf{r}\left( t \right) =$$ $$\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right),$$ $$\alpha \le t \le \beta,$$
then its mass can be computed by the formula
$$m =$$ $${\large\int\limits_\alpha^\beta\normalsize} {\rho \left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) }$$ $${ \sqrt {{{\left( {\large\frac{{dx}}{{dt}}\normalsize} \right)}^2} + {{\left( {\large\frac{{dy}}{{dt}}\normalsize} \right)}^2} + {{\left( {\large\frac{{dz}}{{dt}}\normalsize} \right)}^2}} dt} .$$
If the curve $$C$$ lies in the $$xy$$-plane, then its mass is given by
$$m = {\large\int\limits_C\normalsize} {\rho \left( {x,y} \right)ds}$$
or
$$m =$$ $${\large\int\limits_\alpha^\beta\normalsize} {\rho \left( {x\left( t \right),y\left( t \right)} \right) }$$ $${ \sqrt {{{\left( {\large\frac{{dx}}{{dt}}\normalsize} \right)}^2} + {{\left( {\large\frac{{dy}}{{dt}}\normalsize} \right)}^2}} dt}$$
(in parametric form)
11. Center of mass of a wire
$$\bar x = {\large\frac{{{M_{yz}}}}{m}\normalsize},\;$$ $$\bar y = {\large\frac{{{M_{xz}}}}{m}\normalsize},\;$$ $$\bar z = {\large\frac{{{M_{xy}}}}{m}\normalsize}$$, where
$${M_{yz}} = {\large\int\limits_C\normalsize} {x\rho \left( {x,y,z} \right)ds},\;$$
$${M_{xz}} = {\large\int\limits_C\normalsize} {y\rho \left( {x,y,z} \right)ds}.\;$$
$${M_{xy}} = {\large\int\limits_C\normalsize} {z\rho \left( {x,y,z} \right)ds} .$$
12. Moments of inertia
The moments of inertia of a curve about the $$x$$-axis, $$y$$-axis and $$z$$-axis are given by the formulas
$${I_x} = {\large\int\limits_C\normalsize} {\left( {{y^2} + {z^2}} \right) }$$ $${\rho \left( {x,y,z} \right)ds},\;$$
$${I_y} = {\large\int\limits_C\normalsize} {\left( {{x^2} + {z^2}} \right) }$$ $${\rho \left( {x,y,z} \right)ds},\;$$
$${I_z} = {\large\int\limits_C\normalsize} {\left( {{x^2} + {y^2}} \right) }$$ $${\rho \left( {x,y,z} \right)ds}.$$
13. Area of a region bounded by a closed curve
$$S = {\large\oint\limits_C\normalsize} {xdy} =$$ $$– {\large\oint\limits_C\normalsize} {ydx} =$$ $${\large\frac{1}{2}\oint\limits_C\normalsize} {xdy – ydx}$$

If the closed curve $$C$$ is given in parametric form $$\mathbf{r}\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right),$$ $$\alpha \le t \le \beta ,$$ then the area can be calculated by the formula
$$S = {\large\int\limits_\alpha^\beta\normalsize} {x\left( t \right){\large\frac{{dy}}{{dt}}\normalsize} dt} =$$ $$– {\large\int\limits_\alpha^\beta\normalsize} {y\left( t \right){\large\frac{{dx}}{{dt}}\normalsize} dt} =$$ $${\large\frac{1}{2} \int\limits_\alpha^\beta\normalsize} {\left( {x\left( t \right){\large\frac{{dy}}{{dt}}\normalsize} – y\left( t \right){\large\frac{{dx}}{{dt}}\normalsize}} \right)dt} .$$

1. Volume of a solid formed by rotating a closed curve about the x-axis
$$V = – \pi {\large\oint\limits_C\normalsize} {{y^2}dx} =$$ $$– 2\pi {\large\oint\limits_C\normalsize} {xydy} =$$ $$– {\large\frac{\pi }{2}\oint\limits_C\normalsize} {2xydy + {y^2}dx}$$
1. Work of a field of forces
Work done by a force $$\mathbf{F}$$ on an object moving along a curve $$C$$ is described by the line integral
$$W = {\large\int\limits_C\normalsize} {\mathbf{F} \cdot d\mathbf{r}} ,$$
where $$d\mathbf{r}$$ is the unit tangent vector.

If the object is moved along a curve $$C$$ lying in the $$xy$$-plane, then the work is determined by the following formula
$$W = {\large\int\limits_C\normalsize} {\mathbf{F} \cdot d\mathbf{r}} =$$ $${\large\int\limits_C\normalsize} {Pdx + Qdy} .$$
If the path $$C$$ is specified by a parameter $$t$$ ($$t$$ often means time), then the formula for calculating work becomes
$$W = {\large\int\limits_\alpha^\beta\normalsize} {\Big[ {P\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right){\large\frac{{dx}}{{dt}}\normalsize} }}$$ $$+\;{{ Q\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right){\large\frac{{dy}}{{dt}}\normalsize} }}$$ $$+\;{{ R\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right){\large\frac{{dz}}{{dt}}\normalsize}} \Big]dt},$$
where $$t$$ goes from $$\alpha$$ to $$\beta.$$
If the vector field $$\mathbf{F}$$ is conservative and $$u\left( {x,y,z} \right)$$ is the scalar potential of this field, then the work on an object moving from $$A$$ to $$B$$ can be found by the formula
$$W = u\left( B \right) – u\left( A \right)$$.

1. Ampere’s law
$${\large\oint\limits_C\normalsize} {\mathbf{B} \cdot d\mathbf{r}} = {\mu _0}I$$
The line integral of a magnetic field $$\mathbf{B}$$ around a closed path $$C$$ is equal to the current $$I$$ (times the coefficient $${\mu _0}$$) flowing through the area bounded by the contour $$C.$$
$$\varepsilon = {\large\oint\limits_C\normalsize} {\mathbf{E} \cdot d\mathbf{r}} =$$ $$– {\large\frac{{d\psi }}{{dt}}\normalsize}$$
The electromotive force $$\varepsilon$$ induced around a closed loop $$C$$ is equal to the rate of change of the magnetic flux $$\psi$$ passing through the loop.