Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This comes from the beginning chapter on line integrals in the book Mathematical Methods for Science Students:

Suppose $y=f(x)$ is a real single-valued monotonic continuous function of $x$ in some interval $x_1<x<x_2$. Then if $P(x,y)$ and $Q(x,y)$ are two real single-valued continuous functions of $x$ and $y$ for all points of C, the integrals $$\int_C P(x,y)dx,\quad\int_C Q(x,y)dy$$ and, more frequently, their sum $$\int_C \Bigg\{P(x,y)dx+Q(x,y)dy\Bigg\}$$ are called curvilinear integrals or line integrals, the path of integration C being along the curve $y = f(x)$ from A to B.

Is this a correct definition of a line integral?

It doesn't appear to resemble the wikipedia definition of a line integral over a scalar field, which makes sense to me:

For some scalar field $f : U\subseteq R^n → R$, the line integral along a piecewise smooth curve $C \subset U$ is defined as$$\int_C f ds = \int^b_af(r(t))|r'(t)|dt$$ where r: [a, b] → C is an arbitrary bijective parametrization of the curve C such that r(a) and r(b) give the endpoints of C and a

The function f is called the integrand, the curve C is the domain of integration, and the symbol ds may be intuitively interpreted as an elementary arc length. Line integrals of scalar fields over a curve C do not depend on the chosen parametrization r of C.

share|improve this question
    
the first on is very strange. Of course you can write that down and call it curvilinear integral, but an explanation about what $P$ and $Q$ are supposed to be in that definition would be in order. The second one is a standard definition of what people refer to when talking about line integrals, though usually requesting some smoothness (e.g. $C^1$, but you can do with less) for $r$. –  user20266 Jul 31 '12 at 16:14

2 Answers 2

up vote 4 down vote accepted

The first is a line integral over a vector field (presented quite horribly), defined as

$$\int_a^bF(r(t))\cdot r'(t) dt$$

$F$ represents the formula for the field, and $r$ represents the path. If we write $r$ as $(x(t),y(t))$, then we get that $r'(t)=(\frac{dx}{dt},\frac{dy}{dt})$

If $F(x,y)$ represents the equation defining the vector field, we can write it in component form as $F(x,y)=(P(x,y),Q(x,y))$. so $P$ defines the $x$-component of the vector field at each point $(x,y)$, and $Q$ does the same for the $y$-component. Expanding the dot product:

$$\int_a^b P(x,y)\frac{dx}{dt} dt + Q(x,y)\frac{dy}{dt} dt$$ $$\int_a^b P(x,y)dx + Q(x,y)dy$$

Gives your version. This is useful for determining the work done by a "field type" force (an electromagnetic field, for example) on a moving object.

The second is a line integral over a scalar field. This is just the normal integral really, just extended so it can be defined over any curve, and not just the $x$-axis in the 2D case which is rather limiting.

We call them both line integrals (since we integrate over a curve), but one is over a vector field and the other is over a scalar field, leading to different definitions.

share|improve this answer
    
I feel rather stupid for not seeing this :) –  user10389 Jul 31 '12 at 17:18
    
@user10389 Its not presented well at all. Defining it as the sum of the integrals in each direction (without even saying that's what they represent) works, but it's bizarre. I don't really blame you. You'd generally work with the first version anyway, since you're really integrating over the path (and therefore over $t$). –  Robert Mastragostino Jul 31 '12 at 17:50

The two "line integrals" you are referring to represent completely different things: Let $$\gamma:\quad t\mapsto {\bf z}(t)=\bigl(x(t),y(t)\bigr)\qquad(a\leq t\leq b)$$ be a curve in the $(x,y)$-plane. A graph $x\mapsto\bigl(x,f(x)\bigr)$ $\,(a\leq x\leq b)$ is a special case of this.

When a force field ${\bf F}(x,y):=\bigl(P(x,y),Q(x,y)\bigr)$ is given then the integral $$W:=\int_\gamma {\bf F}\cdot d{\bf z}:=:=\int_a^b {\bf F}\bigl({\bf z}(t))\cdot\dot {\bf z}(t)\ dt\ ,$$ resp., componentwise $$W:=\int_\gamma(P\,dx+Q\,dy):=\int_a^b\Bigl(P\bigl(x(t),y(t)\bigr)\,\dot x(t)+ Q\bigl(x(t),y(t)\bigr)\,\dot y(t)\Bigr)\,dt$$ denotes a work done when the force field moves a cart along $\gamma$ against friction.

On the other hand, when a scalar field $f(x,y)$ is given then the "line integral" $$H:=\int_\gamma f\ ds:=\int_a^b f\bigl({\bf z}(t)\bigr)\ |\dot z(t)|\ dt=\int_a^b f\bigl(x(t),y(t)\bigr)\,\sqrt{\dot x^2(t)+\dot y^2(t)}\ dt$$ denotes a quantity which is related to arc length: Assume that $f(x,y)$ denotes some sort of humidity at the point $(x,y)$. Then $H$ is essentially the total amount of fluid contained in the (physical) thread $\gamma$.

The essential point of the idea of a "line integral" is that the values $W$ or $H$ only depend (a) on ${\bf F}$ resp. $f$ and (b) on the curve $\gamma$ as a "geometrical object", but not on the chosen parametrization of $\gamma$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.