Is this a correct definition of a line integral?

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.

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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

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.

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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 $\gamma$.

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