Line Integrals and Surface Integrals

Can someone please explain what surface integrals and line integrals are measuring?

Is a line integral the arc length along a surface, and a surface integral is the surface area? Also, why is a line integral equal to $0$ on a conservative closed path?

Thank you!

• Line integral are not always equal to zero around a closed path ~ That's the case only if the function in question is conservative (i.e. it has zero curl in a compact domain). May 23 '14 at 0:25
• OK thanks. But what does that mean exactly? May 23 '14 at 0:28
• do you mean line and surface integrals of vector fields, or of scalar fields? "conservative," in your last question, refers to vector fields. in that case, if you interpret the field as a force field, a line integral is the work done by the field around the curve, and a surface integral is the flux of the field through the surface. May 23 '14 at 0:33
• Integrals of scalar-valued functions are hard to conceptualize visually in the general case because your region of integration, at least when first learning multivariable calculus, is almost universally represented in 3-space, which doesn't really leave room to graph that 4th dimension (the output of the scalar-valued function). If your region of integration is in the $xy$-plane however, and you're integrating something of the form $f(x,y)$, you can think of the output of your integral as the volume (or surface area) "below" $z=f(x,y)$. May 23 '14 at 1:23
• @MSIS: You need to clarify exactly what kind of object you're integrating. Let $\mathbf{F}$ be a vector field on a region of $\mathbb{R}^n$. You can integrate the dot product $\mathbf{F}\cdot\mathbf{dr}$ around a curve $c$ contained in that region; this is called a "line integral" of $\mathbf{F}$ along the curve $c$. You can also integrate $\mathbf{F}\cdot\mathbf{dA}$ along a surface $S$, where $\mathbf{dA}$ is related to the normal vector of the surface; this is called a "flux integral." These are different concepts than integrals of plain-old real-valued functions along a curve or surface. Jan 15 '20 at 17:32

There seems to be some confusion here, so first I'll answer the questions and then explain the reasoning.

Is a line integral the arc length along a surface, and a surface integral is the surface area?

No.

Why is a line integral equal to 0 on a closed path?

It isn't. (I've included the original text of your question instead of the edited text here for the sake of the explanation).

Now that that's out of the way, let's start by thinking about the concepts at hand here.

First, an integral. An integral, is, at its core, a summation over a region. Take, for example, $\int_a^b f(x)dx$. Put crudely, it gives you the signed area "under" $f(x)$ by summing up the "heights" of $f(x)$ along the portion of the $x$-axis between $x=a$ and $x=b$.

What, then, are line integrals and surface integrals? These are integrals described by the region over which they are integrating. In the example $\int_a^b f(x)dx$, the region of integration is the $x$-axis from $x=a$ to $x=b$. In the case of a line integral, the region of integration is a curve (yes, this does seem like a bit of a misnomer; it's a rather unfortunate consequence of the new meanings that "linearity" acquires as the level of the math at hand gets higher - for your purposes, thinking of a curve as a twisted line may be the most intuitive method); in the case of a surface integral, the region of integration is a surface.

The line integral and surface integral of $1$ do indeed give you, respectively, the length of the curve and the surface area of the surface that serve as the regions of integration. But there is no such thing as "the arc length along a surface" (there is such a thing as the length of the boundary of a surface, if said boundary exists of course, but arc length is a characteristic of 1-dimensional objects, which surfaces are not), and you should NEVER confuse the idea of the integral of a constant function with the integral itself; nor should you ever confuse the concept of a surface area integral with a surface integral.

Integrals are tools used to study how functions behave in certain regions; line and surface integrals are no exception, they just refer to integrals on specific types of regions. Line and surface integrals which give you arc lengths and surface areas are a more specific subset of line and surface integrals.

Now, going back to:

Why is a line integral equal to 0 on a closed path?

As Ayesha explained in the comments on your post (with reference to, I should note, concepts a bit beyond the scope of your question, e.g. compact domain), "that's the case only if the function in question is conservative." To expound, and to be a bit more precise, that's only if you're taking the line integral of a conservative vector-valued function (also referred to as a conservative vector field).

I can't think of any good, accurate way to recognize, intuitively, whether or not a vector field is conservative (at least by looking at a graphical representation of the vector field) besides saying that there are "rules" governing the distribution of the vectors - but this is a very, very vague and poor description that applies to both non-conservative and conservative vector fields. Perhaps another commenter could offer further insight into this.

The intuitive understanding of why a line integral of a conservative vector field is 0 on a closed path, however, is basically that "summing" up the vectors along a closed path gives you the zero vector; one good example is that of gravity - "what goes up must come down". If you throw an apple up in the air and then catch it exactly where you threw it up, although it's moved, it's returned to its original position (and has the same gravitational potential energy now as it did immediately before you threw it up).

Line Integral:

It is a curved domain! For instance, if you are integrating a function along x-axis, it is a straight line. Here you are integrating the function which is taking values along this straight line (x-axis). Now consider a semi-circle with it's centre at the origin. If you want to integrate the function along this semicircle, you basically have to sum the values the function is taking on the points along the semi-circle.

Integration is just a summation of values (roughly). Now next question is what values. Say $f(x,y) = 1+x^3+y^3$, now if you go along x-axis from -1 to 1, the function values will be different than going along a semicircle of unit radii with origin at center of axis system. Say a mild storm is taking place in your neighbourhood, if you go against the storm you have to work hard, if you go along the storm - you hardly have to do any work. Now considering the work as a function, you can see it depends upon the path you are taking. Same thing here, because my function takes on values all over the $x-y$ plane (let's assume that), you may be interested to integrate along a particular path which may be curved or whatever the case may be.

Line integral along closed curve is not always zero. It will be zero when the function is conservative, which means if an arbitrary vector function $f$ can be expressed as $\vec f = \nabla \phi=\frac {\partial \phi }{\partial x}\vec i + \frac {\partial \phi }{\partial y} \vec j$ considering in $x-y$ plane, where $\phi$ is function of $x,y$.

Now $\oint_c \vec f.\vec dr= \oint_c \nabla \vec \phi. \vec dr = \oint_c (\frac {\partial \phi }{\partial x}dx + \frac {\partial \phi }{\partial y}dy)= \oint_cd \phi$ - from here on you try to get the further steps and prove that it is indeed zero. (Hint: use total derivative theorem and assume $\phi$ is function of $x,y$)

Surface Integral:

This corresponds to 3D surfaces, and the logic flows in the same way as of line integral, here you want to integrate the function values takes at points on the surface. For vector fields it is the flux that you would be integrating.

You can think of a line integral as representing the work done by a force, if you've encountered that concept in earlier studies of calculus. In this case, a particle is moving along the curve $C$ (the one you're integrating over) pushed along by a force of magnitude $f(\textbf{x})$. You are basically integrating the dot product of the unit tangent vector of $C$ with a vector $f(\textbf{x})$, which can be taken to measure how much the path $C$ "lines up" with the vector field.

Surface integrals are basically a generalization of this concept to three dimensions - here, a surface takes the place of the linear path in a line integral. It represents the ${\it flux}$ through the surface.

• Thank you all for your answers! May 23 '14 at 1:16