Having trouble to understand real life applications of surface integrals

Question

I have been revising some maths and in particular I have been focusing on vector calculus. Today I have just finished looking at surface integrals. I do understand how we parametrize a surface integral starting from a region R, but I am having troubles actually understanding how these integrals are put in context when having a scalar function or a vector function.

For instance, a popular example is that you have a surface (for instance a metal sheet) and then you have a function which gives the density of each point of this metal sheet (if I were to plot this function, would it have the same shape of the surface?) Further, I dont understand why calculating the surface integral of the density function gives me the mass (isn't the definition of density mass/volume and hence I would need some kind of volume?)

Thank you for whoever can explain this in simple terms

Answer 1

Several good questions in your post. One question is about what the surface integral of a function computes. As TheHoneyCat says, it helps to consider the integrand as the density of some quantity.

Very often density (for instance, mass density) is measured per unit of volume, but a surface is only two-dimensional. So densities are given per unit of area. No real-world surface is literally infinitesimally thin, but substituting this limit makes the total quantity easier to calculate.

Another twist is that mass is normally positive, so using mass density as a metaphor breaks down when the integrand has negative values. But electric charge is a signed quantity, so we can use that instead. If the integrand represents the charge density (per unit of area) at a point on the surface, the integral is the total charge on the surface.

Because the definite integral is introduced as (and was invented for) a way to compute the area of a curved region, students often get stuck in a "integral=area" or "integral=volume" rut. But really, any product law can become an application of the definite integral. With surfaces, "charge = density times area" is the applicable product law that helps the most with intuition (in my opinion).

Your other question is in this paragraph:

For instance, a popular example is that you have a surface (for instance a metal sheet) and then you have a function which gives the density of each point of this metal sheet (if I were to plot this function, would it have the same shape of the surface?)

I think you're mixing up domains and codomains here. The density function's domain is the surface, and its codomain is real numbers. We don't usually plot or graph functions whose domain is a curved surface. I think there is a way to do that (maybe as a section of the normal bundle), but I don't think it's what you're thinking of.

It is often possible to represent a surface as the graph of some function of two variables (or the union of several such graphs). But the surface is now the image of that parametrization function.

When you compute a surface integral of a function $f$ defined on a surface $S$ parametrized by a function $g$ over a region $R$, you are basically integrating $f(g(x,y))$ over $R$. Points $(x,y)$ in $R$ can locate points on $S$, so we can evaluate $f$ at those points. But this composition alone doesn't take into account the curviness of $S$. The extra terms in the integral make this correction.

Answer 2

I think you will find it helpful to consider what it is you are integrating rather than over what you are integrating...

Let us consider your example... You have you surface $R$, okay. But you also have your density function. Now "density function" is a very generic name, here are different types of density functions:

• The function that indicates the volumetric mass of each point $x \in R$
• The function that indicates the mass per surface unit of each point $x \in R$
• The function that indicates the price per units of time, volume, and frequency of each point $x \in R$

Basically, you can cook up anything... (provided some rigorous condition I shall omit for now) and you only call it "density function" to emphasize that you plan on integrating it at some point...

Let us consider the first one ($f_1$). If $R$ is a surface then you are basically doing a double integration... (it would be triple if $R$ where a volume). Than, you are totally correct when you say that $\int_R f_1(\vec{x}) d^2 x$ doesn't give you the mass of $R$... Simply because $R$ has no mass, it has a mass per unit of length. Physically, you need to stack infinitely many of your sheets until the stack of sheet starts getting a measurable height, and that height times the mass per unit of length will be the mass of the total.

However, if you are using $f_2$ the second possibility, than you sheet does have a mass of its own (and no mass per unit of length). In the end it is (physicaly) a question of how you choose to model reality.

And mathematically, its just a matter of the dimension of what you integrate. If you allow time and frequency to be considered just like units of length (think like a mathematician) than we can see that $f_3$ needs to be integrated on a 5-dimentional object to yield a proper "mass" (well price actually, if I want to respect my initial joke-example)...

I hope this helps you with your original issue. In a word, I believe you already understood what was going on but simply didn't know that people have a tendency to call anything a "density" function.

As a note to future you: this has some deeper mathematical reasons (see measure theory for more) which you might start to guess by wandering how you would model the mass distribution of an object that has both 2D and 3D elements (or worst, a fractal). But there is no need to get into that for your question...

Hope I helped.