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I am studying multivariable calculus, and I have not studied physics in depth yet in my high school career. I'm struggling to understand the real-world uses of line and surface integrals, especially, say, line integrals in a scalar field. I've searched the internet, read three different MV textbooks, cross-posted on Math Stack Exchange (where it was suggested I come to the physics site). But I've not found a good explanation, and math sites suggest that many students are befuddled by this.

One particular aspect I don't understand is what I call the alternative way of making a calculation. For example, at the simplest level, if I'm interested in area, I could take a single integral of $f(x)$ or I could take a double integral of $f(x,y)$ with an integrand of $1$. Both give area. More confusing to me (and I think I'm wrong here) if I want the surface area of a particular surface, I could take a surface integral with an integrand of $1$ times the square root of the sum of the squares of the partials. Is this "surface area" different than the "area" found by taking a double integral over the region with an integrand of $1$? I'm obviously a little confused about a couple points.

I'm trying to prepare a helpful typology for my own use. I'd be grateful if anyone could suggest improvements or additions (esp. about the applications of line integrals). I'm attaching a picture below.

enter image description here

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  • $\begingroup$ Crossposted from math.stackexchange.com/q/3955582/11127 $\endgroup$
    – Qmechanic
    Commented Dec 21, 2020 at 11:18
  • $\begingroup$ What do these integrals compute? $\endgroup$
    – ryang
    Commented May 24, 2022 at 14:13
  • $\begingroup$ - Higher Mathematics, Volume II, V.Smirnov or: - Differential and Integral Calculus, Volume II, N.Piskounov. fulviofrisone.com/attachments/article/476/… archive.org/details/… $\endgroup$
    – The Tiler
    Commented Dec 1, 2022 at 7:36
  • $\begingroup$ archive.org/details/… , page172 $\endgroup$
    – The Tiler
    Commented Dec 1, 2022 at 7:46
  • $\begingroup$ One thing about the line integral of a vector field is that it is actually the line interval of a scalar field: $\int_C \vec{F} \cdot {\rm d}\vec{r} := \int_{t_i}^{t_f} \left[\vec{F}\left(\vec{r}(t)\right) \cdot \frac{{\rm d} \vec{r}} {{\rm d} t} \right] {\rm d} t$, where $\vec{r}(t)$ is the parameterized path representing the curve $C$. So the line interval of a vector field is the scalar line integral of the dot product of the field vector with the tangent vector to the curve, integrated from the beginning to end of the curve. $\endgroup$
    – Ben H
    Commented Dec 1, 2022 at 16:40

2 Answers 2

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I don't have enough reps to comment, so maybe this thing as an answer may not be enough for you.

Anyway, the first thing I want to point out in your table is that the integrand does not need to be 1 in the area, or volume case. There are different values when you do integration in other coordinate systems like spherical, or cylindrical. In those cases the integrand should be the Jacobian of the transformation. example

One example I can give as a line integral in scalar field is the solution of Brachistochrone problem. It is the application of the principle of least action. In that case the integrand is inverse of velocity. The result you will get is the shortest time needed for an object to get from one point to another. example calculation

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Good on you for studying outside the book!

One particular aspect I don't understand is what I call the alternative way of making a calculation. For example, at the simplest level, if I'm interested in an area, I could take a single integral of f(x) or I could take a double integral of f(x,y) with an integrand of 1. Both give area. More confusing to me (and I think I'm wrong here) if I want the surface area of a particular surface, I could take a surface integral with an integrand of 1 times the square root of the sum of the squares of the partials. Is this "surface area" different than the "area" found by taking a double integral over the region with an integrand of 1? I'm obviously a little confused about a couple of points.

A double integral gives you the area enclosed in a planar region, you can't use it for finding the area of a curved surface like a sphere.

For such a task, we need a surface integral. The usual motivation of surface integral goes like setting up a relation between the curved surface's area and a planar area on the plane, from this we can calculate the area of the surface by a double integral.

Now, there are two kind of functions in multivariable /vector calculus: Scalar functions and vector functions. Depending on what you integrate, you have a few kinds of line and surface integrals. For example,

If you have a linear density function $\rho(x,y)$, then the integral of it over a curve in $R^2$ would give you the mass of the curve:

$$ \int \rho(x,y) ds$$

But say instead you had a force field defined over a curve, and a particle is moving on it then the integral of the force over displacement over the curve would give you the work in moving along that curve:

$$ \int \vec{F} \cdot \vec{ds}$$

The dot product is due to the fact that work is defined by the amount of force acting in the same direction as the force.

There is a lot more to say, but you'll be better off reading a textbook page by page. I suggest div, grad curl, and all that with some help from youtube. The book requires some basic double integrals which you can find on a site like a khan academy. The biggest thing is take your time in understanding all these topics, good luck!

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