Geometric Meaning of a Vector Integral

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SUMMARY

The discussion centers on the geometric interpretation of the vector integral \(\int_a^b \vec{r}(t) dt = \left\langle \int_a^b f(t) dt, \int_a^b g(t) dt, \int_a^b h(t) dt\right\rangle\) for the 3-vector function \(\vec{r}(t) = \langle f(t), g(t), h(t) \rangle\). Participants clarify that while one-dimensional integrals represent area, the vector integral traces a line in n-dimensional space by summing infinitesimal changes across each dimension independently. The integral serves as a shorthand for multiple integrals, but lacks a specific geometric meaning when visualized in three dimensions.

PREREQUISITES
  • Understanding of vector functions and notation
  • Familiarity with integral calculus, particularly in multiple dimensions
  • Knowledge of geometric interpretations of integrals
  • Basic principles of complex analysis
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Students of mathematics, physicists, and anyone interested in understanding the geometric implications of vector integrals and their applications in higher-dimensional calculus.

Lolcat Calc
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Hello everyone on these forums. :)

If you would, please consider the 3-vector function r(t) = <f(t),g(t),h(t)>. What sort of geometric meaning can be assigned to the following integral?

\int_a^b \vec{r}(t) dt = \left\langle \int_a^b f(t) dt, \int_a^b g(t) dt, \int_a^b h(t) dt\right\rangle

Or can any meaning can be assigned at all? Please help, I really want to know. We're covering this right now. :)
 
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Nothing? :( Did I phrase the question badly? I really do want to know this.
 
I hope a physicist can reply to this to give some geometric interpretation. But the truth is that I also don't know how to interpet it.

I've seen this used before. But I always thought of it as some kind of shorthand. Instead of writing the three integral, you just write one vectorial integral. This is easier to write.

Note, that this is closely related to the integral of complex numbers:

If f,g:[a,b]\rightarrow \mathbb{R}, then

\int_a^b{(f(t)+ig(t))dt}=\int_a^bf(t)dt+i\int_a^bg(t)dt

This is a very handy definition. Why it is handy can be seen by studying complex analysis where these kind of integrals are essential. I do not know of any intuitive explanation however.

This was a long post. I hope I didn't exceed the character limit... :biggrin:
 
There is no specific geometric meaning.
 
There is geometric meaning of \int_a^by(x)dx, because x-axis is shown on the graph.

The graph of \int_a^b\vec{r(t)}dt only shows x y z without t, so no geometric meaning is shown on this integration.
 
Lolcat Calc said:
Hello everyone on these forums. :)

If you would, please consider the 3-vector function r(t) = <f(t),g(t),h(t)>. What sort of geometric meaning can be assigned to the following integral?

\int_a^b \vec{r}(t) dt = \left\langle \int_a^b f(t) dt, \int_a^b g(t) dt, \int_a^b h(t) dt\right\rangle

Or can any meaning can be assigned at all? Please help, I really want to know. We're covering this right now. :)

Hey Lolcat Calc and welcome to the forums.

In the one dimensional case, we interpret such integrals to be an area. Essentially we are summing up lots and lots of infinitesimal changes and finding the result.

If you extend this to n-dimensions as you have, what you are doing is considering the infinitesimal changes of a point in three dimensions, where the change of each dimension is independent.

The way to visualize this is as follows: consider an initial condition to be a vector in an n-dimensional space. Now consider that the integral takes your initial condition (a point in n-dimensional space) and calculates a lot of infinitesimal additions in each dimension.

What this will do geometrically is trace out a line in n-dimensions if you want to trace the infinitesimals, where the changes in the direction are based on the expression of the integral for each component. The end result is just a vector, but you could visualize the integral as a process of tracing infinitesimal changes through an n-dimensional space.

Lets say you look at a standard dy/dx = bla and integrate it. Let's consider adding a third dimension called z. If we wanted to have the same behaviour we would have the condition dz/dt = 0 with dx=dt and dy/dt = bla. Our initial conditions would be <0,y0,0>.
 
chiro said:
Hey Lolcat Calc and welcome to the forums.

In the one dimensional case, we interpret such integrals to be an area. Essentially we are summing up lots and lots of infinitesimal changes and finding the result.

If you extend this to n-dimensions as you have, what you are doing is considering the infinitesimal changes of a point in three dimensions, where the change of each dimension is independent.

The way to visualize this is as follows: consider an initial condition to be a vector in an n-dimensional space. Now consider that the integral takes your initial condition (a point in n-dimensional space) and calculates a lot of infinitesimal additions in each dimension.

What this will do geometrically is trace out a line in n-dimensions if you want to trace the infinitesimals, where the changes in the direction are based on the expression of the integral for each component. The end result is just a vector, but you could visualize the integral as a process of tracing infinitesimal changes through an n-dimensional space.

Lets say you look at a standard dy/dx = bla and integrate it. Let's consider adding a third dimension called z. If we wanted to have the same behaviour we would have the condition dz/dt = 0 with dx=dt and dy/dt = bla. Our initial conditions would be <0,y0,0>.

Thank you! This makes a lot of sense. Its very helpful :)
 

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