Understanding Line Integrals: Scalar vs Vector Fields

In summary: But again, i don't know how to use it since i am not familiar with what is inside the partial derivatives.In summary, the line integral can be calculated using both a scalar function f and a vector field F. However, it is not true that equation1 (using f) is always equal to equation2 (using F). In fact, there are many examples where they are not equal. This is because the equation given in the book is not always valid, and further investigation is needed to understand the reasons behind this discrepancy.
  • #1
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Well i know that the line integral is
36b2dc0ff136923eb88a33c1f36d7ca8.png
given a scalar function f. equation1

But the line integral is also
cf0ea5b21fdf24a36e4b98844ccd673b.png
given a vector field F. equation2

So, given scalar function f and taking the gradient vector of it in order to turn it into a vector field F. Why is equation1 not equal to equation2?
 
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  • #2
It's not clear to me what you are asking. If [itex]F(r)= \nabla f[/itex] then it is certainly true that [itex]\int_C \nabla F\cdot dr= \int_a^b \nabla F(r)\cdot r'(t)dt[/itex]. But I don't know what you mean by "equation1 equal to equation2".
 
  • #3
Given [itex] \nabla f = F(r)[/itex]
[itex]\int_a^b f(r(t)) |r'(t)|dt \neq \int_a^b F(r(t)) \cdot r'(t)dt[/itex]
Why?
 
  • #4
cshum00 said:
Given [itex] \nabla f = F(r)[/itex]
[itex]\int_a^b f(r(t)) |r'(t)|dt \neq \int_a^b F(r(t)) \cdot r'(t)dt[/itex]
Why?
Why should [itex]\int_a^b f(r(t)) |r'(t)|dt \neq \int_a^b \nabla f(r(t)) \cdot r'(t)dt[/itex]?
 
  • #5
Well, let's try some example:

Let [itex]f(x,y) = xy[/itex]
[itex]F=\nabla f(x,y)=<y,x>[/itex]

Curve C => [itex]y=x, 0<x<1[/itex]
Parametrize [itex]r(t)=<t,t>, 0<t<1[/itex]
[itex]r'(t)=<1,1>[/itex]
[itex]|r'(t)|=\sqrt2[/itex]

[itex]f(r(t))=t^2[/itex]
[itex]F(r(t))=<t,t>[/itex]

If [itex]\int_a^b f(r(t))|r'(t)|dt = \int_a^b F(r(t)) \cdot r'(t)dt[/itex]
[itex]\int_0^1 t^2 \sqrt2dt = \int_0^1 <t,t> \cdot <1,1>dt[/itex]
[itex](\sqrt2/3) t^3 |_0^1 = \int_0^1 2t dt[/itex]
[itex]\sqrt2/3 = t^2 |_0^1[/itex]
[itex]\sqrt2/3 \neq 1[/itex]
 
  • #6
The only reason i have been asking this:
[itex]\int_a^b f(r(t)) |r'(t)|dt \neq \int_a^b \nabla f(r(t)) \cdot r'(t)dt[/itex]
is because the formula given on the book say that they should be equal. However, as i try many examples; i never get the same answer. Here is some general proof i have been trying but never got far.

Let [itex]f(x,y)=f(x,y)[/itex]
[itex]F = \nabla f(x,y)=<f_x, f_y>[/itex]

Let Parametrization of Curve C:
[itex]r(t) = <g(t), h(t)>[/itex]
[itex]r'(t) = <g'(t), h'(t)>[/itex]
[itex]|r'(t)| = [g'(t)^2 + h'^2(t)]^\frac{1}{2}[/itex]

[itex]\int_a^b f(r(t)) |r'(t)|dt = \int_a^b \nabla f(r(t)) \cdot r'(t)dt[/itex]
[itex]\int_a^b f(g(t), h(t))[g'(t)^2 + h'(t)^2]^\frac{1}{2}dt = \int_a^b <f(_x, f_y> \cdot <g'(t), h'(t)>dt[/itex]
[itex]\int_a^b f(g(t), h(t))[g'(t)^2 + h'(t)^2]^\frac{1}{2}dt = \int_a^b g'(t)f_x(g(t), h(t))+ h'(t)f_y(g(t), h(t))dt[/itex]Since limits and variable of integration are the same then
[itex]f(g(t), h(t))[g'(t)^2 + h'(t)^2]^\frac{1}{2} = g'(t)f_x(g(t), h(t)) + h'(t)f_y(g(t), h(t))[/itex]
But since i don't have any idea of what is going on inside those partial derivatives, i don't have a clue how to proceed.

Edit: In addition, the right hand side kind of looks like the product rule of a derivative.
 

What is a line integral?

A line integral is a mathematical concept used in calculus and vector calculus to calculate the total value of a scalar or vector function along a specified curve or path. It is essentially a generalization of the concept of definite integrals from single-variable calculus to multi-dimensional spaces.

What is the purpose of line integrals?

The purpose of line integrals is to calculate the total value of a scalar or vector field along a specified curve or path. This is useful in many applications, such as calculating work done by a force along a specific path, finding the mass of a wire or cable, or determining the flow of a fluid through a particular surface.

How do you calculate a line integral?

To calculate a line integral, you first need to determine the parametric equation of the curve or path in question. Then, you integrate the scalar or vector function along this path with respect to the parameter. This can be done using various techniques, such as the fundamental theorem of calculus, integration by substitution, or integration by parts.

What is the difference between a line integral and a surface integral?

The main difference between a line integral and a surface integral is that a line integral is calculated along a one-dimensional curve or path, while a surface integral is calculated over a two-dimensional surface. Line integrals are also typically used to calculate the total value of a scalar or vector function, while surface integrals are used to find the flux or flow of a vector field through a surface.

What are some real-life applications of line integrals?

Line integrals have many real-life applications, including calculating the work done by a force along a specific path, finding the center of mass of a wire or cable, determining the flow of a fluid through a particular surface, and calculating the voltage or current in an electrical circuit. They are also used in physics and engineering to analyze motion, energy, and other physical phenomena.

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