Evaluating a Line Integral: Exploring the Questions

In summary, the problem is that the line integral is defined in two different ways and the problem format is different from the definition in the book. The first form is in terms of parametric equations and the second is in terms of vector functions. One can always be converted to the other. "ds" is normally the differential of arclength. In this case, with x= t2, y= t, ds= \sqrt{(dx/dt)^2+ (dy/dt)^2}dt= \sqrt{4t^2+ t^2}dt= t\sqrt{5}dt and y is, of course, t. The integral is \int_0^2
  • #1
Saladsamurai
3,020
7

Homework Statement


Evaluate the line integral, where C is the given curve:

[tex]\int_C\,y\,ds \, \,\,\,\,C:\,x=t^2\,\,\,\,y=t[/tex]

for t between 0 and 2 (out of curiousity, how do you make the "greater than or equal 2" in LateX)

Okay, in the book the line integral is defined as [tex]\int_C F\cdot dr=\int_a^bF(r(t))\cdot r'(t)\,dt[/tex]Okay now what is F in this case? What is "ds"? what happened? Why is the problem format so different from the definition?

Can someone help me out conceptually on this one?Thanks!
Casey
 
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  • #2
Hi Casey,

[itex]\geq[/itex] is just \geq :smile:

In this format, [itex]s[/itex] is the arc length of the path that is described by your parametric equations for [itex]x(t)[/itex] and [itex]y(t)[/itex] and so [itex]ds[/itex] represents an infinitesimal change in arc length...look at your notes; how is arc length defined?...how do you express it as a function of x and y?
 
  • #3
Saladsamurai said:

Homework Statement


Evaluate the line integral, where C is the given curve:

[tex]\int_C\,y\,ds \, \,\,\,\,C:\,x=t^2\,\,\,\,y=t[/tex]

for t between 0 and 2(out of curiousity, how do you make the "greater than or equal 2" in LateX)
\ge

Okay, in the book the line integral is defined as [tex]\int_C F\cdot dr=\int_a^bF(r(t))\cdot r'(t)\,dt[/tex]

Okay now what is F in this case? What is "ds"? what happened? Why is the problem format so different from the definition?

Can someone help me out conceptually on this one?


Thanks!
Casey
The first form is in terms of parametric equations. The second is in terms of vector functions. One can always be converted to the other.
"ds" is normally the differential of arclength. In this case, with x= t2, y= t, [itex]ds= \sqrt{(dx/dt)^2+ (dy/dt)^2}dt= \sqrt{4t^2+ t^2}dt= t\sqrt{5}dt[/itex] and y is, of course, t. The integral is
[tex]\int_0^2 t (t\sqrt{5}dt)= \sqrt{5}\int_0^2 t^2 dt[/tex]
 
  • #4
What I am really confused about is which represents my Field and which is my r ?
 
  • #5
HallsofIvy said:
\ge


The first form is in terms of parametric equations. The second is in terms of vector functions. One can always be converted to the other.
"ds" is normally the differential of arclength. In this case, with x= t2, y= t, [itex]ds= \sqrt{(dx/dt)^2+ (dy/dt)^2}dt= \sqrt{4t^2+ t^2}dt= t\sqrt{5}dt[/itex] and y is, of course, t. The integral is
[tex]\int_0^2 t (t\sqrt{5}dt)= \sqrt{5}\int_0^2 t^2 dt[/tex]

hmmmm... isn't [tex]\frac{dy}{dt}=1[/tex]?
 
  • #6
Saladsamurai said:
What I am really confused about is which represents my Field and which is my r ?

Remember that the speed along a curve, is equal to the rate of change of the arc lentgh :

[tex]|\vec{r'}(t)|=|\frac{d \vec{r}(t)}{dt}|=\frac{ds}{dt}[/tex]

And that the velocity along the curve is just speed*tangent vector

[tex]\Rightarrow \vec{r'}(t)dt=\hat{T} \left(\frac{ds}{dt} \right) dt= \hat{T}ds[/tex]

So when you compare that to the form of the path integral in your book, you see that the field in this case could be:

[tex]\vec{F}(\vec{r})=y\hat{T}[/tex]

...Anyways, this is unimportant for this problem: the integral has already been parameterized for you, so all you need to do is express [itex]y[/itex] and [itex]ds[/itex] in terms of [itex]t[/itex] and integrate.

Do you follow?
 

1. What is a line integral?

A line integral is a mathematical concept used to measure the total value of a function along a specific path or curve. It involves breaking down the path into small segments and calculating the value of the function at each segment, then adding them all together to get the total value.

2. Why do we need to evaluate line integrals?

Evaluating line integrals is important in various fields of science and engineering, such as physics, electromagnetics, and fluid mechanics. It allows us to calculate important physical quantities such as work, flux, and circulation, which help us understand and analyze real-world systems.

3. What is the difference between a line integral and a regular integral?

The main difference between a line integral and a regular integral is that a line integral is calculated along a specific path or curve, while a regular integral is calculated over a specific interval on the x-axis. Line integrals also involve vector functions, while regular integrals only involve scalar functions.

4. What are some applications of line integrals in science?

Line integrals have various applications in science, including calculating work done by a force along a curved path, determining the flux of a vector field through a surface, and finding the circulation of a fluid flow. They are also used in vector calculus to solve problems in electromagnetics, fluid mechanics, and other fields.

5. How do you evaluate a line integral?

To evaluate a line integral, you first need to determine the parametric equations of the path or curve. Then, you can use the fundamental theorem of line integrals, Green's theorem, or Stokes' theorem, depending on the type of line integral and the function being integrated. You may also need to use techniques such as substitution or integration by parts to simplify the integral before evaluating it.

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