Curl about an elipse. Line integral of vector field

Click For Summary

Homework Help Overview

The discussion revolves around calculating the area enclosed by a parameterized ellipse using line integrals and Green's Theorem. The original poster presents a problem involving the vector field F = xj and a parameterized curve defined by x = acos(t) and y = bsin(t) for 0 < t < 2π.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Green's Theorem and the relationship between the line integral and the area of the ellipse. There are questions regarding the integration process and the role of constants a and b in the calculations.

Discussion Status

Some participants have provided guidance on how to approach the line integral and its relation to the area. The original poster expresses confusion about integrating the ellipse and the lack of examples in their textbook. Others have noted the known area formula for an ellipse and suggested that the original poster's approach may need reconsideration.

Contextual Notes

There is mention of missing examples in the textbook regarding line integrals of parametric curves, which may be contributing to the confusion among participants. The original poster also notes that the integration process is not straightforward due to the absence of relevant examples.

carstensentyl
Messages
34
Reaction score
0

Homework Statement


It can be shown that the line integral of F = xj around a closed curve in the xy - plane, oriented as in Green's Theorem, measures the area of the region enclosed by the curve. (You should verify this.)

Use this result to calculate the area within the region of the parameterized curve given below.
x = acos(t) y= bsin(t) for 0<t<2pi

The Attempt at a Solution


I tried integrating an ellipse using cartesian limits, but ended up with zero under a radical. I can't think of a way to integrate this in terms of t, since our book has no such example. Using Green's theorem does not put the a and b constants anywhere in the equation, which confuses me...
 
Physics news on Phys.org
Well, you certainly have me confused. It is pretty well known that the area of an ellipse [itex]x^2/a^2+ y^2/b^2= 1[/itex] is [itex]\pi ab[/itex]. Since you haven't shown any work at all, I can't say where you went wrong.

And I can't believe that your text has no examples at all of integrating a vector function [itex]\vec{F}(t)[/itex] over a curve given by the parametric equations [itex]s(t)= x(t)\vec{i}+ y(t)\vec{j}[/itex]. The integral is simply [itex]\int \vec{F}(t)\cdot \vec{ds}(t)[/itex].

In this case, [itex]\vec{ds}= (-a sin(t)\vec{i}+ b cos(t)\vec{j})dt[/itex] and [itex]\vec{F}(t)= x\vec{j}= a cos(t)\vec{j}[/itex]. Take the dot product of those two functions and integrate from t= 0 to [itex]t= 2\pi[/itex].

As for Green's theorem, it says
[tex]{\int\int}{\Omega}\left[\frac{\partial Q(x,y)}{\partial x}- \frac{\partial P(x,y)}{\partial y}\right] dx dy= \int_C P(x,y)dx+ Q(x,y)dy[/itex]<br /> In this case, the function you are integrating around the circumference of the ellipse is just [itex]x\vec{j}[/itex] so P= 0 and Q= x. That means that <br /> [tex]\frac{\partial Q}{\partial x}= 1[/tex]<br /> so you are integrating 1 over the area of the ellipse Of course, that gives the area! a and b appear in the integral on the right when you put in the limits of integration.[/tex]
 
Thanks for the excellent reply. I did some integrating that was similar, but did not think that a line integral could be expressed in terms of t! There is no such example in our book, although I saw one such integral in a packet that our teacher handed us.

Unfortunately, our book does not even have an example of the line integral of a circle!
 
I worked it out and managed to get it. Such a simple answer, but so elusive.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
Replies
12
Views
2K
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K