Area of Region Vector Calculus

Click For Summary

Homework Help Overview

The discussion revolves around the application of Green's theorem in vector calculus to determine the area of a region. The original poster expresses confusion regarding a specific relationship involving area and line integrals, and seeks guidance on how to approach the problem, particularly in relation to parameterization and the normal vector on a curve.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the use of Green's theorem and the implications of different formulations for area. Questions arise regarding the normal vector and its representation in terms of small increments on the curve. There is also a consideration of how the position of the origin affects the area calculation.

Discussion Status

Participants are actively exploring various interpretations of the problem, with some offering insights into the normal vector's formulation. There is a lack of consensus on the best approach, but several lines of reasoning are being examined, indicating a productive discussion.

Contextual Notes

There are indications of missing information regarding the understanding of part (a) of the question, which seems to be impacting the ability to tackle part (b). The discussion also reflects uncertainty about the definitions and relationships involved in the problem.

i_hate_math
Messages
150
Reaction score
2
Originally posted in a technical math section, so no template
I have tried to apply greens theorem with P(x,y)=-y and Q(x,y)=x, and gotten ∫ F • ds = 2*Area(D), where F(x,y)=(P,Q) ===> Area(D) = 1/2 ∫ F • ds = 1/2 ∫ (-y,x) • n ds . This is pretty much the most common approach to an area of region problem. But here they ask you to prove this bizarre relation of Area(D) = 1/2 ∫ F • ds = ∫ (x,y) • n ds. I am clueless what to do.

Without a good understanding of part (a) of the question, I don't know how to approach (b) at all. I know the parameterisation could be x=acost , y=bsint, 0≤b≤2π. It seems easy but I am in desperate need of some guidance.

Thanks heaps for helping!
 

Attachments

  • Untitled copy.png
    Untitled copy.png
    41.3 KB · Views: 521
Physics news on Phys.org
i_hate_math said:
I have tried to apply greens theorem with P(x,y)=-y and Q(x,y)=x, and gotten ∫ F • ds = 2*Area(D), where F(x,y)=(P,Q) ===> Area(D) = 1/2 ∫ F • ds = 1/2 ∫ (-y,x) • n ds . This is pretty much the most common approach to an area of region problem. But here they ask you to prove this bizarre relation of Area(D) = 1/2 ∫ F • ds = ∫ (x,y) • n ds. I am clueless what to do.

Without a good understanding of part (a) of the question, I don't know how to approach (b) at all. I know the parameterisation could be x=acost , y=bsint, 0≤b≤2π. It seems easy but I am in desperate need of some guidance.

Thanks heaps for helping!

For a very short line segment from ##(x_0,y_0)## to ##(x_0 + \Delta x, y_0 + \Delta y)## on the curve ##C##, what would be the normal ##\vec{n}## in terms of ##x_0, y_0, \Delta x, \Delta y##? If ##\Delta s ## is the distance from ##(x_0,y_0)## to ##(x_0 + \Delta x, y_0 + \Delta y)##, what would be the value of ##(x,y) \cdot \vec{n} \Delta s##? If the origin (0,0) is in the interior of the region ##D##, what would you get if you summed over all those small increments like those you just computed above?

Finally, what happens if (0,0) is exterior to ##D##?
 
  • Like
Likes   Reactions: i_hate_math
Ray Vickson said:
For a very short line segment from ##(x_0,y_0)## to ##(x_0 + \Delta x, y_0 + \Delta y)## on the curve ##C##, what would be the normal ##\vec{n}## in terms of ##x_0, y_0, \Delta x, \Delta y##? If ##\Delta s ## is the distance from ##(x_0,y_0)## to ##(x_0 + \Delta x, y_0 + \Delta y)##, what would be the value of ##(x,y) \cdot \vec{n} \Delta s##? If the origin (0,0) is in the interior of the region ##D##, what would you get if you summed over all those small increments like those you just computed above?

Finally, what happens if (0,0) is exterior to ##D##?
n = (dy/ds, −dx/ds) I guess?
 
i_hate_math said:
n = (dy/ds, −dx/ds) I guess?
Why not just ##\vec{n} = (\Delta y, -\Delta dx) / \sqrt{\Delta x^2 + \Delta y^2}##? (That is, if you want a unit normal!)
 
I think ds is equivalent to sqrt(dx +dy)?
 
i_hate_math said:
I think ds is equivalent to sqrt(dx +dy)?
No; it is ##\sqrt{dx^2+dy^2}##.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
12
Views
2K
Replies
1
Views
2K
Replies
5
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
3
Views
2K