Area of Region Vector Calculus

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: 499
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 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}##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top