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Area of Region Vector Calculus

  1. Oct 8, 2016 #1
    • 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!!
     

    Attached Files:

  2. jcsd
  3. Oct 8, 2016 #2

    Ray Vickson

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    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##?
     
  4. Oct 9, 2016 #3
    n = (dy/ds, −dx/ds) I guess?
     
  5. Oct 9, 2016 #4

    Ray Vickson

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    Why not just ##\vec{n} = (\Delta y, -\Delta dx) / \sqrt{\Delta x^2 + \Delta y^2}##? (That is, if you want a unit normal!)
     
  6. Oct 9, 2016 #5
    I think ds is equivalent to sqrt(dx +dy)?
     
  7. Oct 10, 2016 #6

    Ray Vickson

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    No; it is ##\sqrt{dx^2+dy^2}##.
     
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