limits of integration in polar coordinates

Posted on 2003-11-15
Last Modified: 2008-02-01
How to find the limit of integration for an ellipse defined as :
It helps me if you use a detail here plus some use of mathcad or Advanced Grapher or some othersoftware

C: The ellipse 4x^2 + y^2 = 4 in xy-plane, counter clockwise when viewed from the above.
Question by:maimran
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Expert Comment

ID: 9757624
In polar coordinates 4x²  + y²  = 4    then using  dividing by 4 gives

         x²  + y²/4  = 1

ie      x²  + (y/2)²  = 1

    x=cos(theta) ,   y=2sin (theta)

satisfies the relationship and is the polar coordinate reprentation of the ellipse, and the ellipse is traced out by varing theta from 0 to 2Pi, I do not understand what you mean by the rest of the question.


Author Comment

ID: 9757889
OK, I put the whole question here now.

This is from CAlculus and Analytic Geometery by Thomas Fenny 9th edition( page 1122 )

Use the surface integral in Stoke's Theorem to calculate the circulation of the field F around the curve C in the indicated direction.

  F = x^2 i + 2x j + z^2 k
  C: The ellipse 4x^2 + y^2 = 4 in xy-plane, counter clockwise when viewed from the above.

In the solution manual it just use the result as  ( I quote from the solution )
the last line is
= 2(Area of the ellispe ) = 4 PI
it means he use the area as 2 PI but how, I want this.
LVL 31

Accepted Solution

GwynforWeb earned 100 total points
ID: 9759213
The ellipse is

x=cos(s) ,   y=2sin (s)     ,  0 <= s <= 2Pi

As a vector function this becomes

(x(s),y(s)) = (cos(s), 2sin (s)),  0 <= s <= 2Pi

now apply the theorem with the limits 0 to 2Pi and the result comes out quickly. I am sure you can do the rest.


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