Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Green's Theorem with Parameterized Curves

Date: 10/18/2003 at 23:26:14
From: Nicole
Subject: Green's Theorem

My calculus professor worked on a problem in class and I don't
understand it very well. Can you explain how to do this type of problem?

Find the area enclosed by the ellipse parameterized by the curve 

  x = 3cos(x), y = sin(x), for o <= x <= 2pi

by applying Green's theorem to the vector field 

  F = -(1/2)yi + (1/2)xj

I understand what Green's theorem states and how to use it, but the 
parameterizations are confusing me.


Date: 10/19/2003 at 08:12:30
From: Doctor Anthony
Subject: Re: Green's Theorem

Hi Nicole, 

Green's theorem allows you to use double integrals over a region to 
replace a line integral round a closed curve or vice-versa.   

Using Green's theorem, we can write

 INT[P dx + Q dy] = INT INT[(part(dQ/dx) - part(dP/dy))dx dy]
    C                      A

  F = Pi + Qj

    = (-y/2)i + (x/2)j

where  

            P = -y/2                Q = x/2
                                           
  part[dP/dy] = -1/2      part[dQ/dx] = 1/2


So  we get 

  INT[P dx + Q dy] =  INT INT[(1/2 + 1/2)dx dy]
     C                      A

                   =  area of the ellipse

Therefore our problem reduces to integrating round the curve as given
by the line integral on the left.

We parameterize  

    x =  3cos@        y = sin@    

   dx = -3sin@ d@    dy = cos@ d@

    P = -sin@/2       Q = 3cos@/2

and so

 INT[P dx + Q dy] 

              2pi
          = INT[(-sin@/2)(-3sin@ d@) + (3cos@/2)cos@ d@]
               0
                 
                   2pi
          = (3/2)INT[[sin^2(@) + cos^2(@)] d@]
                    0
  
                   2pi
          = (3/2)INT[d@]
                    0

          = (3/2)[@]  from 0 to 2pi

          =  (3/2)[2pi]

          =  3pi

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Calculus

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/