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### 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

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