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### Vector Calculus

```
Date: 02/26/98 at 23:57:40
From: Keith Britton
Subject: Vector Calculus

We're working on Green's Theorem right now in my Calculus class.

The vector field, F = xj, is given.  I'm asked to show that the line
integral of F around any closed curve in the xy-plane, oriented as in
Green's Theorem, is equal to the area of the region enclosed by the
curve.

First, I don't understand the wording of the problem.  Second, if I
pick the unit circle as the arbitrary curve, how do I translate it
into the proper form for the integral in Green's Theorem:

/
| (pdF2/pdx - pdF1/pdy) dxdy
/
R

(I'm using "pdF.." to indicate the partial derivative).

The problem, much to my enjoyment, goes on and requests I use the
results to calculate the area of the ellipse

x^2/a^2 + y^2/b^2 = 1

parametrized by

x = acost, y = bsint;  0 <= t <= 2pi

I think the "dxdy" indicates the area of the curve, but I thought the
only way to compute an area in 3-space was to use a double or triple
integral!  I am really confused on this problem and on the topic of
line integrals in general.

- Keith
```

```
Date: 02/27/98 at 12:30:26
From: Doctor Anthony
Subject: Re: Vector Calculus

Green's Theorem states:

INT.INT[partdP/dy - partdQ/dx]dx.dy  = -INT(path)(P.dx+Q.dy)

That is, a double integral over the plane region can be transformed
into a line integral over the boundary c of the region.

If F = P.i + Q.j  then we are given P = 0 and Q = x

and if we take the path round the circle x^2+y^2 = 1

= -INT[path)[x.dy]          where x = +-sqrt(1-y^2)

= -INT(path)[sqrt(1-y^2)dy]   put y = sin(theta)
dy = cos(theta).d(theta)

= -INT(path)[cos(theta).cos(theta)d(theta)]

= -INT(path)[cos^2(theta).d(theta)]

= -(1/2)INT(path)[1 + cos(2theta).d(theta)]

= -(1/2)[theta + (1/2)sin(2theta)]  (from pi to 0)

= (-1/2)[0 - pi]  = pi/2

If we carry out the same operation again, but take x =-sqrt(1-y^2) and
integrate theta from 0 to pi we get another pi/2.

So the total integral  = pi = the area of the given region.

If we use the double integral version, we have:

I = -INT.INT[partdP/dy - partdQ/dx)dx.dy]

P = 0 and  Q = x so partdQ/dx = 1

I = -INT.INT[0-1)dx.dy

=  INT.INT(region)[dx.dy]

=  area enclosed by c.

= pi

You should be able to apply the same technique now to the ellipse you
quote below, using the line integral round the boundary to show that
area enclosed is pi.ab

-Doctor Anthony,  The Math Forum
Check out our Web site  http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus

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