The Math Forum

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

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 

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

(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   
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- The Math Forum at NCTM. All rights reserved.