Math 294

Lecture 7

Given by Prof. Alan Zehnder on 9/15/95 at 10:10 in Kaufmann Auditorium

More with Green's Theorem
Area and Surface Integrals

In Lecture 6 we showed examples of Green's Theorem and sketched a proof of the theorem. In this lecture, we will give more examples using Green's Theorem. Then we will introduce area and surface integrals.


Let F = (yi - xj) / r^2 where r^2 = x^2 + y^2
We want to calculate the circulation for a closed region C and F.
We choose a region that does not surround the origin, and we use Green's Theorem.

curlF = d/dx( -x / (x^2 + y^2) ) - d/dy( y / (x^2 + y^2) ) = 0

Thus, the circulation equals 0 over any path not surrounding the origin.
Remember, Green's Theorem is restricted to certain F and C.
In this example, it does not apply to any path around the origin, since F is not continuous at the origin.
Therefore, we can only say that the circulation is 0 for any region not containing the origin.
Take a circular path around the origin.
r = a cost i + a sint j, 0 <= t <= 2*pi
T = ( -a sint i + a cost j ) / ||v||

Green's Theorem predicted that the circulation was 0, but as we can see, this is not the case for paths surrounding the origin.

Area and Surface Integrals

Area Integrals

>From these diagrams, we see that dA = dcos
n (dot) p = cos
dA = d n (dot) p
We don't want a negative area so
dA = d ||n (dot) p||
d = dA / ||n (dot) p||
n = grad f / || grad f ||
if the surface is defined by f(x,y,z)=constant, Thus, d = dA || grad f || / || grad f (dot) p ||

So, what is the area of S?


Find the area of the surface defined by z = 1 + x^2 - y^2 and x^2 + y^2 <= b^2
We will solve the integral by projecting the surface onto the x-y plane, so p=k.
f(x,y,z) = x^2 - y^2 - z = -1
grad f = 2xi - 2yj - k
|| grad f || = (4x^2 + 4y^2 + 1)^1/2

Note that R is x^2 + y^2 <= b^2 by the definition of our surface.
Also, to solve the integral, we convert to polar coordinates, substituting x^2 + y^2 = r^2 and dA = r d dr

Surface Integrals

Now let's generalize the area integral to surface integrals - an integral of a function over a surface.

Note that g(x,y,z) in the integral is the value of g on S, i.e., on z = z(x,y)
In other words, you can substitute a function of x and y for g(x,y,z) because you know what z is on the surface. Here g(x,y,z)=g(x,y,z(x,y)), a function of x and y only.
An example will be given next lecture.

Lecture written by Aric Shafran on Sunday, September 24, 1995
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