# 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.

#### Example

Let **F** = (y**i** - x**j**) / 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.
curl**F** = 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?

#### Example

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 = 2x**i** - 2y**j** - **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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