Welcome to a video on the Flux Form
of Green's Theorem.
The goal of the video is to determine
the flux of a vector field in a plane.
So if C is a simple closed smooth curve
oriented counterclockwise that encloses region R in a plane,
where vector field F is equal to angle bracket f comma g,
both having continuous partial derivatives,
then the flux is equal to the line integral
of F dotted with n where n is a unit normal vector
integrated with respects to s or arc length,
which is equal to the double integral
over the region R of div F,
differential A, where div F is equal
to the partial derivative of f with respects to x
plus the partial derivative of g with respects to y.
Again, this is called the flux,
and it measures the flow across the curve C.
So if the flow is outward, the flux will be positive,
and if the flux is negative, the flow will be inward.
This is similar to Green's Theorem except flux integrates
the perpendicular or normal parts of the vector field
while Green's Theorem integrates the tangential part
of the vector field.
So you can think of Green's Theorem measuring
the rotation or twisting of the vector field across C
and the flux measures the flow inward or outward across C.
Let's go ahead and take a look at an example
and we'll also look at it graphically.
Here we want to measure the flux of a vector field
with components negative x, negative y,
across the circle x squared plus y squared equals 16.
And here we'll assume the orientation is counterclockwise
so that we satisfy the conditions needed
to apply the flux form of Green's Theorem.
Let's go ahead and sketch our curve.
So we have a circle with radius four
in a counterclockwise orientation.
Let's take a look at this surface in the vector field given.
So here's our circle with radius four
and here's our vector field, as you can see,
you can see the direction of this vector field
is pointing inward toward the circle
so we should anticipate a negative flux.
Let's go ahead and calculate it.
So we'll have the double integral
over the region where the integrand is the partial
derivative of f with respects to x.
That's going to be negative one,
plus partial derivative of g with respects to y,
that's also going to be negative one, dA.
And because our region is a circle,
let's go ahead and integrate this using polar coordinates.
So we'll have negative two, differential A
is equal to r dr d theta,
and then for the limits of integration,
r will be from zero to four,
and theta will be from zero to two pi.
So we'll have negative two times r squared over two,
that's negative r squared.
It's going to give us negative 16 minus 0,
which will be negative 16.
So subbing in two pi, we're going to get negative 32 pi,
and when theta is 0 we have zero,
so my flux is equal to negative 32 pi.
And again, because the flux is negative,
it does verify our flow should be inward
across this circle.
If it was positive, the flow would be outward.
That's going to do it for this video.
We will revisit this idea of flux
in three dimensions in a future video.
Thank you for watching.