Welcome to our lesson on surface integrals
of the vector field.
Let's first talk about oriented surfaces.
An oriented surface has two distinct sides
with a possible tangent plane at each point on the surface
except on the boundaries.
So here we see a surface in red with a blue tangent plane.
At the point of tendency, there are two possible
unit normal vectors, one pointing up, pictured here in blue,
and one pointing down, not pictured here.
If we use a unit normal vectors pointing up,
it gives the surface an upward orientation.
If we take a look at this plane here,
if we use an unit normal vector n sub two,
it would give this plane a downward orientation.
An example of a nonoriented surface would be
the Mobius Strip, where there's not two sides
to the surface.
Now let's talk about the surface integral of a vector field.
The surface integral is often called the flux integral,
and it measures the flow of F across the surface S.
And sequence with a double integral of F dotted with n
integrated with the specs to S, where S is the surface,
and n is a unit normal vector field to S.
So for surfaces given as a parametric surface,
we use this formula here for n.
And if the surface is given explicitly,
we use this formula here for n.
This video will focus on explicit surfaces.
So let's take a closer look at this formula here,
where big G of x, y and z equals z minus g of x and y
equals zero, n times differential S would give us
this quotient here times differential S,
where differential S has this Jacobian,
we're integrating factor times d A.
So we'll notice here is this simplifies nicely
this magnitude simplifies out,
so n times differential S is just equal to the gradient
of big G times differential A.
And one more thing to notice before we summarize this,
notice that the product of big G, with respects to Z,
is always d one, if the surface is given in this form.
So we often use the following shortcuts
to evaluate the surface integral of a vector field,
given the orientation of the surface.
If z is equal to g of x and y,
and R is in the x y plane,
we can use this double integral over the region R
to evaluate the surface integral if it's oriented upward,
and we can use this double integral over the region R,
if the surface is oriented downward.
And notice the only difference between these two
are the signs of the components as we see here.
Let's take a look at our first example.
Here we want to determine the flux across the given surface
where the vector field is equal to
zero, negative one, negative two,
and the surface is z equals six minus x minus y.
So the first thing we should recognize is that z of x and y
is going to be equal to six minus x minus y.
And before we set this up,
let's take a look at this graphically.
You could think of the vector field
as representing a rain storm.
And you could think of the surface
as the roof of a building.
So the surface integral would represent the flow rate
of the rain on the roof.
So let's go ahead and see what that would be.
Notice it says to use a downward orientation,
so we'll use this double integral over the region R
for our set up.
We'll have the double integral over the region R
and the x, y plane of F, which is
angle bracket zero, negative one, negative two,
dotted with angle bracket the product of g
with respects to x, that's going to be negative one.
The products of g with respects to y,
that'll be negative one.
And the z component will always be negative one,
if it's given in this form.
And then, for differential a, we're going to go ahead
and use d y d x.
And then in order to determine the limits of integration,
we're going to go ahead and sketch the x, y trace
of our surface, so if z was equal to aero,
we could move the x and y terms to the left,
and we'd have x plus y equals six.
So we have an x and a y intercept of six.
So the limits of integration for y,
they'll be from zero all the way to the line,
that will be from zero all the way to a function of x,
so we'll solve this for y to be six minus x.
And the limits of integration for x
will be from zero to six.
Let's go and determine this dot product.
We're going to have zero plus one plus two.
That will be three.
And from here, it will be pretty straightforward.
A three y,
so when y is equal to six minus x,
we'll have three times six and three time a negative x.
It will be 18 minus 3x.
And when y is zero, we have zero.
So we're going to have 18 x minus three halves x squared,
subbing in six for x, we'll have 18 times six
minus three halves times six squared, all minus zero.
This is going to give us 108 minus 54, which is 54.
And, again, we can view that value of 54
as the flow rate of the rain on the given roof.
I think we'll stop here for this video,
and take a look at a second example in part two.
I hope you found this helpful.
Thank you for watching.
I hope you found this helpful.