Welcome back to part two
of surface integral of a vector field.
In part one, we talked about if S is an oriented surface
given by Z equals G of X and Y, and R is in the XY-plane,
based upon the orientation of the surface
affects how we evaluate the surface integral
as a double integral over the region R.
Notice that the orientation is upward.
We take the opposite of the partial derivatives of the G
with respects to X and Y
and the Z component is positive one
and if the orientation is downward,
we take the partial of G with respects to X,
the partial G with respects to Y
and then the Z component would be negative one.
Let's go and take a look at our second example.
We want to determine the flux across the,
where the vector field is equal to
angle bracket X comma Y comma Z
across the surface above the XY-plane
with a unit normal vector oriented upward.
So the first thing we should recognize is that G of X and Y
is equal to nine minus X squared minus Y squared.
Again, before we set this up, let's go ahead
and take a look at this graphically.
Here's our surface in green and our vector field in red.
So the value of our surface integral
will represent the flow rate across this blue surface
and you can see from the graph,
it looks like it will be upward
in the same direction as our orientation
so this should give us a positive flow rate.
So our surface integral is going to be equal to
the double integral over the region R of this dot product.
So for F we're going to have X comma Y
and then for Z we're going to replace that with
nine minus X squared minus Y squared.
Everything needs to be in terms of X and Y
and we'll dot this with the opposite
of the partial derivative of G with respects to X,
well the partial derivative would be negative two X
and the opposite of that would be positive two X
and then for the partial derivative of G with respects to Y,
that would be negative two Y
and then we'll take the opposite of that,
that'll be two Y and the Z component will be one.
Differential A.
Let's go ahead and find this dot product
before we determine our variable's integration
and our limits of integration.
So we'll have two X squared plus two Y squared
and we'll have plus the quantity
nine minus X squared minus Y squared.
So we're going to have X squared plus Y squared
plus nine, and before we decide our limits of integration,
let's go and look at our XY-trace.
So if Z is equal to zero, we would have
X squared plus Y squared equals nine,
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so we have a circle centered at the origin
with a radius of three.
So I think what we'll do here for this one
is use polar coordinates.
Let's go and do the conversion on the next page.
Okay, so we're going to use polar coordinates
X squared plus Y squared is equal to R squared
so we'll have R squared plus nine
and then differential A is going to give us
an extra factor of RDRD theta,
so R is going to be from zero to three based upon our XY-trace
and theta will be from zero to two pi.
So we're going to have R cubed plus nine R.
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So we're going to have R to the fourth over four
plus nine halves R squared,
replacing R with three
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and then when R is zero we'll have zero,
we've got the D theta here.
To save a little bit of time I've already worked this out.
It comes out to 81 fourths plus 81 halves,
that's 243 fourths.
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So I have 243 over four times two pi minus zero,
that'll give us 243 pi over two
as our flow rate across that surface.
Remember, the surface had an upward orientation
and the flow rate was positive
because it was in the same direction
as our orientation.
So the flow rate is up from the surface
given this vector field.
Okay, I hope you found this example helpful.
Thank you for watching and have a good day.