Welcome to a lesson on surface integrals,
where the surface of integration is a parameterized surface.
If we have a smooth surface given by R of U, V,
and F is a continuous function,
then the double integral of F,
integrated over the surface S with respects to S,
is equal to the double integral
in the U, V plane as we see here.
If we write the function F by replacing X
with X of U and V,
Y with Y of U and V,
and Z with Z of U and V,
and differential S is equal to
the magnitude of this cross product,
differential A,
where differential A is either DU, DV, or DV, DU.
And in the the video on area of a parameterized surface,
we discuss where this extra integrating factor came from
when converting from differential S
to differential A,
and it comes from the area of a very small tangent plane.
And before we take a look at an example,
I want to emphasize here
that this video only focuses on surface integrals
where the surface is given as a parameterized surface.
I have a couple other videos that show
how to evaluate a surface integral
when the surface is given as Z equals G of X, Y,
as we see here.
Notice that if this is the case,
differential S is replaced
with this integrating factor,
times differential A.
So let's take a look at our first example.
We want to evaluate this surface integral
using a parametric surface
where our function is X times Y,
and our surface is X squared plus Y squared
equals four,
and Z is on the closed interval from zero to eight,
only in the first octant.
So let's first parameterize our surface.
Because this is a cylinder,
we'll use cylindrical coordinates.
So we'll let X equal two, cosine U.
Y would be two, sine U,
and then Z would just be V,
and because we're in the first octant,
U is going to be on the closed interval
from zero to pi over two,
and V will be on the closed interval from zero to eight.
So let's work on determining the magnitude
of this cross product.
So let's go ahead and find the partial derivative of R
with respects to U.
That's going to be negative two, sine U,
two, cosine U and zero,
then the partial derivative R with respects to V,
that'll just be zero, zero, and one.
So to determine the cross product of these two,
we'll evaluate a three by three determinant
where the first row is I, J, K.
The second row will be negative two, sine U,
two, cosine U, and zero.
Third row will be zero, zero, and one.
Now I'm going to go ahead and evaluate this
using the diagonal method.
So I've already copied the first and second columns
on the end,
and now we'll sum the product
of these three diagonals,
and then we'll subtract the product
of these three diagonals.
That'll give us our cross product.
So we'd have two, cosine U, I,
plus zero, plus zero, minus zero,
minus zero, and minus a negative two, sine U, J.
So that'll be plus two, sine U, J.
So now let's go ahead and determine the magnitude
of this cross product.
So I'll have four, cosine squared U,
plus four, sine squared U.
Now, we factor out the four.
We're going to have cosine squared U plus sine squared U.
That's equal to one.
Well, the square root of four would be equal to two.
So let's go back and take a look at the problem again.
All that work gave us this magnitude equal to two,
and now I'm going to rewrite our function,
X times Y, as a function of U and V,
where X is equal to two, cosine U,
and Y is equal to two, sine U.
So we'll have two, cosine U,
two, sine U,
times two, differential A.
So again, X is equal to two, cosine U.
Y was equal to two, sine U.
Then we had the magnitude was equal to two,
and then for differential A,
let's go ahead and write this as DV, DU.
Remember, V was on the closed interval from zero to eight,
and U was on the closed interval from zero to pi over two
because we're in the first octant.
So remember, our function was equal to X times Y.
That's why we have two, cosine U
times two, sine U.
Okay, so let's go ahead an pull out the eight,
and start to integrate.
Integrate with respects to V is just going to give us
V times cosine U, sine U.
Replacing V with eight,
we'll have eight, cosine U, sine U,
minus zero,
so we'll have 64, cosine U, sine U.
Now here we'll perform our substitution.
Let's go ahead and let W equal sine U.
So DW is going to be equal to cosine U, DU,
which is good because all of this
will be DW,
and this'll be W to the first.
So that'll give us W to the second
divided by two,
or 64 times 1/2, W to the second,
where W is sine U.
And then the sine of pi over two is one.
One squared is one,
minus sine of zero is zero.
So it looks like we have a value of 32
for our surface integral
using a parameterized surface.
I think we'll stop here on this video,
and we'll take a look at the second example
in part two.
Thank you for watching.