Welcome back to a second example of surface
integrals where the surface is a parameterized surface.
So, if our surface is given as the vectored valued function
r of u and v, we can rewrite the double integral of f,
integrated over the surface, S, with respects to S,
as a double integral over the region R in the uv plane
by rewriting the function, f, as a function of u and v
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 would be dudv or dvdu.
And, again, we did discuss where this extra integrating
factor came from in the video on the area
of a parameterized surface.
Let's go and take a look at our second example.
We're going to evaluate the given double integral,
where f is equal to x squared plus y squared,
and our surface is a unit sphere above the xy plane.
So, because our surface is a sphere it'll be pretty easy
to parameterize this surface using spherical coordinates.
We'll just let phi equal u, and theta equal v.
So, x is going to be equal to sin u, cosine v.
Y is going to be equal to sin u, sin v.
And z is going to be equal to cosine u.
And, u is going to be on the closed interval from zero
to pi over two since we're above the xy plane.
And theta, or v, will be on the closed interval
from zero to two pi.
Now, let's start working on this integrating factor.
So, now we'll find the partial derivatives of r.
So, here we'll have cosine u, cosine v,
cosine u, sin v,
and here we'll have negative sin u.
And now for the partials with respects to v
we're going to have, this will be negative sin v,
so I'll have negative sin u, sin v.
Here we'll have sin u, cosine v,
and here we'll have zero.
Now, we'll determine the cross product
of these two vectored valued functions.
To do that we'll evaluate the three by three determinant.
We're going to go ahead and use the diagonal method,
and I've already set that up to save some time.
So, again, here is the three by three determinant,
but because I'm using the diagonal method I copied
column one and column two again.
Now, I'll sum the product of these blue diagonals,
one, two, three,
and then I'll subtract the product of
these three diagonals in green.
And for that sake of time I've already worked this out.
It comes out to sin squared u,
cosine v,
sin squared u, sin v,
and sin u, cosine u.
So, you may want to pause the video and work this out.
That's what it should come out to.
And, now, we need to determine the magnitude
of this cross product.
I did show this part in another video.
You end up having to factor the radicand twice.
You end up with the square root of sin squared u,
which is equal to sin u.
So, all of this gives us the magnitude of the cross product.
Let's go back to the initial problem just for a moment.
So, all of this is equal to sin u.
And, now, we have to rewrite our function,
f equals x squared plus y squared,
using x equals sin u, cosine v,
and y equals sin u, sin v.
So, let's go ahead and do that now.
We're going to have sin squared u,
cosine squared v,
there's our x squared,
plus y squared, is going to be sin squared u,
sin squared v.
Again, remember our function, f of xyz,
was equal to x squared plus y squared.
So, here's x squared, here's y squared.
This magnitude here is equal to sin u.
And then we'll use the order of the dudv,
where u is on the interval from zero to pi over two,
and v is on the interval from zero to two pi.
Now, you need to simplify this radicand.
Let's see what we can do here.
If you factor out sin squared u,
we'll have cosine squared v plus sin squared v,
that's going to be equal to one.
So, we have sin squared u, times sin u.
That's going to be sin to the third u, dudv.
Let's go ahead and perform a trig substitution here.
Let's go and replace two of the factors,
sin squared u with one minus cosine squared u.
Now, let's go ahead and distribute the sin u.
So, we'll have sin u minus sin u,
cosine squared u.
Now, if we let w equal cosine u,
dw is going to be equal to negative sin u, du.
Notice we have a negative sin u, du here.
So, this second part simplifies just w squared.
Integrand respects to u we're going to have negative cosine u,
remember, this is going to be w to the second,
so we'll have 1/3 w cubed,
or plus 1/3 cosine cubed u.
So, when u is pi over two these will both be zero,
and then we'll have minus.
And then when u is zero we're going to have negative one
plus, this will be 1/3, so we have negative 2/3.
Got the dv in here.
So, we have a positive 2/3.
Let's go ahead and finish this on the next page.
If we have 2/3 v,
so we have 2/3 times two pi, minus zero.
That's going to give us four pi over three.
So, this example was pretty involved.
I did skip a little bit of work due to time,
but I hope you found this example helpful.