# Surface integral ex3 part 3: Top surface | Multivariable Calculus | Khan Academy

- [Instructor] We're now in the home stretch.
We just need to evaluate this third surface integral,
which is over this top part of our little chopped cylinder
right over here.
So let's try to think of a parameterization,
and let me just copy and paste this entire drawing,
just so that I can use it down below
as we parameterize it.
So let me copy it, copy, and then go all the way down here
and let me paste it.
Let me paste it, okay, that is our shape again, our surface.
And then we go to the layer that I wanted to get on.
And then let me start, let's start evaluating it.
So what we wanna care about is the integral
over surface three of z d s.
And surface three here, we see that the x and y values,
and since you take on all the possible x and y values,
inside of the unit circle, including the boundary,
and then the z values are gonna be a function
of the x values.
We know that this plane,
that this top surface right over here, S3,
it is a subset of the plane z.
Z is equal to one minus x.
It's the subset that's kind of above the unit circle
in the xy-plane, or kind of the subset
that intersects with our cylinder and kind of chops it.
So let's think about x and y's first.
Let's think about it in terms of polar coordinates
because that's probably the easiest way to think about it.
I'm gonna redraw kind of a top view.
So I'm gonna redraw a top view, so that is my y-axis
and this is my x-axis.
X-axis.
And the x's and y's can take on any value,
they said you have to fill the unit circle.
So if you were to kind of project this top surface down
onto the xy-plane, you would get this orange surface,
that bottom surface, which looked like this.
It was essentially the unit circle just like that.
And I'm gonna redraw it a little bit neater than that.
I can do a better job, so that would be, all right.
So let me draw the unit circle as neatly as I can.
So there's my unit circle.
And so we can have one parameter.
We can have one parameter that essentially says
how far around the unit circle we're going.
Essentially that would be our angle
and let's use theta, 'cause that's, well just for fun.
We haven't used theta as a parameter yet.
That's theta, but if we had x's and y's
as just a function of theta and we had a fixed radius,
that would essentially just give us the points
on the outside of the unit circle.
But we need to be able to have all of the xy's
that're outside and inside the unit circle.
So we actually have to have two parameters.
We need to not only vary this angle,
but we also need to vary the radius.
So we would wanna trace out the outside of that unit circle
and maybe we'd wanna shorten it a little bit.
And then trace it out again.
And then shorten it some more.
And then trace it out again.
And so you wanna actually have a variable radius as well
so that you can have how far out you're going.
You could call that r.
So for example if r is fixed
and you change your ranges of theta,
then you would essentially get all of those points
right over there and you would do that for all of the r's
and from r zero all the way to r one
and you would essentially fill up the entire unit circle.
Let's do that, so r is going to go between zero and one.
R is going to be between zero and one.
And our theta, our theta's gonna go all the way around.
So our theta is going to go between zero and two pi.
This is, lemme write this down,
I wrote zero instead of theta.
Our theta is going to be greater than or equal to zero,
less than or equal to two pi.
And now we're ready to parameterize it.
X of r and theta is going to be equal to,
so whatever r is, it's going to be r cosine theta.
So x is going to be r cosine theta.
Y is going to be r sine theta.
That's going to be the y value.
R sine theta.
And now z is essentially just a function of x.
Z is going to be equal to one minus x,
but x is just r cosine theta.
So there you have it.
We have our parameterization
of this surface right over here.
The x's and y's can take all the values on the unit circle,
but then the z is up here based on a function of,
well based on really on a function of x.
It's one minus x.
So that would give us all of the possible points
right over here on the surface.
You pick an x and y and then the z is going to pop us
right here someplace on that surface.
And we can write it as a position vector function.
Instead of calling that position vector function r,
I will call it, I don't know, let's call it,
I'm just gonna pick a random letter here,
let's call it p for position vector function.
And so p, our surface p is, we can write it as,
actually no why don't I just call it surface three.
So surface three, surface three,
I'll do it in that same purple color too,
surface three as a position vector function,
as a function of theta and r, maybe I'll write r and theta
'cause that's how I think of things.
R and theta is going to be equal to r cosine theta i
plus r sine of theta j plus one minus r cosine theta,
need to get some real estate here,
one minus r cosine of theta k.
One minus r cosine of theta k.
And now we are ready to start doing all of the business
of evaluating the actual surface integral.
So the first thing we need to do is take the cross product
of this, the partial of this with respect to r,
and the partial of this with respect to theta.
Let's just get down to business.
Let's take the cross product, let's take the cross product
and so we have our i unit vector, we have our j unit vector,
and we have our k unit vector.
And this might get a little bit involved
but we'll try our best to just work through it.
Give myself a little bit more space.
And so the partial of this with respect to r,
so let's take the partial of this with respect to r,
I'll do it in blue, the partial of this with respect to r,
the partial of this with respect to r is just cosine theta i
so this is just cosine theta,
I said I was gonna do it in blue and that's not blue,
so this is going to be cosine, cosine theta.
The partial of this with respect to r is sine theta.
Sine theta.
And the partial of this term right over here
with respect to r is negative cosine theta.
Negative cosine theta.
Now let's take the partial with respect to theta.
The partial of this with respect
to theta is negative r sine of theta.
Negative r sine of theta.
The partial of this with respect to theta is r cosine theta.
R cosine theta.
And the partial here, this is zero,
and then this would be negative r sine theta.
Negative r, oh no let me be careful.
This is going to be, you have a negative r
so the derivative of cosine theta with respect to theta is
negative sine theta.
So the negatives are gonna cancel out
and so you're going to have r, r sine, r sine theta.
And now we can actually evaluate that this determinant
to figure out the cross product
of the partial of this with respect to r
and the partial of this with respect to theta.
I'm not writing it down,
just to kind of save some real estate here.
And so we have, actually maybe I will write it down
just to be clear what we're doing.
The partial of s three with respect to r crossed
with the partial of s three
with respect to theta is equal to,
now our i component is gonna be
sine theta times r sine theta,
so that's going to be r sine squared theta,
minus r cosine theta times negative cosine theta.
So that's plus r cosine squared theta.
All of that times i, and you already,
a simplification might be popping out here at you,
and then you have minus the j component.
The j component is gonna be cosine theta times r sine theta.
So it's r cosine theta sine theta.
And then we're gonna subtract from that.
See the negative signs cancel out.
So you're gonna subtract r sine theta cosine theta,
or r cosine theta sine theta.
Well this is interesting
because these are the negatives of each other.
R cosine theta sine theta minus r cosine theta sine theta.
This just evaluates to zero.
So we have no j component.
And then finally for our k component,
we have cosine theta times r cosine theta.
So we have r cosine squared theta
minus r sine theta times sine theta.
Or minus negative r sine theta times sine theta.
So this would give you a negative
but we're gonna have to subtract
until it gives you a positive.
So plus r sine squared theta k.
And so this simplifies quite nicely
because this is going to be equal to,
this term up here you can factor out an r.
This is r times the sine squared theta
plus cosine squared theta, which is just,
that part simplifies to one so that's just r times i.
So this is equal to r times i.
And we do the same thing over here.
This also simplifies.
This is actually the same thing.
This also simplifies to r.
So this whole thing simplifies, lemme write it this way,
this is also r sine squared theta plus cosine squared theta.
Also simplifies to r.
So you have r times k, and so if you want the magnitude
of this business, so let me make it clear,
so the magnitude, we'll go back to the magenta,
the magnitude of, I don't feel like rewriting it all,
just copy and paste it.
Edit copy and paste.
The magnitude of all of this business is going to be
equal to the square root of this squared,
which is r squared, plus this squared, which is r squared,
which simplifies to, this is two r squared.
So you take the square root of both of those,
you get the square root of two times r.
So this is equal to the square root of two times,
it would be the absolute value of r
but we know that r only takes on positive values.
R only takes on positive values
so it's the square root of two times r,
which is very nice because now we can evaluate d s.
D s is going to be this business times d r d theta.
So let's do it.
So our surface integral,
the thing that we were dealing with from the beginning,
that thing right over there.
Our surface integral, the surface s three, of z d s is now
equal to, so I'm gonna use different colors
for the different variables of integration,
so one on the outside, and then I'll do one on the inside,
I'll do the inside one in pink.
Z is equal to one minus r cosine theta.
So the z is equal to one minus r cosine theta.
It involves both so I'll use a different color.
Minus r cosine theta.
And then I just have to integrate
relative to both of the variables.
One minus r cosine theta, oh no I have to do the times d s.
D s is this thing, it's this thing,
times d theta d, d r.
So let's see, let's write this out.
Times square root of two r, so let's write that down,
so this times the square root of two r,
times the square root of two,
and we can write the square root of two out front
since it's a constant so lemme just do that,
simplifies things.
Square root of two times r, and now d theta d r.
Or we can write d r d theta, either way.
So let's do that.
D, let's do d r d theta.
So d r d theta, we could do it either way,
it's gonna be about the same level of complexity.
And so first we're going to integrate with respect to r,
and lemme do the colors the same way, actually.
So d r, d theta.
If I've got colors I might as well use them.
And actually I just realized that I'm way out of time.
So actually, let me continue this in the next video.
We'll set up the boundaries of integration
and then just evaluate it.