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# Surface integral ex3 part 2: Evaluating the outside surface | Multivariable Calculus | Khan Academy

Where we left off in the last video,
we were focused on surface two, this blue kind of outside,
or I guess the blue side of our little kind of chopped cylinder
that we were dealing with.
And we found a decent parametrization.
And then given that parametrization,
we were able to come up with dS for that surface,
for surface two.
It just simplified, all this business simplified to 1.
So it just equalled du dv.
And so now we are ready to evaluate the surface integral.
This surface integral right over here.
We're ready to evaluate the surface integral
over surface two of zdS.
And it's going to be equal to a double integral over u and v.
So let's write this.
I'm going to do two different colors
for the different variables of integration.
So yellow for one and maybe purple for the other.
And we're taking the integral of z.
And in our parametrization, z is equal to v.
So this right over here is the same thing as v.
So we can write v right over there.
And we already saw that dS is the same thing as du dv.
Or we could even write that is dv du.
We could just switch the order right over there.
And I'm going to choose to integrate with respect to dv
first, to do dv on the inside integral,
and then do du on the outside integral.
And the reason why I'm choosing to integrate with respect to v
first is based on the bounds of our parameters.
v is bounded on the low end by 0.
But on the high end, it's bounded
by essentially a function of u.
Its upper bound changes.
Because you see right over here, depending
where we are, depending on what our x value is, essentially
we have a different height that we need to get to.
And since it's a function of u, we
can integrate with respect to v. Our boundaries are going
to be 0 and 1 minus cosine of u.
This, all of this business in magenta,
will give us a function of u.
And then we'll be able to integrate with respect to u.
And u just goes from 0 to 2 pi.
So that will give us a nice straightforward number,
assuming all of this works out OK.
And so this is simplified to a straight-up double integral.
And now we're ready to compute.
And so let me write the outside part.
The outside part is from 0 to 2 pi, is du.
And so the inside part, the antiderivative of v
is v squared over 2.
And we're going to evaluate that from 0 to 1 minus cosine of u.
And so this is going to be equal to, once again,
the outside integral is 0 2 pi.
I'll write du right over here.
And so this is going to be equal to all of this.
Let me just write 1/2.
And, actually, I can even write the 1/2 out here.
I'll just write 1/2 times 1 minus cosine u squared.
Well that's just going to be 1 squared minus 2 times a product
of both of these, so minus 2 cosine of u.
Actually, let me give myself a little bit more real estate
here.
1 minus 2 cosine of u plus cosine of u
squared minus this thing evaluated at 0,
which is just going to be 0.
So we just get that right over there.
And then we have du.
And so now we can evaluate this.
We can integrate this with respect to u.
So let's do that.
So the answer-- and I'll just-- let
me just take the 1/2 on the outside
just to simplify things.
So we have the 1/2 out here.
And so if you take the antiderivative
of this with respect to u, you still have this 1/2 out front.
So this is equal to 1/2.
And we're going to take the antiderivative.
So let's do it carefully.
Actually let me just simplify it so it's easier
to take the antiderivative.
So it's going to be 1/2 times the integral.
I'll break this up into three different integrals.
1/2 times the integral from 0 to 2 pi
of 1 du, which is just du minus 2 times the integral from 0
to 2 pi of cosine of u du.
That's this term right over here.
Plus the integral from 0 to 2 pi of cosine squared u.
And cosine squared u, it's not so
easy to take the antiderivative of that.
So we'll use one of our trig identities.
I always forget the formal name.
I just think of it as the one that takes us
from cosine squared to cosine of 2u.
So this trig identity, this thing right over here
is the same thing.
This comes straight out of our trigonometry class.
This is 1/2 plus 1/2 cosine of 2 theta,
or I should say [? dot ?] theta, cosine of 2u.
So this last integral right over here,
I can rewrite it as, I'll just rewrite in that same color,
actually.
1/2 plus 1/2 cosine of 2u.
And then we have our final du.
And now let me close the brackets.
And all of that is times 1/2.
So this thing right over here, cosine of squared u,
just a trig identity, takes us to that.
Now this is pretty easy to evaluate.
The antiderivative of this is just
going to be u evaluated at 2 pi and 0.
So it, essentially-- let me just write it out.
This part right over here is just
u evaluated from 0 to 2 pi.
It's 2 pi minus 0.
It just gets evaluated and we get 2 pi.
So out front, you have your 1/2 and then times 2 pi.
And then this right over here, the antiderivative,
this is going to be equal to minus 2
times the antiderivative of cosine of u.
Well that's just sine of u evaluated from 0 to 2 pi.
Well sine of 2 pi is 0.
Sine of 0 is 0.
So this whole thing is going to evaluate to 0.
So we could say minus 0.
And then we take the antiderivative
of this right over here.
The antiderivative of this is going
to be-- the antiderivative of 1/2 is 1/2 u.
And the antiderivative of 1/2 cosine of 2u,
well if we had a 2 out front here, then
that would be the derivative of sine of 2u.
But we don't have a 2 out here.
But we can add a 2.
Let me actually write it this way.
We can add a 2 right over here, and then divide by a 2.
We can add a 2-- and that's a little bit too confusing.
Let me make it very clear.
1/2 cosine of 2u is equal to 1/2 times 1/2--
let me write it this way-- is equal to 1/4 times 2 cosine 2u.
These are the exact same quantities.
And the reason why I wrote it this way is this is clearly--
this is the derivative of sine of 2u.
So when you take the antiderivative of this,
it's the same thing as plus-- do it
in that same color-- plus 1/4 sine of 2u.
And we're going to evaluate that from 0 to 2 pi.
And you could confirm for yourself.
You take the derivative of this.
You do the chain rule.
You get the 2 out front.
2 times 1/4 gives you 1/2.
And the derivative of sine of 2u with respect to 2u
is cosine of 2u.
Now we need to evaluate this at 2 pi and at 0.
When you evaluate it at 2 pi, you get have 1/2 times 2 pi,
which is pi.
So this is plus pi plus 1/4 times sine of 2 times 2 pi,
sine of 4 pi.
That's just going to be 0.
So this is going to evaluate to 0.
And then minus 1/2 times 0.
And then 1/4 sine of 2 times 0.
This is all going to be 0 when you put 0 there.
So this whole thing of value is just 2 pi.
And we're in the home stretch at least for surface two.
And I'll switch back to surface two's color now,
now that we're near the end.
So this, the surface integral for surface two
is just going to be 1/2 times 2 pi minus 0 plus pi.
So it's 1/2 times 3 pi, which is equal to-- we
can have our drum roll, or I guess our mini drum roll
since we're not really done with the entire problem right now.
But it's equal to 3 pi over 2.
So we're making pretty good headway.
This was 0.
And now this part right over here is 3 pi over 2.
And in the next video we will try to tackle surface three.