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## Surface integrals

# Surface integral ex3 part 2

## Video transcript

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.