Line Integral Example 2 (part 1)

Line integral over a closed path (part 1)

So let's say I have a function
of x and y; f of x and y is

equal to x plus y squared.

If I try to draw that,
let's see if I can have

a good attempt at it.

That is my y axis-- I'm going
to do a little perspective here

--this is my x axis-- I make do
the negative x and y axis,

could do it in that direction
--this is my x axis here.

And if I were to graph this
when y is 0, it's going to be

just a-- let me draw it in
yellow --is going to be just a

straight line that looks
something like that.

And then for any given
actually, we're going to

have a parabola in y.

y is going to look
something like that.

I'm just going to it in
the positive quadrant.

It's going to look
something like that.

It'll actually, when you go
into the negative y, you're

going to see the other half of
the parabola, but I'm not going

to worry about it too much.

So you're going to
have this surface.

it looks something like that.

Maybe I'll do to another
attempt at drawing it.

But this is our ceiling we're
going to deal with again.

And then I'm going to have
a path in the xy plane.

I'm going to start at the
point 2 comma 0. x is

equal to 2, y is 0.

And I'm going to travel, just
like we did in the last video,

I'm going to travel along a
circle, but this time the

circle's going to
have of radius 2.

Move counter clockwise
in that circle.

This is on the xy plane,
just to be able to

visualize it properly.

So this right here's
a point 0, 2.

And I'm going to come
back along the y axis.

This is my path; I'm going to
come back along the y access

and then so I look a left here,
and then I'm going to take

another left here in and
come back along the x axis.

I drew it in these
two shades of green.

That is my contour.

And what I want to do is I want
to evaluate the surface area

of essentially this little
building that has the roof of f

of xy is equal to x plus y
squared, and I want to find the

surface area of its walls.

So you'll have this wall right
here, whose base is the x axis.

Then you're going to have this
wall, which is along the curve;

it's going to look something
like kind of funky wall on

that curved side right there.

I'll try my best effort to
try to-- it's going to be

curving way up like that
and then along the y axis.

It's going to have like a half
a parabolic wall right there.

I'll do that back wall
along the y axis.

I'll do that in orange,
I'll use magenta.

That is the back wall
along the y axis.

Then you have this front
wall along the x axis.

And then you have this weird
curvy curtain or wall-- do that

maybe in blue --that goes along
this curve right here, this

part of a circle of radius 2.

So hopefully you get
that visualization.

It's a little harder; I'm
not using any graphic

program at this time.

But I want to figure out
the surface area, the

combined surface area
of these three walls.

And in very simple notation we
could say, well, the surface

area of those walls-- of this
wall plus that wall plus that

wall --is going to be equal to
the line integral along this

curve, or along this contour--
however you want to call it

--of f of xy,-- so that's x
plus y squared --ds, where ds

is just a little length
along our contour.

And since this is a closed
loop, we'll call this a

closed line interval.

And we'll sometimes see
this notation right here.

Often you'll see that
in physics books.

And we'll be dealing
with a lot more.

And we'll put a circle
on the interval sign.

And all that means is that the
contour we're dealing with is a

closed contour; we get back
to where we started from.

But how do we solve this thing?

A good place to start
is to just to find

the contour itself.

And just to simply it, we're
going to divide it into three

pieces and it essentially just
do three separate

line integrals.

Because you know, this isn't
a very continuous contour.

so the first part.

Let's do this first part of
the curve where we're going

along a circle of radius 2.

And that's pretty easy to
construct if we have x-- let me

do each part of the contour in
a different color, so if I do

orange this part of the contour
--if we say that x is equal 2

cosine of t and y is equal to 2
sine of t and if we say that

t-- and this is really just
building off what we saw on the

last video --if we say that t--
and that this is from t is a

greater than or equal to 0 and
is less than or equal to pi

over 2 --t is essentially going
to be the angle that

we're going along the
circle right here.

This will actually
describe this path.

And if you know, how I
constructed this is little

confusing, you might want
to review the video on

parametric equations.

So this is the first
part of our path.

So if we just wanted to find
the surface area of that wall

right there, we know we're
going to have to find

dx, dt and dy, dt.

So let's get that out
of the way right now.

So if we say dx, dt is going to
be equal to minus 2, sine of t,

dy, dy is going to be equal to
2 cosine of t; just the

derivatives of these.

We've seen that
many times before.

So it we want this orange
wall's surface area, we can

take the integral-- and if any
of this is confusing, there are

two videos before this where we
kind of derive this formula

--but we could take the
integral from t is equal to 0

to pi over 2 our function of x
plus y squared and

then times the ds.

So x plus y squared will
give the height of

each little block.

And then we want to get the
width of each little block,

which is ds, but we know that
we can rewrite the ds as the

square root-- give myself some
room right here --of dx of the

derivative of x with respect to
t squared-- so that is minus 2

sine of t squared --plus
the derivative of y with

respect to t squared, dt.

This will give us the orange
section, and then we can worry

about the other two walls.

And so how can we
simplify this?

Well, this is going to be equal
to the integral from 0 to pi

over 2 of x plus y squared.

And actually, let me write
everything in terms of t.

So x is equal to 2 cosine of t.

So let me write that down.

So it's 2 cosine of t plus y,
which is 2 sine of t, and we're

going to square everything.

And then all of that times
this crazy radical.

Right now it looks like a hard
antiderivative or integral to

solve, but I we'll find
out it's not too bad.

This is going to be equal to
4 sine squared of t plus

4 cosine squared of t.

We can factor a 4 out.

I don't want to forget the dt.

This over here-- let me just
simplify this expression so I

don't have to keep
rewriting it.

That is the same thing is
the square root of 4 times

sine squared of t plus
cosine squared of t.

We know what that
is: that's just 1.

So this whole thing just
simplifies to the square

root of 4, which is just 2.

So this whole thing simplifies
to 2, which is nice for

solving our antiderivative.

That means simplifying
things a lot.

So this whole thing simplifies
down to-- I'll do it over here.

I don't want to waste too much
space; I have two more walls to

figure out --the integral from
t is equal to 0 to pi over 2.

I want to make it very clear.

I just chose the simplest
parametrization I

could for x and y.

But I could have picked
other parametrizations, but

then I would have had to
change t accordingly.

So as long as you're consistent
with how you do it, it

should all work out.

There isn't just one
parametrization for this curve;

it's kind of depending on how
fast you want to go

along the curve.

Watch the parametric functions
videos if you want a little

bit more depth on that.

Anyway, this thing simplifies.

We have a 2 here; 2 times
cosine of t, that's

4 cosine of t.

And then here we have 2 sine
squared sine of t squared.

So that's 4 sine squared of t.

And then we have to multiply
times this 2 again, so

that gives us an 8.

8 time sine squared of t, dt.

And then you know, sine squared
of t; that looks like a

tough thing to find the
antiderivative for, but we can

remember that sine squared of,
really anything-- we could say

sine squared of u is equal
to 1/2 half times 1

minus cosine of 2u.

So we can reuse this identity.

I can try the t here; sine
squared of t is equal to 1/2

times 1 minus cosine of 2t.

Let me rewrite it that way
because that'll make the

integral a lot easier to solve.

So we get integral from 0 the
pi over 2-- and actually I

could break up, well I won't
break it up --of 4 cosine of

t plus 8 times this thing.

8 times this thing; this
is the same thing as

sine squared of t.

So 8 times this-- 8 times 1/2
is 4 --4 times 1 minus cosine

of 2t-- just use a little trig
identity there --and

all of that dt.

Now this should be reasonably
straight forward to get

the antiderivative of.

Let's just take it.

The antiderivative of this
is antiderivative of cosine

of t; that's a sine of t.

The derivative of
sine is cosine.

So this is going to be 4 sine
of t-- the scalars don't affect

anything --and then, well let
me just distribute this 4.

So this is 4 times 1 which
is 4 minus 4 cosine of 2t.

So the antiderivative of 4 is
4t-- plus 4t --and then the

antiderivative of minus
4 cosine of u00b5 t?

Let's see it's going
to be sine of 2t.

The derivative of sine of
2t is 2 cosine of 2t.

We're going to have to have a
minus sign there, and put a 2

there, and now it
should work out.

What's the derivative
of minus 2 sine of t?

Take the derivative of
the inside 2 times

minus 2 is minus 4.

And the derivative of sine
of 2t with respect to

2t is cosine of 2t.

So there we go; we've figured
out our antiderivative.

Now we evaluate it
from 0 the pi over 2.

And what do we get?

We get 4 sine-- let me write
this down, for I don't want to

skip too many --sine of pi over
2 plus 4 times pi over 2--

that's just 2 pi minus 2 sine
of 2 times pi over 2 sine of

pie, and then all of that minus
all this evaluated at 0.

That's actually pretty
straightforward because

sine of 0 is 0.

4 times 0 is 0, and sine of
2 times 0, that's also 0.

So everything with the
0's work out nicely.

And then what do we have here?

Sine of pi over 2-- in my head,
I think sine of 90 degrees;

same thing --that is 1.

And then sine of pi is
0, that's 180 degrees.

So this whole thing
cancels out.

So we're left with 4 plus 2 pi.

So just like that we were able
to figure out the area of this

first curvy wall here,
and frankly, that's

the hardest part.

Now let's figure out the
area of this curve.

And actually you're going to
find out that these other

curves as they go along the
axes are much, much, much

easier, but we're going to
have to find different

parametrizations for this.

So if we take this curve
right here, let's do a

parametrization for that.

Actually, you know what?

Let me continue this in the
next video because I realize

I've been running
a little longer.

I'll do the next two walls and
then we'll sum them all up.