# Line integral example 2 (part 1) | Multivariable Calculus | Khan Academy

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
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.