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### Course: Multivariable calculus > Unit 4

Lesson 1: Line integrals for scalar functions# Line integral example 2 (part 1)

Line integral over a closed path (part 1). Created by Sal Khan.

## Want to join the conversation?

- at the beginning of the vid. When y=0 why isn't the line along the x-axis? My intuition says it should be..(8 votes)
- It is along the x-axis, it's just going upwards in the z-direction making it look like it's in the x-y plane.(13 votes)

- I can't understand the initial line that Sal draws when y = 0 in the beginning of video, from which the parabolic surface is shown to arise. Should it not be the x axis itself. In that case we will not have the curtain between parabolic surface and x axis.(4 votes)
- When y = 0 then f(x,y) = x.

f(x,y) is theof the graph along the z axis.**height**

The first line is z=f(x,y)=x+0², or, z=x, which is a line that rises up above the xy plane at a 45 degree angle and is positioned directly over the x axis (since the x axis is where y=0). When x=0, z=0, when x=1, z=1, when x=2, z=2.

That means therea curtain along the x axis whose height, z is given by z=x.**is**(7 votes)

- I don't understand the draw, he draws a straight yellow line and after that a bunch of parabolas going up in the Z direction, a yellow circle and two lines for the travel of integration but after that¿ how is that he knows that the orange wall has that strange form?¿and how does he knows the form of the violet and yellow walls?(3 votes)
- Because he defined them. That is what Sal is doing in the beginning. He defines the boundries of the problem as 0<= x <=2, 0<+y<+2 and the arc from (2,0) to (0,2). That is the Line we want to integrate around ( in three pieces.)(2 votes)

- Correct me if I'm wrong, but if you're evaluating a closed loop line integral in a conservative vector field, then the answer will always be zero regardless of path?(3 votes)
- at12:16why does Sal evaluate one of the Sin(t)'s at Pi all of a sudden (instead of Pi/2)? I got for that problem:

2Pi + 2 and sal has 4 + 2Pi. Is this a mistake?(3 votes)- The second expression with a sine was sin(2t), which, evaluated at pi/2, is sin(2pi/2)=sin(pi)=0(1 vote)

- 5:08why it can change x = 2cost ? is it defined so that x(t) = 2cost ? where this relationship is defined ?(2 votes)
- This is parametrization.

Here t is defined as the angle from the x axis on the quarter circle path. Always x squared plus y squared will equal 4. As on the unit circle, x and y are proportional to cos(theta) and sin(theta, respectively.

As he says in the video, it may also help to review parametrization. A single variable t is allowed to vary, and x and y move with it at the same time, tracing out a path that represents the way they are dependent on one another.

Hope this helps! I'm happy to offer more clarification if needed.(2 votes)

- I can never tell in this whole series if the dS and other "d"s are supposed to be the ordinary differential lower case "d" or the partial derivative symbol ∂... which are they? Does it matter very much?(2 votes)
- It doesn't matter when integrating since you only integrate with respect to one variable at a time anyway. You just need to be careful about it when taking derivatives since the result will differ if you do a partial or total derivative.(2 votes)

- At about7:28, you mentioned that the squareroot of 4 is just 2. Isn't it important to remember that it could also be -2?(2 votes)
- He was looking for the distance , and distance is always positive.(2 votes)

- Upon dwelling on the idea of closed paths and their line integrals, I thought " If the region in question is finite and the path closed, there must be some way of finding the volume"; that is my question: how would one find the volume of a region bounded by f(x,y) with the perimeter described by the path c?(2 votes)
- This is what is Known as Green's Theorm, and it Discussed in the Next Section.

Basically if you have a Closed, Bounded Area bounded by C, the Volume could be Expressed By a Double Integral With D. Where D would be the Area The Curve Makes.

Integral_C Pdx + Qdy = DoubleIntegral_D dQ/dx - dP/dy dA(2 votes)

- At9:08, why does Sal take the antiderivative of the sin^(2)t? Isn't that only necessary when finding the final antiderivative?(2 votes)

## Video transcript

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.