If we're just dealing with two dimensions, and we want to find
the area under a curve, we have good tools in our toolkit
already to do it, and I'll just remind us of our tools.
so let's say, that's the x-axis, that's the y-axis, let
me draw some arbitrary function right here, and that's
my function f of x.
And let's say we want to find the area between x is equal to
a, so that's x equal to a, and x is equal to b.
We saw this many, many, many videos ago.
The way you can think about it, is you take super small widths
of x, or super small changes in x.
We could call them delta x's, but because they're so small,
we're going to call them a dx.
Super, infinitesimally small changes in x.
And then you multiply them times the value of f
of x at that point.
So you multiply it times the height at that point, which
is the value of f of x.
So you get f of x times each of these infinitesimally small
bases, that'll give you the area of this infinitesimally
narrow rectangle right there.
And since each of these guys are infinitely small, you're
going to have an infinite number of these rectangles
in order to fill the space.
You're going to have an infinite number
of these, right?
And so the tool we use was the definite integral.
The definite integral is a sum, is an infinite sum of these
infinitely small areas, or these infinitely
small rectangles.
And the notations that we use, they would go from a b.
And we've done many videos on how do you evaluate
these things.
I just want to remind you, conceptually,
what this is saying.
This is conceptually saying, let's take a small change in x,
multiply it times the height at that point, and you're going to
have an infinite number of these, because these x's are
super small, they're infinitely small, so you're going to have
an infinite number of those.
So take an infinite sum of all of those, from x is equal
to a to x is equal to b.
And that's just our standard definite integral.
Now what I want to do in this video is extend this, broaden
this a little bit, to solve, I guess it maybe could say a
harder or a broader class of problems.
Let's say that we are, let's go to three dimensions now.
And I'll just draw the x-y plane first.
Maybe I'll keep this, just to kind of make the analogy clear.
I'm going to kind of flatten this, so we
have some perspective.
So let's say that this right here is the y-axis, kind of
going behind the screen.
You can imagine if I just pushed on this
and knocked it down.
So that's the y-axis, and that is my x-axis right there.
And let's say I some path in the x-y plane.
And in order to really define a path in the x-y plane, I'll
have to parameterize both the x and y variables.
So let's say that x is equal to, let me switch colors.
I'm using that orange too much.
Let's say that x is is equal to some function of some parameter
t, and let's say y is equal to some other function of that
same parameter t, and let's say we're going to start, we're
going to have t go from, t is going to be greater than or
equal to a, and then less than or equal to b.
Now this will define a path in the x-y plane, and if this
seems confusing, you might want to review the videos on
parametric equations.
But essentially, when t is equal to a, you're going to
have x is equal to, so t is equal to a, you're going to
have x is equal to g of a, and you're going to have
y is equal to h of a.
So you're going to have this point right here, so maybe it
might be, I don't know, I'll just draw a random point here.
When t is equal to a, you're going to plot the
coordinate point g of a.
That's going to be our x-coordinate.
This is g of a, right here.
And then our y-coordinate is going to be h of a.
Right?
You just put t is equal to a in each of these equations,
and then you get a value for x and y.
So this coordinate right here would be h of a.
And then, you would keep incrementing t larger and
larger, until you get to b, but you're going to get a series of
points that are going to look something like that.
That right there is a curve, or it's a path, in the x-y plane.
And you know, you're saying, how does that relate
to that right now?
What are we doing?
Well, let me just write a c here, for saying, that's our
curve, our that's our path.
Now, let's say I have another function that associates every
point in the x-y plane with some value.
So let's say I have some function, f of x y.
What it does is associate every point on the x-y
plane with some value.
So let me plot f of x y.
Let me make a vertical axis here.
We could do a different color.
Call it the f of x y axis, maybe we could even call it
the z-axis, if you want to.
But some vertical axis right there.
And for every point, so if you give me an x and a y, and put
into my f of x y function, it's going to give you some point.
So I can just draw some type of a surface that
f of x y represents.
And this'll all become a lot more concrete when I do
some concrete examples.
So let's say that f of x y looks something like this.
I'm going to try my best to draw it.
I'll do a different color.
Let's say f of x y.
Some surface.
I'll draw part of it.
It's some surface that looks, let's say it looks
something like that.
That is f of x y.
And remember, all this is, is you give me an x, you give me a
y, you pop it into f of x y, it's going to give me some
third value that we're going to plot in this vertical
axis right here.
I mean, example, f of x y?
It could be, I'm not saying this is a particular case,
it could be x plus y.
It could be f of x y.
These are just examples.
It could be x times y.
If x is 1, y is 2, f of x y will be 1 times 2.
But let's say when you plot, for every point on the x-y
plane, when you plot f of x y you get this surface up
here, and we want to do something interesting.
We want to figure out, not the area under this curve, this
was very simple when we did it the first time.
I want to find the area if you imagine a curtain, or a fence,
that goes along this curve.
You can imagine this being a very straight linear path,
going just along the x-axis from a to b.
Now we have this kind of crazy, curvy path that's going
along the x-y plane.
And you can imagine if you drew a wall, or curtain, or a fence
that went straight up from this to my f of x y, let me do my
best effort to draw that.
Let me draw it.
So it's going to go up to there, and maybe this point
corresponds to there.
And when you draw that curtain up, it's going to intersect
it something like that.
Let's say it looks something like that.
So this point right here corresponds to that
point right there.
So if you imagine, you have a curtain, f of x y is the roof,
and this is a, what I've drawn here, this curve, this kind of
shows you the bottom of a wall.
This is some kind of crazy wall.
And let me say, this point it corresponds to, well, actually,
let me draw it little bit different.
This point will correspond to some point up here, so when you
trace where it intersects, it will look something maybe
like that, I don't know.
Something like that.
And I'm trying my best to help you visualize this.
So maybe I'll shade this in to make it a little solid,
let's say f of x y is little transparent.
You can see.
But you have this curvy-looking wall here.
And the whole point of this video is, how can we figure out
the area of this curvy-looking wall, that's essentially the
wall or the fence that happens if you go from this curve
and jump up, and hit the ceiling at this f of x y?
So let's think a little bit about how we can do it.
Well, if we just use the analogy of what we did
previously, we could say, well look.
Let's make a little change in distance of our curve.
Let's call that ds.
That's a little change in distance of my
curve, right there.
And if I multiply that change in distance of the curve times
f of x y at that point, I'm going to get the area of that
little rectangle right there.
Right?
So if I take the ds, my change in my, you can imagine the arc
length of this curve at that point, so let me write, you
know, ds is equal to super small change in arc length of
our path, or of our curve.
That's our ds.
So you can imagine, the area of that little rectangle right
there, along my curvy wall, is going to be ds, I'll make it a
capital S, ds times the height at that point.
Well, that's f of x y.
And then if I take the sum, because these are infinitely
narrow, these ds's have infinitely small width, if I
were take the infinite sum of all of those guys, from t is
equal to a to t is equal to b, right, from t is equal to a, I
keep taking the sum of those rectangles, to t is equal
to b, right there, that will give me my area.
I'm just using the exact same logic as I did up there.
I'm not being very mathematically rigorous, but I
want to give you the intuition of what we're doing.
We're really just bending the base of this thing to get a
curvy wall instead of a straight, direct wall
like we had up here.
But you're saying, Sal, this is all abstract, and how can I
even calculate something like this, this makes no sense to
me, I have an s here, I have an x and a y, I have a t,
what can I do with this?
And let's see if we can make some headway.
And I promise you, when we do it with a tangible problem,
the end product of this video is going to be a little
bit hairy to look at.
But when we do it with an actual problem, it'll actually,
I think, be very concrete, and you'll see it's not too
hard to deal with.
But let's see if we can get all of this in terms of t.
So first of all, let's focus just on this ds.
So let me re-pick up the x-y axis.
So if I were to reflip the x-y, let me switch colors,
this is just getting a little monotonous.
So if I were to reflip the x-y axis like that, actually, let
me do that with that same green, so you know we're
dealing with the same x-y axis.
So that's my y-axis, that is my x-axis.
And so this path right here, if I were to just draw it straight
up like this, it would look something like this.
Right?
That's my path, my arc.
You know, this is when t is equal to a, so this is t is
equal to a, this is t is equal to b.
Same thing, I just kind of picked it back up so
you can visualize it.
And we say that we have some change in arc length, let's
say, let me switch colors.
Let's say that this one right here.
Let's say that's some small change in arc length, and
we're calling that ds.
Now, is there some way to relate ds to infinitely
small changes in x or y?
Well, if we think about it, if we really-- and this is all a
little bit hand-wavy, I'm not being mathematically rigorous,
but I think it'll give you the correct intuition-- if you
imagine this is, you can figure out the length of ds if you
know the length of these super small changes in x and
super small changes in y.
So if this distance right here is ds, infinitesimally small
change in x, this distance right here is dy,
infinitesimally small change in y, right?
Then we could figure out ds from the Pythagorean Theorem.
You can say that ds is going to be, it's the hypotenuse of this
triang.e It's equal to the square root of dx squared
plus dy squared.
So that seems to make things a little bit, you know, we can
get rid of the ds all of a sudden.
So let's rewrite this little expression here, using this
sense of what ds, is really the square root of dx
squared plus dy squared.
And I'm not being very rigorous, and actually it's
very hard to be rigorous with differentials, but intuitively
I think it makes a lot of sense.
So we can say that this integral, the area of this
curvy curtain, is going to be the integral from t is equal to
a to t is equal to b of f of x y, instead of writing ds, we
can write this, times the square root of dx squared
plus dy squared.
Now we at least got rid of this big capital S, but we still
haven't solved the problem of, how do you solve something, you
know, an integral, a definite integral that looks like this?
We have it in terms of t here, but we only have it in
terms of x's and y's here.
So we need to get everything in terms of t.
Well, we know x and y are both functions of t, so we can
actually rewrite it like this.
We can rewrite it as from t is equal to a, to t is equal to b.
And f of x y, we can write it, f is a function of x, which is
a function of t, and f is also a function of y, which is
also a function of t.
So you give me a t, I'll be able to give you an x or y, and
once you give me an x or y, I can figure out what f is.
So we have that, and then we have this part right here.
I'll do it in orange.
Square root of dx squared plus dy squared.
But we still don't have things in terms of t.
We need a dt someplace here in order be able to
evaluate this integral.
And we'll see that in the next video, when I do
a concrete problem.
But I really want to give you a sense for the end product, the
formula we're going to get at the end product of this
video, where it comes from.
So one thing we can do, is if we allow ourselves to
algebraically manipulate differentials, what we can do
is let us multiply and divide by dt.
So one way to think about it, you could rewrite, so let me
just do this orange part right here.
Let's do a little side right here.
So if you take this orange part, and write it in pink, and
you have dx squared, and then you have plus dy squared, and
let's say you just multiply it times dt over dt, right?
That's a small change in t, divided by a small change in t.
That's 1, so of course you can multiply it by that.
If we're to bring in this part inside of the square root sign,
right, so let me rewrite this.
This is the same thing as 1 over dt times the square root
of dx squared plus dy squared, and then times that dt.
Right?
I just wanted to write it this way to show you I'm
just multiplying by 1.
And here, I'm just taking this dt, writing it there, and
leaving this over here.
And now if I wanted to bring this into the square root sign,
this is the same thing, this is equal to, and I'll do it very
slowly, just to make sure, I'll allow you to believe that I'm
not doing anything shady with the algebra.
This is the same thing as the square root of 1 over dt
squared, let me make the radical a little bit bigger,
times dx squared plus dy squared, and all of
that times dt, right?
I didn't do anything, you could just take the square root of
this and you'd get 1 over dt.
And if I just distribute this, this is equal to the square
root, and we have our dt at the end, of dx squared, or we could
even write, dx over dt squared, plus dy over dt squared.
Right? dx squared over dt squared is just dx over dt
squared, same thing with the y's.
And now all of a sudden, this starts to look
pretty interesting.
Let's substitute this expression with this one.
We said that these are equivalent.
And I'll switch colors, just for the sake of it.
So we have the integral.
From t is equal to a.
Let me get our drawing back, if I-- from t is equal to a to t
is equal to b of f of x of t times, or f of x of t and f of,
or and y of t, they're both functions of t, and now instead
of this expression, we can write the square root of, well,
what's dx, what's the change in x with respect to, whatever
this parameter is?
What is dx dt?
dx dt is the same thing as g prime of t.
Right? x is a function of t.
The function I wrote is g prime of t.
And then dy dt is same thing as h prime of t.
We could say that, you know, this function of t.
So I just wanted to make that clear.
We know these two functions, so we can just take their
derivatives with respect to t.
But I'm just going to leave it in that form.
So the square root, and we take the derivative of x with
respect to t squared, plus the derivative of y with respect to
t squared, and all of that times dt.
And this might look like some strange and convoluted formula,
but this is actually something that we know how to deal with.
We've now simplified this strange, you know, this
arc-length problem, or this line integral, right?
That's essentially what we're doing.
We're taking an integral over a curve, or over a line,
as opposed to just an interval on the x-axis.
We've taken the strange line integral, that's in terms of
the arc length of the line, and x's and y's, and we've put
everything in terms of t.
And I'm going to show you that in the next video, right?
Everything is going to be expressed in terms of t,
so this just turns into a simple, definite integral.
So hopefully that didn't confuse you too much.
I think you're going to see in the next video that this, right
here, is actually a very straightforward
thing to implement.
And just to remind you where it all came from, I think I
got the parentheses right.
This right here was just a change in our arc length.
That whole thing right there was just a
change in arc length.
And this is just the height of our function at that point.
And we're just summing it, doing an infinite sum of
infinitely small lengths.
So this was a change in our arc length times the height.
This is going to have an infinitely narrow width, and
they're going to take an infinite number of these
rectangles to get the area of this entire fence, or
this entire curtain.
And that's what this definite integral will give us,
and we'll actually apply it in the next video.