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## Multivariable calculus

### Course: Multivariable calculus > Unit 5

Lesson 9: Proof of Stokes' theorem# Stokes' theorem proof part 7

Using Green's Theorem to complete the proof. Created by Sal Khan.

## Want to join the conversation?

- I got a confused at3:15when using the multivariable chain rule on Q(x, y, z) when z is a function of x and y. It looks like you have Partial Q/Partial x = Partial Q/Partial x + Partial Q/Partial z times Partial z/Partial x. What confuses me is whether or not Partial Q/Partial x on the left side of the equation is the same thing as the Partial Q/Partial x on the right side of the equation. Even though they are written the same (one uses the symbol for partial derivative and the other uses subscript notation), it somehow "feels" like they are not the same. And if they were, we would subtract them from each side of the equation and have Partial Q/Partial z times Partial z/Partial x equaling zero and that doesn't make sense either. Can someone straighten me out? Thanks a lot!(8 votes)
- Thanks for responding. That seems very reasonable; I couldn't imagine that those two partials could be the identical same thing. I wonder, is there some kind of formal or fancy mathematical notation to write the left and right partials in ∂Q/∂x = ∂Q/∂x + ∂Q/∂z * ∂z/∂x

that clearly represents their inherent differences and avoids this ambiguity? Thanks!!(6 votes)- I would think of it as

∂/∂x Q( x, y, z(x) ) = ∂/∂x Q( x, y, z ) + ∂/∂z(x) Q( x, y, z(x) ) * ∂/∂x z(x)

Note the changes between z and z(x)(10 votes)

- Sal keeps on saying that we are proving Stokes' theorem for the special case where there are continuous second derivatives of z( x, y ) along with z having to be a function of x and y. Does Stokes' theorem not apply at all when these conditions are not met or is there a more general way to prove Stokes' theorem where this doesn't have to be true?(9 votes)
- I think there is a more general way to prove this(1 vote)

- This seems to be a pretty straightforward argument (Proof of Stoke's Theorem): calculate both sides and show that they're equivalent; however, I was wondering if there was a good way to build intuition behind Stoke's Theorem? Sal's "intuition of Stoke's Theorem" video seems rather abstract, and not very applied, any recommendations for secondary source?(5 votes)
- Isn't there some video with the proof of the multivariable chain rule?(5 votes)
- Greens theorem is not a special case of Stokes Theorem? So why would you prove sth using the same theory you want to prove? pls tell me if im wrong :) .... but still the proof was nice, accurate and awesome(1 vote)
- You're right that Green's theorem is a special case of Stoke's theorem. However, we proved Green's theorem independently of Stoke's theorem, so it is perfectly valid to use it to prove Stoke's theorem. This is actually a very common technique in mathematics, using a theorem to prove its generalization.(6 votes)

- how can we practically imagine the stokes theorem in daily life?(2 votes)
- At3:36, shouldn't ∂Q/∂x be equal to ∂Q/∂z * ∂z/∂x ,
**not**∂Q/∂x + ∂Q/∂z * ∂z/∂x ?(1 vote)- Sal is correct. ∂/∂x (Q) = ∂Q/∂x + ∂Q/∂z * ∂z/∂x. This is because of the multivariable chain rule. Note that Q = Q(x, y, z) = Q(x, y, z(x, y)) where we do not only want this x: Q(x, y, z(
**x**, y)). For this x, ∂/∂x (Q) = ∂Q/∂z * ∂z/∂x. We also want to consider how Q changes with respect to x without respect to z. We need to consider this x: Q(**x**, y, z(x, y)). For this x, ∂/∂x (Q) = ∂Q/∂x. By the multivariable chain rule, we add these changes to get ∂/∂x (Q) = ∂Q/∂x + ∂Q/∂z * ∂z/∂x.

See https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/multivariable-chain-rule/v/multivariable-chain-rule (video) or https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/differentiating-vector-valued-functions/a/multivariable-chain-rule-simple-version (article).(3 votes)

- Is it possible to prove the theorem without setting r(for c and c1) as a parametric vector(just r=xi+yj+z(x,y)k,without t)?(1 vote)
- Can you please do a proof using the formal definition of curl? That way proof can also provide an explanation as to why Stokes’ Theorem actually makes sense.(1 vote)

## Video transcript

Where we left off
in the last video, we'd expressed our line
integral around the boundary of our surface. So this is f.dr where our
path is this boundary right over here-- this path c. We had expressed it
in terms of a line integral around this boundary--
around path c1-- which is the boundary of
region R. And the reason why this is going
to be valuable to us is now we can directly apply
Green's Theorem to this to turn this into a double
integral over this region right over here-- the region
that it is actually bounding. So let's actually do that. We're just applying
Green's Theorem here. So Green's Theorem. And actually, let me box
this off right over here. This was just
something that we wrote to remind ourselves
in the last video. So let me just box it
off right over here. But when you apply
Green's Theorem, you get that this is going
to be the exact same thing as the double integral
over the region that c1 bounds-- that
region, R, that's in our xy plane-- of the
partial of this with respect to x of this business. So I'll do that in
that same green color. Q plus R times the partial
of z with respect to y minus the partial
with respect to y of P plus R times the partial
of z with respect to x. And then, dA, little
differential of our region. dA. And so let's actually
calculate this right here. You take the partials of
each of these expressions. And we'll see that if we
expand them and then simplify, we'll get something very
similar-- or actually, hopefully, identical--
to this right over here. And that'll show that this line
integral for this special case is the same thing as this
surface integral, which will prove Stokes' theorem
for this special case. So let's do it. So let's just apply the
partial derivative operator. So first we want to take the
partial derivative with respect to Q. And we need to remind
ourselves-- and we did this way up here where we first thought
about it-- we saw that P, Q, and R, they're each
functions of x, y, and z. So we're assuming that's
how they're represented. And if z was not a
function of x, then, if we were to take the partial
of Q with respect to x, we would just write it as a
partial of Q with respect to x. But we know that we've
assumed that z is itself a function of x and y. So if we're taking the partial
with respect to x over here, we have to think
first-- Well, how can Q directly change
with respect to x? And then, how can it change
due to something else changing due to x? And so that other thing that
could change due to x is z. y is independent of x, but
z is a function of x. So let's keep that in mind. So we're really going to do
the multi-variable chain rule. So when we try to take the
derivative of this part, of this whole function
with respect to x, we have to think about--
How will Q change directly with respect to x? And to that, we have to add how
Q could change due to changes in other variables
due to changes in x. And the only other
variable that Q is a function of that could
change due to a change in x is z. So Q could also change to z
because z has changed due to x. So the operator here is
the partial with respect to x plus the partial
of Q with respect to z times the partial
of z with respect to x. If we rewrote Q so that
z was substituted with x's and y's-- because it is a
function of x and y's-- then we would just have to write this
first term right over here. But we're assuming
this is expressed as a function of x, y, and z. And z itself is a function of x. And so that's why we had to use
the multi-variable chain rule. Now, let's move
to the next part. And both of these might
have some x's in them. So we have to use the
product rule right over here. So first, we can take the
derivative of r with respect to x. And then, multiply
that times z sub y. Then, we have to take
the derivative of z sub y with respect to x and
multiply that times R. So this is going to be plus. If we take the derivative
of this with respect to x, same exact logic-- R
can change directly due to x, and it can change due to y. And that could change due
to z and multiply that times how z could change due to x. Once again, you could view this
as the multi-variable chain rule in action here. But of course, we
take the derivative of the first term
times the second term. And I'll do the second
term in magenta. So the partial of z with
respect to y plus the derivative of the second term--
which is the partial of z with respect to y, and
then taking the partial of that with respect to
x, which we could just write as that--
times the first term. So that's the
partial with respect to x of all of this
business right over here. And then, we need to
subtract the partial of this with respect to y. And we're going to use
the exact same logic. So then, we're
going to subtract-- and I'll put it in
parentheses like that. So P could change
directly due to y. I will circle P. Let
me do it with a color that I haven't used yet. P could change
directly due to y. So we could say the partial
of P with respect to y, but it could also change due
to z changing because of y. So plus the partial
with respect to z times the partial of z
with respect to y-- and I'll do R maybe
in that same color-- plus the derivative of R. Well, we already figured
that, but actually now it's with respect to x,
not with respect to y. You have to be careful. So it's going to be the
partial of R with respect to y plus the partial
of R with respect to z times the partial of z
with respect to y times z sub x plus-- now,
we take the derivative of the second term
times the first term. The derivative of the partial
of z with respect to x, then with respect to y,
is going to be z. And then we're going to
multiply that times R. And now, let's see if
we can expand this out. And hopefully, things simplify. And just a reminder,
I'm just working on the inside of
this double integral. I'll rewrite the double
integral and the dA once I get this all
cleaned up a little bit. So let me rewrite
it a little bit. So this is equal to
the partial of Q. I'll try to color-code
it the same way. And this is really just
algebra at this point. The partial of Q with
respect to x plus the partial of Q with respect to z times
the partial of z with respect to x plus the partial
of R with respect to x times the partial
of z with respect to y. And then plus the partial
of R with respect to z times the partial
of z with respect to x times the partial
of z with respect to y. And then we have this
term right over here-- which I'll just do in
purple-- plus the partial of z with respect to y, and
then respect to x, times R. And now, we're going to subtract
all of this business right over here. I'll do it in the blue. Minus the partial
of P with respect to y minus the partial of
P with respect to z times the partial of z
with respect to y. And then we're going
to subtract from that. Minus the partial
of r with respect to y times-- we want to
distribute this-- times the partial of z with respect
to x minus the partial of R. This gets a little bit tedious. But hopefully, it'll get
us to where we need to go. The partial of R with respect
to z times the partial of z with respect to y times the
partial of z with respect to x. And then, finally, this
term right over here minus this-- because we have
that negative out there-- minus the partial
of z with respect to y, then with respect to x,
and then with respect to yR. Now, let's see if we
can simplify things. So the first thing-- this
and this look to be the same. We just can commute the order
in which we actually multiply. But these are the
exact same term. So that is going to
cancel out with that. And because we
assumed way up here that we have continuous second
derivatives of the function z-- z is a function
of x and y, that that is equal to that-- we can
now say that these two right over here are going to
be the negatives of each other or that they are
going to cancel out. And so this simplified
things a good bit. And let's see if
I can group terms in a way that might
start to make sense. And actually, I'm
going to try to see if I can make them
similar to this. So I have all the terms that
have a z sub x and the z sub y and then the rest of them. So z sub x I'll do in blue. So you have the terms
that have a z sub x in it. So you have this
term right over here and this term right over here. And we can factor
out the z sub x. And we get the partial
of z with respect to x times the partial
of Q with respect to z minus the partial
of R with respect to y. And then, let's do
the-- and I want to get the colors
the same way, too. I did yellow next. Plus, I have all the
terms of the partial of z with respect to y. So it's that term and
that term right over here. So it becomes plus the
partial of z with respect to y times the partial
of R with respect to x minus the partial
of p with respect to z. And then we have
these last two terms. And I used the color
green up there, so I'll use the
color green again. So for these two
terms, I'll just write plus the partial
of Q with respect to x minus the partial
of P with respect to y. So our double integrals
have kind of-- I can't really say
simplified-- but we can rewrite it like this. And we don't want to forget
this was all a simplification of our double integral
over the region dA. This is what we have
been able to using Green's theorem and the
multi-variable chain rule and whatever else. We've been able to
say that that line integral around the
boundary of our surface is the same thing as this. And now, we can compare that to
what our surface integral was. So let's see if I have space. So copy. And then, let me see if I have
space up here to paste it. Well, it doesn't look like
I actually have much space to paste it, although
I'll try anyway. So if I paste it, you see
that they are identical. They are identical. Our line integral is
identical to this. We get the exact same thing. So our line integral,
f.dr, around this path c simplified to this and our
surface integral simplified to this. So using the
assumptions we had, they both simplified
to the same thing. So now we know, for
this special case, our line integral is equal
to our surface integral. And we are done.