
#1
May2405, 08:29 PM

P: 4,780

Ok so im reviewing multivariable now that i have some time; (why is it taking me so long to grasp some of these concepts!? ) anyways, and Im reading the proof of stokes theorem. The book I use is Stewart, but it seems to be ripped off word for word from swokowski, which in turn rippes off S.L salas and Einar Hille, (maybe each new author contiuned the publishing over time?).
Here is whats throwing me a curve, at one point, they show that the [tex]curl \vec{F} \cdot d\vec{S} = \int_c \vect{F} \cdot d\vec{r}[/tex] They assume that [tex] \vec{F}=P\hat{i}+Q\hat{j}+R\hat{k}[/tex] I will just put the proof up to avoid confusion when I refer to it: (1) [tex] \int_c \vec{F} \cdot d\vec{r} = \int^b_a (P\frac{dx}{dt} +Q\frac{dy}{dt}+ R\frac{dz}{dt}) dt [/tex] (2) [tex] ' ' = \int^b_a [P \frac{dx}{dt} +Q \frac{dy}{dt}+ R( \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt})] dt [/tex] (3) [tex]' ' = \int^b_a [(P+R\frac{\partial z}{\partial x})\frac{dx}{dt} + (Q+R\frac{\partial z}{\partial y})\frac{dy}{dt}] dt [/tex] (4) [tex]' '=\int_{c1} (P+R\frac{\partial z}{\partial x})dx + (Q+R\frac{\partial z}{\partial y})dy [/tex] (5) [tex]' '=\int\int_D[\frac{\partial}{\partial x}(Q+R\frac{\partial z}{\partial y})\frac{\partial}{\partial y}(P+R\frac{\partial z}{\partial x}]dA [/tex] By Green's Theorem. Then using the chain rule again and remembering that P,Q, and R are functions of x,y and z, and that z is a function itself of x and y, we get: (6) [tex] \int_c \vect{F} \cdot d\vec{r} = \int\int_D [( \frac{\partial Q}{\partial x} + \frac{\partial Q}{\partial z} \frac{\partial z}{\partial x}+\frac{\partial R }{\partial x}\frac{\partial z }{\partial y }+ \frac{\partial R }{\partial z }\frac{\partial z }{\partial x }\frac{\partial z }{\partial y }+ R \frac{\partial^2 z }{\partial x \partial y }) ( \frac{\partial P }{\partial y }+ \frac{\partial P }{\partial z } \frac{\partial z }{\partial y }+ \frac{\partial R }{\partial y } \frac{\partial z }{\partial x } + \frac{\partial R }{\partial z } \frac{\partial z }{\partial y } \frac{\partial z }{\partial x }+R \frac{\partial^2 z }{\partial y \partial x })]dA [/tex] TOO MUCH TYPING! For some reason I cant see what I typed?? Aha, "" marks are reserved for something thats what it is. YES! the last line worked finally, aye! Ok, time to get back on track. I will walk through each step, 16 followed by the last line and describe what they are doing. If Im wrong along the way, let me know. I will also explain what part is throwing me off. (1) This is just standard notation derived previously when doing the line integral of a vector field. The reason for the dt outside the parthensis is because we have x as a function of t; x=f(t), and when doing a line integral we want to integrate P W.R.T dx, not dx/dt(dx/dt is the speed at which x changes, but we are interested in the change of x only, not the change in its speed), which is why the extra dt outside the parenthesis to cancel out the dt, thus integrating W.R.T dx. I.E dx/dt*dt = dx Similar arguments hold for the dy/dt and dz/dt. (2) This part is fine too, becuase z is a function of x and y. Furthermore, x and y are functions of t. So when we do the total derivative of z, we get the junk inside the parenthesis. This makes sense, since it was derived earlier in the book. We did the linearization of the TANGENT PLANE and arrived at an equation for dz. We divided this entire equation by dt, and took the limit, and we get the result inside (" "). (3) Easy, Easy, Easy, just move things around and factor out differentials (4) Seems like the trick played here is that they UNPARAMETERIZED the function with respect to t to just x and y again. (5) Now they just apply Green's theorem for a vector function, which they can do because they use the planar curve c1 which lies on the projection plane of the surface on the xy plane. So it only varies in x and y. Enter Confusion: Performing the partial derivative is messing me up. Since it is the same for all the partials, lets just deal with the first part in the brackets []( the minus " " terms are the same procedure of differentiation.) Now we have the partial derivative of Q W.R.T x. Now Q is a function of x AND a function of z. Of course Z is itself a function of x. But how did they get the part: [tex] \frac{\partial Q}{\partial x} + \frac{\partial Q}{\partial z} \frac{\partial z}{\partial x} [/tex]. I understand you have to take the derivative of the x part and the z part. But how did they arrive at this equation, because this equation does not reseble the linearization i.e dz=partialf/partialx(dx)+partialf/partialy(dy)? What proof can I turn to, so that I can say ah yes, this is why you do the derivative this way. 



#2
May2405, 08:37 PM

P: 2,223

Your missing a couple lines there, #4 and 5.




#3
May2505, 04:50 AM

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It's just an application of the chain rule. See 'The Chain Rule (General version)' on page 793. You might as well consider x and y as functions x(x,y), y(x,y) and use that y is constant w.r.t. x (it doesn't really depend on x). It looks weird, but in this way you can readily use the form presented in the book if it's not immediately obvious:
[tex]\frac{\partial}{\partial x} Q(x(x,y),y(x,y),z(x,y))=\frac{\partial Q}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial Q}{\partial y}\frac{\partial y}{\partial x}+\frac{\partial Q}{\partial z}\frac{\partial z}{\partial x}=[/tex] [tex]\frac{\partial Q}{\partial x}\cdot 1+\frac{\partial Q}{\partial y}\cdot 0+\frac{\partial Q}{\partial z}\frac{\partial z}{\partial x}=\frac{\partial Q}{\partial x}+\frac{\partial Q}{\partial z}\frac{\partial z}{\partial x}[/tex] 



#4
May2505, 07:45 AM

P: 4,780

The Chain Rule, death to anyone that breaks the rule!Funny you should say that, I was thinking about it in terms of what you said last night right before i went to bed but i wasnet sure. The thing that I was not sure about was the second part of the fraction where you have dx/dx, dy/dx and dz/dx. Is the reason you have partial x/ partial x and not dx/dx, is that you made x a function of x and y? 



#5
May2505, 08:04 AM

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One might use a pedantic notation here: 1. Let us have a function [tex]Q'(x',y',z')[/tex] 2. Let [tex](x',y',z')\in\mathbb{R}^{3}[/tex] be related to [tex](x,y)\in\mathbb{R}^{2}[/tex] as follows: x'=X(x,y)=x,y'=Y(x,y)=y,z'=Z(x,y) 3. We may now define a function Q(x,y) as follows: [tex]Q(x,y)=Q'(X(x,y),Y(x,y),Z(x,y))[/tex] 4. We also define: [tex]\vec{x}'=(x',y',z'), \vec{X}(x,y)=(X(x,y),Y(x,y),Z(x,y))[/tex] 5.Thus, we have: [tex]\frac{\partial{Q}}{\partial{x}}=(\frac{\partial{Q'}}{\partial{x'}}\frac {\partial{X}}{\partial{x}}+\frac{\partial{Q'}}{\partial{y'}}\frac{\part ial{Y}}{\partial{x}}+\frac{\partial{Q'}}{\partial{z'}}\frac{\partial{Z} }{\partial{x}})\mid_{\vec{x}'=\vec{X}(x,y)}[/tex] 6. Now, one might ask oneself if pedantic notation is really worthwhile.. 



#6
May2505, 09:44 AM

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#7
May2505, 09:47 AM

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No, I just use [tex]\vec{x}'[/tex] to designate a point in [tex]\mathbb{R}^{3}[/tex]
Any such point can be represented with the aid of a vector having three components; the components of the vector [tex]\vec{x}'[/tex] are the scalars x',y',z'. 



#8
May2505, 10:01 AM

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i recommend you write out a proof yourself for integration around a rectangle. it is very easy, following from precisely two facts: FTC and Fubini. I.e. Fubini reduces the two dimensional integral to one dimensional integrals, and then FTC does those.
Then all other cases are obtained from that by the chain rule, which is a separate issue, having nothing to do with Stokes theorem, only with how to generalize it technically. 



#9
May2505, 10:27 AM

P: 4,780

before you had x'=X(x,y), so x' was a scalar function. Now you have x' as a vector. How come you changed it into a vector function. I thought x' is only the scalar component in the x direction.




#10
May2505, 10:31 AM

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Is [tex]\vec{x}'[/tex] the same symbol as x'???
Doesn't look so for me, at least. I could equally well have called my vector [tex]\vec{v}'=(x',y',z')[/tex] if you are happier with that. 



#11
May2505, 10:37 AM

P: 4,780

I see what your saying now. It looks as if [tex]\vec{x'}[/tex] and [tex] \vec{X'} [/tex] are the same thing. Why was it necessary to use both of them in the proof? Oh, and i think [tex] \vec{s'} [/tex] would have been a little bit clearer. Calling the position vector by x confused me with the x component of a vector. 



#12
May2505, 10:47 AM

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With [tex]\vec{x}'[/tex], I designate a POINT (element) in [tex]\matbb{R}^{3}[/tex]
With (x,y) (or, if you like, [tex]\vec{x}[/tex] (no apostroph/hyphen here!!)), I designate a POINT (element) in [tex]\mathbb{R}^{2}[/tex] The statement [tex]\vec{x}'=\vec{X}'(x,y)[/tex] says that we have a mapping [tex]\vec{X}':\mathbb{R}^{2}\to\mathbb{R}^{3}[/tex] i.e, a point (x,y) in the plane (the input value) is mapped onto a point (x',y',z') (the output value) in [tex]\mathbb{R}^{3}[/tex] in accordance with some rule. 



#14
May2505, 12:11 PM

P: 4,780

x'=X(x,y)=x,y'=Y(x,y)=y,z'=Z(x,y)=z Here you are reusing the values of x, and y when you call x'=X(x,y)=x. I guess the thing here is to remember that the x=X(x,y) is not the same as the variable x inside the parthensis. It has a double meaning that i should be VERY carefull about.




#15
May2505, 12:22 PM

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Our "machine", or, function, X has input (x,y), and spits out the xvalue of (x,y) as the output (i.e, X can be regarded as the operation of projecting (x,y) onto the xaxis.)
That is, we have specified a rule so that given input, we may calculate output. This output is then what x' is set equal to. 



#16
May2505, 12:35 PM

P: 4,780

Right, just to be clear, the x's I was reffering to are: X(x,y), the x inside the (x,y) and the x in X(x,y)=x on the right side of the equal sign. Here we are reusing the variable x for two different purposes. I dident mean the capital X. (sorry if I was not clear on that.)




#17
May2505, 12:37 PM

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It was I who wasn't clear about it from the start (sorry about that), but I think you've got it.




#18
May2505, 12:48 PM

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this stuff is trivial, or ought to be if done right. Stewart et al are just clogging it up with notation.
Look, if W is a one form, (i.e. something like Pdx + Q dy), then the "curl" is just dW = dP^dx + dQ^dy which is expanded out by the rule that dP = dP/dx dx + dP/dy dy, and dx^dx = 0 = dy^dy, and dy^dx = dx^dy, as usual. so dW = (dQ/dx  dP/dy)dx^dy. Then greens theorem, which is repeated integration plus FTC, says the integral of W over the boundary of the rectangle equals the integral of dW over the rectangle. then the so called "Stokes theorem" just says this remains true for surfaces parametrized by a rectangle. thats all. and this is just the chain rule. I.e. let f(s,t) = (x(s,t), y(s,t), z(s,t)), be a map from the rectangle R in the s,t plane into x,y,z, space. Then by definition, the boundary of f(R), is f of the boundary of R. and if W is a one form in x,y,z space, then curl of the pullback f*W, of W to s,t space, is the pullback of the curl of W. (this is the chain rule). i.e. f*(dW) = d(f*W) Hence the integral of W over the boundary of f(R), equals the integral of f*W over the boundary of R, which by Green equals the integral of d(f*W) over R, which equals the inetgral of f*(dW) over R, which by definition equals the integral of d(W) over f(R). If we write dot product for integral and b for boundary, we get just this: <W,b(f(R))> = <W,f(bR)> = <f*W, b(R)> = <d(f*W),R> = <f*(dW),R> = <dW,f(R)>. done. I.e. the first equation is true because f(b(R)) = b(f(R)). the second by definition of how to integrate over a parametrized curve, the third is true by greens theorem, the fourth is true by the chain rule which implies that f*(dW) = d(f*W), and the last is true by definition. thats it. 


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