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?).(adsbygoogle = window.adsbygoogle || []).push({});

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! :yuck:

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! :tongue2:

Ok, time to get back on track.

I will walk through each step, 1-6 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.

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# The Chain Rule, death to anyone that breaks the rule!

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