:grumpy: I cant figure this out, help me NOW! :tongue2: Just kidding.(adsbygoogle = window.adsbygoogle || []).push({});

So anyways, heres the question:

Part of stokes theorem has the following in it:

[tex] \frac {\partial } {\partial x} ( Q + R \frac{\partial z}{\partial y} ) [/tex]

Which is written as:

[tex] \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 ^2z }{\partial x \partial y} [/tex]

I guess they got the above anwser after doing the following, but tell me if not.

[tex] \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} + \frac { \partial R }{\partial x} \frac { \partial x}{\partial x}\frac { \partial z }{\partial y}+ \frac { \partial R }{\partial y} \frac { \partial y}{\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 ^2z }{\partial x \partial y} [/tex]

I see that dx/dx is equal to 1, so it does not appear in the anwser. Does the second term drop, because [tex] \frac {\partial y} { \partial x } =o [/tex].

They show this by rewriting it as follows: [tex] \frac {\partial } { \partial x } y =o [/tex].

I dident realize you could separate the partial operator from the numerator term like that.

Could a possible reason be that y is an independent variable, so a change in x, another independent variable, has no effect on y, which is why it is equal to zero? In other words, the change in y has nothing to do with the change in x, which is why the fraction is zero.

And I guess for the R, term, the same follows, except for the fact that they use two chain rules, one for many variables, and one for the calc2 version.

uv= u'v+v'u

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# Stokes Theroem

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