# Stokes Theroem

1. Jan 23, 2005

### Cyrus

:grumpy: I cant figure this out, help me NOW! :tongue2: Just kidding.

So anyways, heres the question:

Part of stokes theorem has the following in it:

$$\frac {\partial } {\partial x} ( Q + R \frac{\partial z}{\partial y} )$$

Which is written as:

$$\frac { \partial Q }{\partial x} + \frac { \partial Q }{\partial z} \frac { \partial z }{\partial x} + \frac { \partial R }{\partial x} \frac { \partial z }{\partial y}+ \frac { \partial R }{\partial z}\frac { \partial z }{\partial x}\frac{ \partial z }{\partial y}+ R \frac { \partial ^2z }{\partial x \partial y}$$

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

$$\frac { \partial Q }{\partial x}\frac { \partial x }{\partial x} +\frac { \partial Q }{\partial y}\frac { \partial y }{\partial x}+ \frac { \partial Q }{\partial z} \frac { \partial z }{\partial x} + \frac { \partial R }{\partial x} \frac { \partial x}{\partial x}\frac { \partial z }{\partial y}+ \frac { \partial R }{\partial y} \frac { \partial y}{\partial x}\frac { \partial z }{\partial y}+ \frac { \partial R }{\partial z}\frac { \partial z }{\partial x}\frac{ \partial z }{\partial y}+ R \frac { \partial ^2z }{\partial x \partial y}$$

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

They show this by rewriting it as follows: $$\frac {\partial } { \partial x } y =o$$.

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

2. Jan 24, 2005

### Galileo

Yeah, $\frac{\partial y}{\partial x}=0$ here, because (although not stated in your question): x=x(t), y=y(t), z=g(x(t),y(t)). So y is no function of x.

For the term with the R, they just used the product rule. Derivation looks good.

BTW: It doesn't matter whether you write $\frac{\partial f}{\partial x}$ or $\frac{\partial}{\partial x}f$
to denote the partial derivative of f with respect to x: