How Is Stokes Theorem Applied to Partial Derivatives?

[Cyrus]

I can't figure this out, help me NOW! Just kidding.

So anyways, here's 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

 

[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:

 

[M98Ranger]

Yes, that is correct. In this case, the partial derivative of y with respect to x is equal to zero because y is an independent variable and is not affected by changes in x. Similarly, the partial derivative of z with respect to x is not affected by changes in y or z, so it is also equal to zero. This simplifies the expression and allows us to use the chain rule to solve for the remaining terms. The two chain rules used are for multivariable functions and for the second partial derivative (calc2 version). I hope this helps clarify the solution to the problem. Keep practicing and you will become more comfortable with solving Stokes Theorem problems!