Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Stokes Theroem

  1. Jan 23, 2005 #1
    :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:

    [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
     
  2. jcsd
  3. Jan 24, 2005 #2

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    Yeah, [itex]\frac{\partial y}{\partial x}=0[/itex] 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 [itex]\frac{\partial f}{\partial x}[/itex] or [itex]\frac{\partial}{\partial x}f[/itex]
    to denote the partial derivative of f with respect to x:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Stokes Theroem
  1. Gauss and Stokes (Replies: 4)

  2. Stoke's theorem (Replies: 1)

  3. Stokes theorem (Replies: 1)

  4. Stokes theorem (Replies: 2)

Loading...