Solving Stokes Theorem: A Guide

In summary, the conversation discusses the use of the partial derivative in Stokes theorem and how it is applied in a specific equation. It also explains the concept of an independent variable and how it relates to the partial derivative. The conversation concludes with a note on notation for denoting the partial derivative.
  • #1
Cyrus
3,238
16
:grumpy: I can't figure this out, help me NOW! :tongue2: Just kidding.

So anyways, here's 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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  • #2
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:
 
  • #3



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!
 

1. What is Stokes Theorem and when is it used?

Stokes Theorem is a mathematical theorem that relates the surface integral of a vector field over a surface to the line integral of the same vector field along the boundary of the surface. It is used in calculus and vector calculus to solve problems involving vector fields and curved surfaces.

2. What are the requirements for applying Stokes Theorem?

In order to use Stokes Theorem, the surface must be a smooth, closed, and oriented surface. The vector field must also be continuously differentiable in the region enclosed by the surface.

3. What are the steps for solving a problem using Stokes Theorem?

The first step is to determine if the given surface and vector field meet the requirements for applying Stokes Theorem. Then, you must parameterize the surface and calculate the surface normal vector. Next, calculate the curl of the vector field and set up the line integral along the boundary of the surface. Finally, evaluate the line integral to find the solution.

4. Can Stokes Theorem be applied to any surface and vector field?

No, Stokes Theorem can only be applied to smooth, closed, and oriented surfaces and continuously differentiable vector fields. If the surface or vector field does not meet these requirements, another method must be used to solve the problem.

5. What are some real-world applications of Stokes Theorem?

Stokes Theorem has many applications in physics and engineering, including fluid dynamics, electromagnetism, and heat transfer. It is also used in computer graphics and 3D modeling to calculate the flow of vector fields over curved surfaces.

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