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Please explain parametric surface in the book

  1. Apr 11, 2013 #1
    This is about change of parameter from (u,v) plane to (x,y) plane. If you read the begining, it said σ is a smooth parametric surface on a region R. It went on and talk about continuous and all in region R shown in Fig 15.4.10 (a). But that is not correct. If you read on, this is about mapping of (u,v) plane of Fig (a) to (x,y) plane in Fig (b)!!! It is obvious that σ is a region in (x,y) plane shown in Fig (b). Seems like the book is wrong.

    It should be ##\vec r=\vec r(x(u,v), y(u,v))=\hat x x(u,v)+\hat y y(u,v)+\hat z z(u,v)## which is a vector value function in (x,y) plane shown in Fig (b).

    Attached is the scan of the book, please ignore all my scribbles, just read the book.



    [PLAIN]http://i46.tinypic.com/2byywh.jpg[/PLAIN]



    Am I missing something?
     
  2. jcsd
  3. Apr 11, 2013 #2

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    The function would be ##\vec r=\vec r(u,v)=\hat x x(u,v)+\hat y y(u,v)+\hat z z(u,v)##.
    This is not necessarily a vector value function in (x,y) plane.
    That becomes apparent when considering a surface that is bent in such a way that it has multiple values for a specific (x,y).

    For instance ##\vec r(u,v)=\hat x \cos u \cos v+\hat y\cos u \sin v+\hat z \sin u##.
    This is a unit sphere.

    The function is called a map to some surface.
    Convoluted surfaces cannot be mapped with 1 such function.
    That's when we need an atlas of maps.
    Mapping the surface of the earth is typically done with such an atlas.
     
  4. Apr 12, 2013 #3
    Thanks for the response, I don't see what you are driving at. The vector value function ##\vec r=\vec r(u,v)=\hat x x(u,v)+\hat y y(u,v)+\hat z z(u,v)## has x,y and z components, not u and v components as can be seen with ##\hat x,\;\hat y,\;\hat z##.

    Yes you can have multiple values of u and v for one specific (x,y). But what is the problem. Like you gave the example of ##\vec r(u,v)=\hat x \cos u \cos v+\hat y\cos u \sin v+\hat z \sin u##.
    Yes, the value of (x,y) is the same when ##\phi = 2n\pi## etc. This is no different from a vector value function ##\vec r(\theta)=\hat x \cos \theta +\hat y \sin \theta## that draw out a unit circle. As ##\theta## increase beyond ##2\pi##, the tip of the vector value function just keep tracing out the unit circle over and over!!!

    It is perfectly legal to have two separate values of u and v to point to the same exact location in the (x,y) plane. Problem arise ONLY if the same values of u and v points to more than one point in the (x,y) plane.


    But that's not my question. My question is the book said the surface σ is in region R which is not true. If you read the scanned page, the book called the region σ in the (x,y) plane. The only way that can be true is if the book call both region in (u,v) and (x,y) plane σ. But this is very thin.
     
    Last edited: Apr 12, 2013
  5. Apr 12, 2013 #4

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    I was triggered by you saying that r "should be" a function of x and y.
    It isn't. It's a function of u and v.

    It doesn't say that σ is in region R.
    It says that σ is on region R, meaning σ is defined as a mapping from R to (x,y,z) space.
     
  6. Apr 12, 2013 #5
    Yes, it's a function of u and v, what I said in the original post is ##\vec r## is a vector value function in the (x,y) plane. I'm glad that clear this up, I was worry that I got it all wrong.

    Is that how the math book describe? That σ is on region R, meaning σ is defined as a mapping from R to (x,y,z) space? Wow, I find it very hard to grapes the meaning. How do you describe if it is in the region R? And what is the meaning of "on" exactly. I always have a very hard time reading the definition in all the calculus books.

    Thanks
     
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