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Parametric Surface

  1. Aug 5, 2011 #1
    1. The problem statement, all variables and given/known data

    This is a part of a bigger problem I am working on for my calculus 3 class. There is a parametric surface: [itex]^{\vec{}}r(u,v)=<u+v,u-v,1-2u>[/itex]
    It represents the plane through points (1,0,0), (0,1,0) (0,0,1). As part of the problem, I need to set up a surface integral (specifically through this parametrization) and evaluate it. Now, what is the proper way of finding the upper and lower value for u and v?

    2. Relevant equations

    see above

    3. The attempt at a solution
    I tried plugging the values x,y and z from the above given points into the following system: [itex]^{}X(u,v)=u+v, Y(u,v)=u-v, Z(u,v)=1-2u[/itex]

    As result, I obtained 3 pairs of u and v: [itex]^{}u=1/2,v=1/2 (for (1,0,0));u=0,v=0 (for (0,0,1)); u=1/2,v=-1/2 (for (0,1,0)) [/itex]

    Is it mathematically correct to send the double integral over the parametric region for [itex]^{}u\epsilon[0,1/2][/itex] and [itex]^{}v\epsilon[-1/2,1/2] [/itex] ?

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    3. The attempt at a solution
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    3. The attempt at a solution
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    3. The attempt at a solution
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    3. The attempt at a solution
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    3. The attempt at a solution
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    3. The attempt at a solution
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  2. jcsd
  3. Aug 5, 2011 #2

    Dick

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    Homework Helper

    No, u in [0,1/2] and v in [-1/2,1/2] is a four sided polygon. What you want is a three sided polygon. As you've already figured out, the vertices are (u,v)=(1/2,1/2), (0,0) and (1/2,-1/2). That's a triangle, right? The upper and lower limits can't both be constants.
     
  4. Aug 6, 2011 #3
    thank you :) You are right. I have actually figured this one out last night. I temporarily made an explicit (x,y) parametrization to the surface and saw that its identical to the one given through (u,v). It was easy from there. I guess my question is whether there is any formal method on finding boundaries of (u,v) when you are given a surface in x,y,z and a complicated parametrization? What if the case is not as obvious as the one I had above?
     
  5. Aug 6, 2011 #4

    HallsofIvy

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    You have [itex]\vec{r}= <u+ v, u- v, 1- 2u>[/itex] and your vertices are (1,0,0), (0,1,0), and (0,0,1).

    For the first point, you must have u+ v= 1, u- v= 0. Adding the two equations, 2u= 1 so u= 1/2. Then, of course, v= 1/2 also. Check the last coordinate: 1- 2(1/2)= 0. If that last didn't check, the point would not be on this surface.

    For the second point, you must have u+ v= 0, u- v= 1. Adding the two equations give 2u= 1 so u= 1/2 but now v= -1/2. Again, 1- 2(1/2)= 0.

    For the third point, you must have u+ v= 0, u- v= 0. Those two equations give the obvious u= 0, v= 0 and, of course, 1-2(0)= 1.

    That is, your triangle has vertices, in the "(u, v) plane" (0, 0), (1/2, -1/2), and (1/2, 1/2).

    The line from (0, 0) to (1/2, 1/2) is v= u. The line from (0, 0) to (1/2, -1/2) is v= -u.
    You need u from 0 to 1/2 and, for each u, v from -u to u.

    And your differential, in terms of du and dv is?
     
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