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  1. anonymous
    • one year ago
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    For any level surface S given by \[\phi(x,y,z)=c\] Then the region R that is it's projection on xy plane is always given by \[\phi(x,y,0)=c\] ?

  2. anonymous
    • one year ago
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    That seems true .

  3. anonymous
    • one year ago
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    Is this a true false question?

  4. anonymous
    • one year ago
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    Nope, it's a doubt that I have

  5. anonymous
    • one year ago
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    we can do an example

  6. anonymous
    • one year ago
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    Idk, can you think of a counter example??

  7. anonymous
    • one year ago
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    can't think of a counterexample

  8. anonymous
    • one year ago
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    hmmm let's ask someone else @IrishBoy123

  9. anonymous
    • one year ago
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    maybe he can tell :P

  10. IrishBoy123
    • one year ago
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    let's test it first with a simple surface, a plane, which i shall just make uo \[\pi: 2x + 3y + 4z = 5\] \[\phi(x,y,z) = 2x + 3y + 4z [= const]\] \[\phi(x,y,0) = 2x + 3y =5\] you might think of this in terms of the line of intersection of plane \(\pi\) and the xy plane. in practise the projection is everywhere. the projection can be found from the surface integral formulation \[\iint_{R} \frac{1}{|\hat n \bullet \hat k|} \ dx \ dy\] \[= \frac{1}{4} \iint_{R} \ dx \ dy\]

  11. IrishBoy123
    • one year ago
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    maybe we need to understand what projection means "intersection"? then yes. just set z to zero for intersection projection, no

  12. IrishBoy123
    • one year ago
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    another example paraboloid \(z = x^2 + y^2 - 4\) \(\phi(x,y,z) = x^2 + y^2 - z = const \ \ [ie \ 4]\) \(\phi(x,y,0) = x^2 + y^2 = 4\) that's the circle on the xy plane, ie the *intersection*

  13. anonymous
    • one year ago
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    \[\frac{1}{4}\int\limits_{0}^{\frac{2}{5}} \int\limits_{0}^{\frac{5-2x}{3}}dydx=\frac{1}{12}\int\limits_{0}^{\frac{2}{5}}(5-2x)dx=\frac{1}{12}[5x-x^2]_{0}^{\frac{2}{5}}\] Yeh sorry my bad, what I mean is that the equation we can form from \[\phi(x,y,0)\] is what we will use as a limit when we transform from S to R

  14. anonymous
    • one year ago
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    What I'm trying to say is something like \[\phi(x,y,0)\] will always give a curve in xy plane that bounds R

  15. IrishBoy123
    • one year ago
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    that looks good to me!

  16. IrishBoy123
    • one year ago
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    i think that's right

  17. anonymous
    • one year ago
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    So there's no reason to draw a graph every single time, we can just set z=0 and get our curve's equation, then we can double integrate within the limits

  18. IrishBoy123
    • one year ago
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    well, i always draw. always. seeing is believing. thing is you can get the line of intersection but you might still want to know if it is above or below the xy plane loads of people on here don't draw and they seem fine. to me, it's madness :-) and by draw, i mean rough sketch with key numbers. i use drawings to get the limits right, for example. you know, switch you integral from dydx to dxdy? i'll do that with my sketch. the integral http://www.wolframalpha.com/input/?i=%5Cfrac%7B1%7D%7B4%7D%5Cint%5Climits_%7B0%7D%5E%7B%5Cfrac%7B2%7D%7B5%7D%7D+%5Cint%5Climits_%7B0%7D%5E%7B%5Cfrac%7B5-2x%7D%7B3%7D%7Ddydx

  19. anonymous
    • one year ago
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    Good insight, this time I'll have less questions for my exams because it's university exams, in schools we had like 30 questions, but this time there will be around 7 questions of high weightage and each question would require more depth and explanation, so it might just help to draw a graph actually !!

  20. IrishBoy123
    • one year ago
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    ah! 2/5 in limit should be 5/2 ?!?! i spotted that from my sketch :-))

  21. anonymous
    • one year ago
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    Oops, \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\] Not \[ax\]!!

  22. anonymous
    • one year ago
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    25/48

  23. anonymous
    • one year ago
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    Once again thanks!!

  24. IrishBoy123
    • one year ago
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    good thread!!

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