anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
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 \[\phi(x,y,0)=c\] ?
anonymous
  • anonymous
That seems true .
anonymous
  • anonymous
Is this a true false question?

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anonymous
  • anonymous
Nope, it's a doubt that I have
anonymous
  • anonymous
we can do an example
anonymous
  • anonymous
Idk, can you think of a counter example??
anonymous
  • anonymous
can't think of a counterexample
anonymous
  • anonymous
hmmm let's ask someone else @IrishBoy123
anonymous
  • anonymous
maybe he can tell :P
IrishBoy123
  • IrishBoy123
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 \[\iint_{R} \frac{1}{|\hat n \bullet \hat k|} \ dx \ dy\] \[= \frac{1}{4} \iint_{R} \ dx \ dy\]
IrishBoy123
  • IrishBoy123
maybe we need to understand what projection means "intersection"? then yes. just set z to zero for intersection projection, no
IrishBoy123
  • IrishBoy123
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*
anonymous
  • anonymous
\[\frac{1}{4}\int\limits_{0}^{\frac{2}{5}} \int\limits_{0}^{\frac{5-2x}{3}}dydx=\frac{1}{12}\int\limits_{0}^{\frac{2}{5}}(5-2x)dx=\frac{1}{12}[5x-x^2]_{0}^{\frac{2}{5}}\] 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
anonymous
  • anonymous
What I'm trying to say is something like \[\phi(x,y,0)\] will always give a curve in xy plane that bounds R
IrishBoy123
  • IrishBoy123
that looks good to me!
IrishBoy123
  • IrishBoy123
i think that's right
anonymous
  • anonymous
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
IrishBoy123
  • IrishBoy123
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
anonymous
  • anonymous
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 !!
IrishBoy123
  • IrishBoy123
ah! 2/5 in limit should be 5/2 ?!?! i spotted that from my sketch :-))
anonymous
  • anonymous
Oops, \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\] Not \[ax\]!!
anonymous
  • anonymous
25/48
anonymous
  • anonymous
Once again thanks!!
IrishBoy123
  • IrishBoy123
good thread!!

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