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anonymous
 one year ago
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anonymous
 one year ago
ques

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For 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
 one year ago
Best ResponseYou've already chosen the best response.0Is this a true false question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nope, it's a doubt that I have

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we can do an example

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Idk, can you think of a counter example??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can't think of a counterexample

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm let's ask someone else @IrishBoy123

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0maybe he can tell :P

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1let'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
 one year ago
Best ResponseYou've already chosen the best response.1maybe we need to understand what projection means "intersection"? then yes. just set z to zero for intersection projection, no

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1another 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
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{4}\int\limits_{0}^{\frac{2}{5}} \int\limits_{0}^{\frac{52x}{3}}dydx=\frac{1}{12}\int\limits_{0}^{\frac{2}{5}}(52x)dx=\frac{1}{12}[5xx^2]_{0}^{\frac{2}{5}}\] 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
 one year ago
Best ResponseYou've already chosen the best response.0What I'm trying to say is something like \[\phi(x,y,0)\] will always give a curve in xy plane that bounds R

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1that looks good to me!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1i think that's right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.1well, 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%7B52x%7D%7B3%7D%7Ddydx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Good 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
 one year ago
Best ResponseYou've already chosen the best response.1ah! 2/5 in limit should be 5/2 ?!?! i spotted that from my sketch :))

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oops, \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\] Not \[ax\]!!
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