Question

ques

Answer 1

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
\[\phi(x,y,0)=c\]
?

Answer 2

That seems true .

Answer 3

Is this a true false question?

Answer 4

Nope, it's a doubt that I have

Answer 5

we can do an example

Answer 6

Idk, can you think of a counter example??

Answer 7

can't think of a counterexample

Answer 8

hmmm let's ask someone else @IrishBoy123

Answer 9

maybe he can tell :P

Answer 10

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

\[\iint_{R} \frac{1}{|\hat n \bullet \hat k|} \ dx \ dy\]
\[= \frac{1}{4} \iint_{R} \ dx \ dy\]

Answer 11

maybe we need to understand what projection means

"intersection"?

then yes. just set z to zero for intersection

projection, no

Answer 12

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*

Answer 13

\[\frac{1}{4}\int\limits_{0}^{\frac{2}{5}} \int\limits_{0}^{\frac{5-2x}{3}}dydx=\frac{1}{12}\int\limits_{0}^{\frac{2}{5}}(5-2x)dx=\frac{1}{12}[5x-x^2]_{0}^{\frac{2}{5}}\]
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

Answer 14

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

Answer 15

that looks good to me!

Answer 16

i think that's right

Answer 17

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

Answer 18

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

Answer 19

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 !!

Answer 20

ah!

2/5 in limit should be 5/2 ?!?!

i spotted that from my sketch :-))

Answer 21

Oops,
\[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\]
Not
\[ax\]!!

Answer 22

25/48

Answer 23

Once again thanks!!

Answer 24

good thread!!