## anonymous one year ago ques

1. 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 $\phi(x,y,0)=c$ ?

2. anonymous

That seems true .

3. anonymous

Is this a true false question?

4. anonymous

Nope, it's a doubt that I have

5. anonymous

we can do an example

6. anonymous

Idk, can you think of a counter example??

7. anonymous

can't think of a counterexample

8. anonymous

hmmm let's ask someone else @IrishBoy123

9. anonymous

maybe he can tell :P

10. 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 $\iint_{R} \frac{1}{|\hat n \bullet \hat k|} \ dx \ dy$ $= \frac{1}{4} \iint_{R} \ dx \ dy$

11. IrishBoy123

maybe we need to understand what projection means "intersection"? then yes. just set z to zero for intersection projection, no

12. 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*

13. 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

14. 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

15. IrishBoy123

that looks good to me!

16. IrishBoy123

i think that's right

17. 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

18. 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

19. 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 !!

20. IrishBoy123

ah! 2/5 in limit should be 5/2 ?!?! i spotted that from my sketch :-))

21. anonymous

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

22. anonymous

25/48

23. anonymous

Once again thanks!!

24. IrishBoy123