# Restrict a function in 3D view

Drako shared this question 2 years ago
Answered

Hello,

I need restrict this function:

1. z=1-x^2

in the intervals

1. 0<=x<=1
2. 0<=y<=2

But, I get a error when I use the function if.

please, help me

## Comments (7) 1

Hi,

z=If[0<=x<=1&&0<=y<=2,1-x^2]

works for me... what exactly did you type in?

Cheers,

Zbynek 1

Thanks, its work, my error was the logic connection "and". I didn't know the correct form.  1

Will the If() command work in 3D only when the function is stated explicitly? I have tried

If(0<=x<=85, (x-z)^2 + y^2 = z^2) and

If(0<=x<=85 && z>=0, (x-z)^2 + y^2 = z^2)

but get an input error. I don't require any restriction on y. 1

you can not use a condition on z for 3D surface at this moment

in your case you can try if(0<=x,(x^2+y^2)/(2x)) 1

Thanks mathmagic. So z has to be expressed explicitly in terms of x & y. Is it possible to restrict the surface by restricting the values of z, say 76<z<114? 1

first define a(x,y)=(x^2+y^2)/(2x) then if(76<=a(x,y)<=114,a(x,y))

I think better if you can parametrize the function and use surface() because you can have a problem with discontinuity 1

in your case you have horizontal circles centred in (z,0,z)

surface(u+ucos(v),u sin(v),u,76,114,v,0,2 pi)

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