# Restrict a function in 3D view

Drako shared this question 4 years ago

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.

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)

1

You can do it like me, restricting first in an axis (could be x) and then you need to restrict again the function but on the other axis.

For example:

e(x, y) = If(0 ≤ x ≤ 2π, sen(x) + sen(y) + sen(x + y))

f(x,y) = If(0 ≤ y ≤ 2π, e(x, y))This will restrict the domain on X and Y from 0 to 2pi in both axis.

1

es más corto, y creo que más fácil

If(0 ≤ x ≤ 2π && 0 ≤ y ≤ 2π , sen(x) + sen(y) + sen(x + y))

&& es .AND. es decir el y logico