# Intersection of surfaces

1. Dec 10, 2007

### astrokat11

z=x^2+y^2 and x^2+y^2+z^2=2...I need to find the intersection of these two surfaces. Would I just substitute z=x^2+y^2 into the equation of the sphere to find the curve of intersection? But when I do that I get an equation with fourth powers and I don't know what kind of curve that makes.

2. Dec 10, 2007

### astrokat11

1. The problem statement, all variables and given/known data
z=x^2+y^2 and x^2+y^2+z^2=2...I need to find the intersection of these two surfaces. Would I just substitute z=x^2+y^2 into the equation of the sphere to find the curve of intersection? But when I do that I get an equation with fourth powers and I don't know what kind of curve that makes.

2. Relevant equations

3. The attempt at a solution

3. Dec 10, 2007

### mathman

I am not sure what you are looking for, but let u=x^2 and v=y^2, and you will get an equation of second degree in u and v, so the curve is a conic in (u,v) space. How that looks in the original I can only guess.

4. Dec 10, 2007

### quasar987

Is that so bad, that you do not know how to draw the curve? You will have found the answer nonetheless.

Interesting remark: if you set x'=x^2, y'=y^2, you get the equation of a conic. So the curve is "a conic squared"!

5. Dec 11, 2007

### coomast

I would suggest to make a sketch of the problem first. Then you would "see" that the intersection is a circle. Try to find the properties of the circle, i.e. the radius and the center point. This can be done by inserting the equation of the parabolic cone into the one of the sphere.

If you are looking for an equation of the circle, there are several. Firstly, stating that the curve is the intersection of the two equations you gave is a valid one. Secondly a circle in space can be presented also as an intersection of a cylinder and a plane perpendicular to the axes of the cylinder. In this case it would be a cylinder with it's axis the Z-axes and a radius you have obtained and a plane z=1 if I'm not mistaken. There are possibilities in using cylinder coordinates etc. Up to you to see what's most appropriate. If anything is not clear just post...

 It seems that you have posted this question twice. It is not the intention to do this astrokat11. Anyway, I seems I was right on z=1 plane.

Last edited: Dec 11, 2007
6. Dec 11, 2007

### HallsofIvy

Yes, that gives $x^2+ y^2+ (x^2+ y^2)^2= 2$. Do not multiply out that last square! As astrokat11 suggested, if you let $u= x^2$ and $v= y^2$ you get $(u+ v)^2+ u+ v= 2$. Now let p= u+v, and we have simply $p^2+ p= 2$ so p= u+ v= -2 or p= u+ v= 1. That's a "degenarate" conic- two straight lines. Going back to x and y, $x^2+y^2= -2$, which, of course, is impossible or $x^2+y^2= 1$, a circle.
One of your equations is z= $x^2 +y^2$= 1 so this is a circle in the z= 1 plane with center (0, 0, 1) and radius 1.

Last edited by a moderator: Dec 11, 2007