How Do I Find the Intersection of the Surfaces z=x^2+y^2 and x^2+y^2+z^2=2?

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Discussion Overview

The discussion revolves around finding the intersection of the surfaces defined by the equations z=x^2+y^2 and x^2+y^2+z^2=2. Participants explore various methods to derive the curve of intersection, including substitution and transformations, while grappling with the implications of the resulting equations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests substituting z=x^2+y^2 into the sphere's equation to find the intersection but expresses confusion over the resulting fourth-degree equation.
  • Another participant proposes using variables u=x^2 and v=y^2, indicating that this leads to a second-degree equation in u and v, suggesting the intersection is a conic in that space.
  • A different viewpoint questions the necessity of knowing how to draw the curve, implying that finding the equation is sufficient.
  • One participant suggests sketching the problem to visualize the intersection as a circle and discusses properties such as radius and center point, while also mentioning alternative representations involving cylinders.
  • Another participant reiterates the substitution method and simplifies the resulting equation, concluding that it leads to a degenerate conic or a circle in the z=1 plane.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the intersection, with some suggesting it is a circle while others explore the concept of a degenerate conic. The discussion remains unresolved regarding the best approach to visualize or represent the intersection.

Contextual Notes

There are limitations in the discussion regarding assumptions made about the nature of the curves and the transformations applied. The dependency on specific definitions of conics and the implications of degenerate cases are not fully resolved.

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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.
 
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Homework Statement


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.



Homework Equations





The Attempt at a Solution

 
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.
 
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"!
 
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...

[Edit] 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:
astrokat11 said:

Homework Statement


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.

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