Solving an Equation: Is it a Paraboloid or Cone?

[Amaelle]

Homework Statement

what is the nature of this object?

Relevant Equations

z=sqrt(2-x^-y^2)

Good day
while solving some integral I met with the following equation
z=sqrt(2-x^-y^2) that looks like a paraboloid?!
I thought first that it might be a cone!
any insights?
thank you!

 

[Frabjous]

Try rearranging the equation to (2-x)=y²+z² and look at it in the plane z=0 or y=0.

 

[Amaelle]

you mean 2-x^2?
if I do as you said x=+-sqrt(2-z^2) ?!

 

[Frabjous]

Your original equation does not have the square.

Rewrite the equation as
x²+y²+z² = 2
Do you recognize it?

 

[Steve4Physics]

Homework Statement:: what is the nature of this object?
Relevant Equations:: z=sqrt(2-x^-y^2)

Good day
while solving some integral I met with the following equation
z=sqrt(2-x^-y^2) that looks like a paraboloid?!
I thought first that it might be a cone!
any insights?
thank you!

You have given your original equation as $z = \sqrt {2 - x^{-y^2}}$

Was this a mistake? Maybe you meant $z = \sqrt {2 - x^2 - y^2}$.

 

[Frabjous]

You have given your original equation as $z = \sqrt {2 - x^{-y^2}}$

I didn‘t read that at all. I really need to cut back on the hallucinogens

 

[Amaelle]

You have given your original equation as $z = \sqrt {2 - x^{-y^2}}$

Was this a mistake? Maybe you meant $z = \sqrt {2 - x^2 - y^2}$.

yes the last one was what I meant thank you!

 

[Frabjous]

yes the last one was what I meant thank you!

Then my post #4 applies.

 

[Amaelle]

Your original equation does not have the square.

Rewrite the equation as
x²+y²+z² = 2
Do you recognize it?

Now yes
it's an ellipsoid thank you very much

 

[Frabjous]

Now yes
it's an ellipsoid thank you very much

True, but it is also a sphere.

 

[Mark44]

it's an ellipsoid

Following up on @caz's comment, it's better to think of this as a sphere rather than an ellipsoid. A sphere is a special case of ellipsoids, in which both foci are located at the center of the sphere.

From the equation you can find the center of the sphere and its radius. Owing to the square root in the original equation, you get only part of the sphere.

 

[PeroK]

It's a hemisphere,to be precise. As $z$ is only positive.

Just saw that's already been said.

 

[Amaelle]

Following up on @caz's comment, it's better to think of this as a sphere rather than an ellipsoid. A sphere is a special case of ellipsoids, in which both foci are located at the center of the sphere.

From the equation you can find the center of the sphere and its radius. Owing to the square root in the original equation, you get only part of the sphere.

thank you very much