0
$\begingroup$

So, I'm trying to solve a problem where we have an intersection of two regions, one of them being $ 0 < z < 2-(x^2 + y^2)$. At the first glance, I thought that that was the top part of the circumference defined by the equation above, added with a cylinder of equation $x^2 + y^2 = 2$. But then I realized that if it was the answer, $z$ would be between $x^2 + y^2$ and $(2-(x^2+y^2))$.

So, now I have no idea of what region is this…

$\endgroup$

2 Answers 2

1
$\begingroup$

$x^2+y^2 < 2-z$, so for a fixed $z$ you have circles of radius up to $\sqrt{2-z}$ at height $z$. In other words a paraboloid, flipped upside down. See:

https://www.wolframalpha.com/input/?i=x%5E2%2By%5E2%3D2-z

$\endgroup$
0
$\begingroup$

hint

If we put $$x=r\cos(t) \;,\; y=r\sin(t)$$ then

$$0<z<2-(x^2+y^2)$$ means that the point $(x,y,z)$ is above the plane $ z=0$ and the surface $ z=2-r^2$ is a paraboloid of revolution.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .