Inequalities considering surfaces

Question

I have one general question considering the surface area. If we have for example body bounded with $z^2\le x^2+y^2, z\le 2-x^2-y^2$. The first inequality tells us we are looking outside of the cone, and the second inside of paraboloid. But in solution they are first looking at the part that is inside of cone and paraboloid between $z=0,z=1$(and I don't understand why because we should only look outside of cone?). Second area according to solutions is symmetric to this one just below $z=0$. And the third(one that I would only consider if I had to work on this) is outside of cones and inside of paraboloid. So, my main question is why are we looking those two areas that are inside cones, when the given inequality should only be outside of cone?

Answer 1

We are talking about surface area so it is always on the surface of at least one of the intersecting solids. Now take the surfaces in your question with given inequalities,

$z^2\le x^2+y^2$

$z\le 2-x^2-y^2$

Equating both, we have $z^2 +z - 2 = 0 \implies z = 1, -2$.

First let's consider only the cone above $z-$ plane and its intersection with the paraboloid. Also the sketch shows you that you have part of the paraboloid inside the cone and part of the cone inside the paraboloid. So which surface area do you find? For that, let's see the inequalities.

You have surface of paraboloid inside the cone for $z \geq 1$. But at any given $z$, if you are inside the cone, clearly, $x^2 + y^2 \lt z^2$ which is not what the inequality suggests.

Now let's take the surface of the cone inside the paraboloid which is for $0 \leq z \leq 1$. As you are on the surface of the cone, the cone inequality (read equality) is obviously met and as we are inside the paraboloid surface, at any given $z$, clearly $x^2 + y^2 \lt 2 - z$ which meets the other inequality as well.

The same logic applies for the cone below $z-$plane. But you must note that for this cone, while there is finite surface of cone inside the paraboloid, for $z \leq -2$, the paraboloid is completely inside the cone surface and hence there is unbound surface area of the paraboloid inside the cone.

Last but not least the bound surface area above $z-$plane and below are not equal and it is not symmetric with respect to the $z-$ plane.