Finding limits - function of three variables 
Find limits of a function $f:\mathbb{R}^3\rightarrow \mathbb{R}$ given by formula $f(x,y,x)=x+y+z$ on set $M=\left\{ (x,y,z)\in\mathbb{R}^3:x^2+y^2\le z\le 1 \right\}$. Does $f$ reaches all its limits?

To answer the last question I need to know if $M$ is a closed and bounded set. If it is then $f$ reaches all its limits by Bolzano-Weierstrass theorem. But I'm not sure if it is closed (bounded it is I think it just simply follows from inequalities, but probably it isn't precise proof). Previously I had only sets like $\left\{(x;y)\in\mathbb{R}^2: 4x^2+y^2\le 25\right\}$ and a reason for closedness was rather simple. I always used to state given set as an inverse image by continuous function of closed set. But in situation with $M$ I can't find similar explanation.
Finding limits seems to be hard too. I always used partial derivatives to find extremes in the interior of set and parameterization to explore boundary of set.
But here I'm even not sure what will be interior and what will be boundary of $M$.
Can anybody help?
 A: $M$ is like a cone with the vertex in $(0,0,0)$ and the base in the circumference $\{x^2+y^2=1, z=1\}$, then $M$ is a compact set in $\mathbb{R}^3$. The function is lineal with gradient $(1,1,1)$ and $M$ is a convex set because the surface $z=x^2+y^2$ is a revolution paraboloid. The paraboloid  has normal $(2x,2y,-1)$ in any point $(x,y,z)$. The minimum is  in the point with $\displaystyle\frac{2x}{1}=\frac{2y}{1}=\frac{-1}{1}$, that is $\displaystyle (-\frac{1}{2},-\frac{1}{2},\frac{1}{2})$ . The maximum is obtained for $z=1$. The projection of the gradient over the plane $z=1$ is $(1,1,0)$. Then the maximum is obtained for $x=y$, and hence $x=y=\frac{\sqrt 2}{2}$. The maximum value is $f(\frac{\sqrt 2}{2},\frac{\sqrt 2}{2},1)=\sqrt 2 +1$.
A: Here’s an approach that minimizes the calculus and relies mostly on the geometry.
Let $R$ be the region in the $xz$-plane bounded by the curves $z=x^2$ and $z=1$; $M$ is the solid of revolution obtained by revolving $R$ about the $z$-axis. $R$ is a closed set in the plane, so $M$ is closed, and the minimum and maximum are attained.
The function $f(x,y,z)=x+y+z$ is constant on planes normal to the line $x=y=z$ and increases linearly along that line, so the minimum and maximum values of $f$ on $M$ will occur at points where such a plane just touches the surface of $M$; there’s no need to look for them in the interior of $M$. Since for fixed $x$ and $y$, $f(x,y,z)$ increases as $z$ increases, the maximum must occur on the top face of $M$. On that face it will occur where $x+y$ is maximal, which is evidently in the plane $x=y$, so it must occur at $\left(\frac1{\sqrt2},\frac1{\sqrt2},1\right)$. (If that’s not entirely clear, project to the $xy$-plane and observe that the maximum must occur where a line of the form $x+y=c$ is tangent to the circle $x^2+y^2=1$ in the first quadrant.) The value of $f$ at this point is $1+\sqrt2$.
Similarly, the minimum must also occur in the plane $x=y$, so we’re looking for $z\in[0,1]$ that minimizes $2x+z$, where $z=2x^2$. (Recall that the minimum must be on the surface of $M$, and it clearly isn’t on the top.) This is just the problem of minimizing $2x+2x^2=2x(1+x)$ for $0\le 2x^2\le 1$, or $|x|\le\frac1{\sqrt2}$, which doesn’t even require any calculus: it occurs at the vertex of the parabola, which is on the axis, $x=-\frac12$, halfway between the zeroes at $x=0$ and $x=-1$. Thus, the function $f$ takes its minimum on $M$ at $\left(-\frac12,-\frac12,\frac12\right)$. The value of $f$ at this point is $-\frac12$.
A: Your set $M$ is defined by inequalities of the form $g_i(x,y,z)\geq0$ with continuous $g_i:{\mathbb R}^3\to{\mathbb R}$, therefore it is closed; and all $(x,y,z)\in M$ satisfy $x^2+y^2\leq 1$ as well as $0\leq z\leq 1$, therefore $M$ is bounded. (In fact $M$ looks like an inverted sugar cone.) It follows that any continuous function $f:M\to{\mathbb R}$ assumes a global minimum as well as a global maximum on $M$. 
These two extremal points $(\xi,\eta,\zeta)$ are lying (a) in the interior of $M$, or (b) on the parabolic part $z=x^2+y^2$ of the boundary of $M$, or (c) in the interior of the top surface $z=1$ of $M$, or (d) on the bounding circle $\phi\mapsto(\cos \phi,\sin \phi,1)$ of the top surface.
In the case (a) one would necessarily have $\nabla f(\xi,\eta,\zeta)=0$. As $\nabla f(x,y,z)\equiv(1,1,1)$ there is no extremal point of this kind.
For (b) we look at the pullback $$f_1(x,y):=f(x,y,x^2+y^2)=x+y+x^2+y^2$$ of $f$ to the paraboloid and its two-dimensional gradient $\tilde f_1(x,y)=(1+2x,1+2y)$. This gradient vanishes at $\bigl(-{1\over2},-{1\over2}\bigl)$ which is a point in the interior of the unit circle. The corresponding point $p$ on the boundary $\partial M$ is $p=\bigl(-{1\over2},-{1\over2},{1\over2}\bigr)$, and one has $f(p)=-{1\over2}$.
For (c) we look at the pullback $$f_2(x,y):=f(x,y,1)=x+y+1$$
of $f$ to the top surface $z=1$. As the two-dimensional gradient $\tilde f_2(x,y)\equiv(1,1)$ is nonzero there are no extremal point of this kind.
For (d) we look at the pullback
$$f_3(\phi):=f(\cos\phi,\sin\phi, 1)=\cos\phi+\sin\phi+1=\sqrt{2}\sin\bigl(\phi+{\pi\over4}\bigr)+1$$
of $f$ to the boundary circle of the top surface. The function $f_3$ assumes its minimum $1-\sqrt{2}>-{1\over2}$ for $\phi=-{3\pi\over4}$ and its maximum $1+\sqrt{2}$ for $\phi={\pi\over4}$. The latter $\phi$-value corresponds to the point $q=\bigl({1\over\sqrt{2}},{1\over\sqrt{2}},1\bigr)\in\partial M$.
All in all we have found that $f$ takes on $M$ the minimal value $-{1\over2}$ at the point $p$ and the maximal value $1+\sqrt{2}$ at the point $q$.
