On describing a solid using cylindrical coordinates Can someone help me with this question?
Consider the solid A defined by  

Describe the solid A in cylindrical coordinates.
The solution is
$$(x,y,z)=(r\cosθ ,r\sinθ ,z): r \in [0,3], \space θ \in [π/3,4π/3], \space z \in [-1,2-r] \qquad$$
My questions are:
Why is $z \in [-1,2-r]$ ?  I initially thought that $z \in [-1,2]$. 
There's another way to calcute $r$ without doing  $(-1-2)^2 = r^2 \iff r(\max) = 3$
so $r \in [0,3]$?
 A: In cylindrical coordinates, $x^2 + y^2 = r^2$. Thus, the first inequality becomes
$$\left(z - 2\right)^2 \ge r^2 \tag{1}\label{eq1}$$
Moving $r^2$ to the left & factoring the difference of squares gives
$$\left(z - 2 + r\right)\left(z - 2 - r\right) \ge 0 \tag{2}\label{eq2}$$
This requires that one of the factors on the LHS is $0$, both are positive or both are negative. First, for being $0$, we get $z = 2 - r$ or $z = 2 + r$. Since $z \le 2$ & $r \gt 0$, the second one only holds when $r = 0$. For both being positive, $z \gt 2 - r$ and $z \gt 2 + r$. Since $r \ge 0$ and $z \le 2$, the second one never holds. For both being negative, this requires $z \lt 2 - r$ and $z \lt 2 + r$. Note the first condition implies the second. Overall, this gives that $z \le 2 - r$. Combine this with the range restriction of $-1 \le z \le 2$ gives $z \in [-1,2-r]$.
As for the range of $r$, note that from \eqref{eq1} that the maximum value of $r$ is given by the maximum value of $\mid z - 2 \mid$. Within the range of $-1 \le z \le 2$, this occurs at $z = -1$, as you stated, giving that the maximum value of $r$ is $3$. Also, since there is no restriction on how small $r$ may be, the lower bound would be $0$. Thus, overall $r \in [0,3]$. As for there being any other way of calculating this, note that in one way or another you will need to use the inequality to determine the maximum value of $r$, so as far as I know, there's no significantly different method of doing that.
