How to draw a region where z coordinates is bounded separately in terms of both y and x in cylindrical coordinates? Draw the region R with the cylindrical coordinates  

$ 0 \leq \theta \leq \pi/2$ ,  $ 0\leq r \leq 1, 0 \leq z \leq
 \sqrt{1-r^2*cos^2(\theta)}, 0 \leq z \leq \sqrt{ 1-r^2*sin^2(\theta)
 }$.
  I can't visualize how to draw R.

I know that there definitely exists a region where z is both a function of y and a function of x but I don't understand how it can be that z is both a function of y and x. How do we even draw this region given that z depends on both y and x but we can't force z to be expressed in terms of both y and x in one equation or inequality?.

First of all, these are inequalities, not equalities. While they do translates to 
$0 \le z \le \sqrt{1-x^2}$ and $0 \le z \le \sqrt{1-y^2}$, the first does not make $z$ a function of $x$ and the second does not make $z$ a function of $y$. For that you would need $z = \sqrt{1-x^2}$, or $z = \sqrt{1-y^2}$.
Instead, the first inequality indicates that the region $R$ lies inside the cylinder $x^2 + z^2 = 1$, and also lies inside the cylinder $y^2 + z^2 = 1$. I.e. $R$ is inside the intersection of these two solid cylinders. What portion of this intersection is $R$ is obtained from the other two inequalities $z \ge 0$ and $0 \le \theta \le \pi / 2$.
