91Ó°ÊÓ

We are given the range of z-values as: $$ 0 \leq z \leq 8-2 r. $$ By rearranging this inequality, we find the range for r in terms of z: $$ 0 \leq r \leq \frac{8 - z}{2}. $$ This inequality represents a half-circular cone with its vertex at the origin and the base in the positive z direction since r is always non-negative and is restricted by a varying maximum value as we move along the z-axis.

We know that the vertex of our cone is at the origin (r=0, z=0), and its base lies in the positive z-direction. Let's start by sketching the base of the cone, for which z = 8: $$ r = 8 - z = 8 - 8 = 0, $$ Giving us r = 4. Thus, the base of the cone has radius 4 at z=8. Next, we will sketch the side of the cone. Let's observe what happens at \(\theta = 0\) for some arbitrary z-value in the range 0 < z < 8. For this purpose, let's take, for example, z = 4: $$ r = \frac{8 - 4}{2} = 2. $$ Thus, at z = 4, we have a circle of radius 2 along the positive x-axis (\(\theta = 0\)) when viewed from above. In addition, as \(\theta\) changes, we observe the circular symmetry of the cone since there is no dependence on \(\theta\). As a result, every point at z = 4 will have a circle of radius 2. Now, let's sketch the cone in cylindrical coordinates. Plot the circles obtained along the z-axis, with the vertex at the origin (r=0, z=0) and the base at z = 8, having a radius of 4. Connect these circles with a conical surface, considering the circular symmetry along the \(\theta\) direction.