91Ó°ÊÓ

The given conditions for \((r, \theta, z)\) are: $$0 \leq z \leq 8 - 2r$$ Since there are no restrictions on the \(\theta\) coordinate, it can take any value between \(0\) and \(2\pi\). This means that the region will be centered around the \(z\)-axis and have some kind of axial symmetry. The restriction on the \(z\) coordinate depends on the value of \(r\), as \(z\) must be between \(0\) and \(8 - 2r\). This gives us information on the shape and size of the region in the \(rz\)-plane.

To determine the shape of the region in the \(rz\)-plane, we can analyze the condition given on the \(z\)-coordinate. When \(r = 0\), the condition becomes \(0 \leq z \leq 8\), which represents a line segment along the \(z\)-axis, stretching from \(z = 0\) to \(z=8\). As \(r\) increases, the upper bound of \(z\) decreases (since \(z \leq 8 - 2r\)), meaning that the shape of the region in the \(rz\)-plane will be a triangle. The region will have a vertex at \((r, z) = (0, 8)\) and a base along the \(r\)-axis, from \(r = 0\) to \(r = 4\). This can be determined by setting \(z\) to \(0\) in the constraint: $$0 = 8 - 2r \implies r = 4$$

To sketch this region in cylindrical coordinates, we can begin by drawing the \(rz\)-plane triangle determined in Step 2. Then, we can use the axial symmetry from the unrestricted \(\theta\) value to sketch the full three-dimensional region. The region will be a cone centered around the \(z\)-axis, with a vertex at the origin (because the base of the triangular region in the \(rz\)-plane is along the \(r\)-axis). The height of the cone will be \(8\) (since the vertex of the triangle in the \(rz\)-plane is at \((0, 8)\)). Additionally, the base of the cone will have a radius of \(4\) (since the base of the triangle in the \(rz\)-plane stretches from \(r = 0\) to \(r = 4\)). To finalize the sketch, draw the cone in three-dimensional space centered around the \(z\)-axis, labeling the appropriate dimensions noted above.