Question

In Exercises 61 and \(62,\) sketch the solid that has the given description in cylindrical coordinates. $$ 0 \leq \theta \leq 2 \pi, 2 \leq r \leq 4, z^{2} \leq-r^{2}+6 r-8 $$

Short answer

After checking the boundaries for \(r\) and \(θ\), and examining the upper limit for \(z\), it is observed that the specified figure could not exist as no real solutions for \(z\) exist.

Step-by-step solution

Bounding the Radius

Given that 2 \(\leq\) \(r\) \(\leq\) 4, the figure will lie between two concentric cylinders of radii 2 and 4 respectively. This highlights that the radial distances are at least 2 and at most 4 from the z-axis.

Determining the Azimuthal Angle

The next thing is to examine the azimuthal angle given as 0 \(\leq\) \(θ\) \(\leq\) \(2 π\). This highlights that the figure will cover the entire \(360^°\) around the z-axis.

Bounding the Height

According to cylindrical coordinates, \(z^{2} \leq -r^{2} + 6r - 8\) is the equation for the upper limit of the height \(z\). As \(z^{2}\) cannot be negative, the negative discriminant of the quadratic equation suggests that no real solutions for \(z\) exist if \(2 \leq r \leq 4\). Therefore, no part of the paraboloid lies within this range of \(r\) and no solid figure can be drawn.

Dealing with Negative Discriminant

A negative discriminant for the cone equation where 2 \(\leq\) \(r\) \(\leq\) 4 declares that the specified figure cannot exist. However, this does not make the equation meaningless. The discriminant being negative just means that the figure does not exist for the given range of \(r\) which only encapsulates for \(r\) \(\leq\) 2, and \(r\) \(\geq\) 4.