Cylindrical Coordinates — Question 6

Question 6 – Solution

Strategy. Convert the disk inequality and enforce the convention $r\ge 0$.

Step 1: Convert The inequality becomes $$r^2\le 4r\cos\theta.$$ With $r\ge 0$, this is $0\le r\le 4\cos\theta$. The upper bound must be nonnegative, so $\cos\theta\ge 0$.

Step 2: Bounds One nonoverlapping description is $$\boxed{-\frac\pi 2\le\theta\le\frac\pi 2,\quad 0\le r\le 4\cos\theta,\quad 1\le z\le 5}.$$

Step 3: Geometry Completing the square gives $(x-2)^2+y^2\le 4$: a radius-$2$ disk centered at $(2,0)$, extended from height $1$ to $5$. Negative $r$ is unnecessary because directions with $\cos\theta<0$ would duplicate points already represented using nonnegative $r$ and opposite angles.