Cylindrical Coordinates — Question 10

Question 10 – Solution

Strategy. Vertical consistency requires lower height not exceed upper height.

Step 1: Compare heights $$r^2\le 6-r\quad\Longleftrightarrow\quad r^2+r-6\le 0\quad\Longleftrightarrow\quad(r+3)(r-2)\le 0.$$ With $r\ge 0$, this requires $0\le r\le 2$.

Step 2: Combine radial restrictions The stated planar bound is $r\le 2+\cos\theta$, which ranges from $1$ to $3$. Both conditions must hold, so $$\boxed{0\le r\le\min\{2,2+\cos\theta\}}.$$ Equivalently, the upper bound is $2$ when $\cos\theta\ge 0$ and $2+\cos\theta$ when $\cos\theta<0$.

Step 3: Final description $$\boxed{0\le\theta<2\pi,\quad 0\le r\le\min\{2,2+\cos\theta\},\quad r^2\le z\le 6-r}.$$ The height surfaces meet at $r=2$. For angles on the right half-plane the height condition truncates the planar radius; on the left half-plane the limacon-type boundary is already smaller than $2$.