Cylindrical Coordinates — Question 9

Question 9 – Solution

Strategy. Convert the sphere to $r^2+z^2=45$ and combine it with the cone relation.

Step 1: Solve Substitute $r=2z$: $$(2z)^2+z^2=45\quad\Longrightarrow\quad 5z^2=45.$$ Since $z\ge 0$, $\boxed{z=3}$ and $\boxed{r=6}$.

Step 2: Angular freedom No condition restricts $\theta$, so $$\boxed{(r,\theta,z)=(6,\theta,3),\quad 0\le\theta<2\pi}.$$ In Cartesian form the points are $(6\cos\theta,6\sin\theta,3)$.

Step 3: Locus They form the horizontal circle $$\boxed{x^2+y^2=36,\qquad z=3}.$$ Verification gives $36+9=45$ and $r=6=2(3)$.