Answer
(a) See proof below.
(b) see graph.
(c) $\theta=5.13$ radians
Work Step by Step
(a) Follow the steps of the previous problem, $C=6\theta=2\pi r$ which gives $r=\frac{3\theta}{\pi}$.
Use the Pythagorean Theorem, $h^2+r^2=6^2$ which gives $h=\sqrt {36-r^2}=\sqrt {36-\frac{9\theta^2}{\pi^2}}$ and the volume of the cone is $V=\frac{1}{3}\pi r^2h=\frac{\pi}{3}\times\frac{9\theta^2}{\pi^2}\sqrt {36-\frac{9\theta^2}{\pi^2}}=\frac{9\theta^2}{\pi^2}\sqrt {4\pi^2-\theta^2}$ with a domain of $0\lt\theta\lt 2\pi$
(b) see graph.
(c) Based on the graph, a maximum of $V=87.06cm^3$ happens when $\theta=5.13$ radians
