To solve the problem, we'll begin by understanding how the conical cup is formed and how the radius $r$ and height $h$ of the cone depend on the angle $\theta$ of the sector cut out from the circular piece of paper.

Step 1: Determine the Radius $r(\theta)$ of the Cone's Base

Given:

• The original radius of the circular piece of paper is $R = 6$ cm.
• When a sector with central angle $\theta$ is cut out, the remaining arc of the circle becomes the circumference of the base of the cone.

The circumference $C$ of the base of the cone is given by: $C = 2\pi r$ This is equal to the length of the remaining arc: $C = (2\pi - \theta)R$ Thus, we have: $2\pi r = (2\pi - \theta) \cdot 6$ $r(\theta) = \frac{(2\pi - \theta) \cdot 6}{2\pi} = 6 - \frac{3\theta}{\pi}$

Step 2: Determine the Height $h(\theta)$ of the Cone

The height $h$ can be determined using the Pythagorean theorem. The slant height $l$ of the cone is the radius of the original circle: $l = R = 6 \text{ cm}$ The relationship between the height $h$, radius $r$, and slant height $l$ is: $h = \sqrt{l^2 - r^2}$ Substituting $r(\theta)$ from Step 1: $h(\theta) = \sqrt{6^2 - \left(6 - \frac{3\theta}{\pi}\right)^2}$ $h(\theta) = \sqrt{36 - \left(36 - 36\cdot\frac{\theta}{\pi} + \frac{9\theta^2}{\pi^2}\right)}$ $h(\theta) = \sqrt{36\left(\frac{\theta}{\pi} - \frac{\theta^2}{\pi^2}\right)}$

Step 3: Determine the Volume $V(\theta)$ of the Cone

The volume $V$ of a cone is given by: $V = \frac{1}{3}\pi r^2 h$ Substituting $r(\theta)$ and $h(\theta)$: [ V(\theta) = \frac{1}{3}\pi \left(6 - \frac{3\theta}{\pi}\right)^2 \cdot \sqrt{36\left(\frac{\theta}{\pi} - \frac{\theta^