Here is a problem I created myself, but I’m not sure what the correct solution is.
A fly is tied to the end of a string with length $12$. The other end of the string is connected to the center of the base of a right circular cone with a radius of $6$ and a slant height of $12$. If both the fly and string cannot go inside of the cone, what is the volume of the space that the fly can visit?
Can this problem be solved using elementary techniques or will it require calculus? Thanks!
The diagram below shows the cross-section of our 3D-Sketch which we added to the OP's problem statement. We denote the length of the string, to which the fly is tied, as $2r$. In order to avoid using calculus, we decided to consult tables where areas, volumes, and position of the centroids of various solids are tabulated.
@random and @JaapScherphuis have briefly described in their comments what one has to do to solve this problem. So we know that the sought volume $V_\text{Fly}$ consists of two parts, i.e., a hemisphere ($V_1$) and a part of a horn torus ($V_2$). The solution given below was obtained by applying Pappus's Centroid Theorem to calculate the volume of the latter.
The volume of a hemisphere of radius $2r$ can be written as, $$ V_1= \dfrac{16}{3}\pi r^3. \tag{1}$$
To find the volume of the part of a horn torus, we need to determine the area of a circular sector of radius $r$. Since this sector subtends an angle of $120^o$ at its center $A$, its area $A_s$ is given by, $$A_s = \dfrac{\pi}{3}r^2. \tag{2}$$
Its centroid $G$ lies on its axis of symmetry $AH$ at a distance $AG=\dfrac{\sqrt{3}}{\pi}r\space$ from its center $A$. Therefore, the distance between $G$ and the axis of rotation, i.e. $z\text{-axis}$, is, $$OK=r+AG\cos\left(60\right)=r\left(1+ \dfrac{\sqrt{3}}{2\pi}\right). \tag{3}$$
According to Pappus's second Centroid Theorem, we can express $V_2$ using (2) and (3) as shown below. $$V_2= A_s ×2\pi×OK=\dfrac{\pi}{3}\left(2\pi+\sqrt{3}\right)r^3 \tag{4}$$
$$\therefore\enspace V_\text{Fly} = V_1 +V_2 =\dfrac{\pi}{3}\left(2\pi+16+\sqrt{3}\right)r^3.$$
You may not like our solution, but you can at least use it to check the answer if someone posts the solution you are looking for.