We will have the following:
First, we can see that the volume of the sphere and the regular cone are given by:
[tex]V_s=\frac{4}{3}\pi r^3[/tex]
And:
[tex]V_c=\frac{1}{3}\pi r^2\ast h[/tex]
Now, since the volume of the cone is inscribed by a maximum stablished by the sphere, we know that the maximum height for the cone will be equal to the radius of the sphere, so, we re-write the volume of the cone:
[tex]V_c=\frac{1}{3}\pi r^2\ast r\Rightarrow V_c=\frac{1}{3}\pi r^3[/tex]
i. Now, we determine the ratio of both volumes as follows:
[Cone to sphere]
[tex]\begin{gathered} r=\frac{(1/3)\pi r^3}{(4/3)\pi r^3}\Rightarrow r=\frac{(1/3)}{(4/3)} \\ \\ \Rightarrow r=\frac{1\ast3}{4\ast3}\Rightarrow r=\frac{1}{4} \end{gathered}[/tex]
So, the ratio of the volumes of the cone to the sphere will be of 1:4.
ii. And the expression in terms of r for the volume inside the sphere but outside the cone will be:
[tex]V=\frac{4}{3}\pi r^3-\frac{1}{3}\pi r^3\Rightarrow V=\pi r^3[/tex]
So, the expression is:
[tex]V=\pi r^3[/tex]
