A right circular cone $\mathbb{C}$ has altitude $40$ and a circular base
of radius $30$ inches. A sphere $\mathbb{S}$ is inscribed in $\mathbb{C}$. Compute the volume of the region inside $\mathbb{C}$ which is above $\mathbb{S}$.
The answer is $\boxed{300\pi}$.
First, it is easy to find the length of the slant is $50$. Then, we claim that the radius of the sphere is $15$. This can be computed by taking a vertical cross-section that passes the apex. The triangle's area is $1200$ and its semi-perimeter is $80$. Hence, the inradius will be $1200\div 80 = 15$. Computing inradius using area and semi-perimeter is discussed in the book Geometry Techniques .
Then, using similar triangle, it can be determined that the vertical distance from the apex to the tangent point on the side is $16$.
Now, let's put this cross-section in the horizontal direction and set its apex as the origin and its line of symmetry as the $x$-axis. Then the to-be-determined volume equals the difference between a cone which is constructed by rotating the highlighted triangle and the dome which is constructed by rotating the arc.
The volume of the cone equals $\frac{1}{3}\cdot\pi\cdot 12^2 \cdot 16 = 768\pi$.
The equation of the circle is
$$(x-25)^2 + y^2 = 15^2 \implies y^2 = 15^2 - (x-25)^2$$
Therefore the volume of the dome is given by
$$\pi\int_{10}^{16}(15^2 - (x-25)^2dx=\pi\left.\left(15^2x -\frac{1}{3}(x-25)^3\right)\right|_{10}^{16}=468\pi$$
It follows that the final result is $768\pi-468\pi=\boxed{300\pi}$.