analytic geometry
I do understand that when a plane cuts a cone it does in two lines, but where did the other two lines come from? The solution(see pic) says that "the two cones intersect in four lines", now how is that possible? It should have been two lines as a plane is cutting a cone(only one and not two).
I'll assume the following:
$$\text{Height of cone } = h \\ \text{Radius of the cone at its base } = R \\ \text{Distance from the cone axis to the slicing plane } = d,\quad 0\leq d\leq R.$$
Each horizontal slice of the solid is a circular segment. These segments vary in size due to the change of radius as we move vertically up or down the cone. Let's suppose at any such height $z$ the circle has radius $r$ and the segment subtends an angle $\theta$ at the circle centre (see the picture at the link below). Then, simple trigonometry/geometry shows that
$$\cos\frac{\theta}{2} = \frac{d}{r}, \\ \text{and the thickness of these slices will be: } dz = \dfrac{h}{R}dr.$$
(Actually, $r$ decreases as $z$ increases but this is handled by flipping the limits of integration below on $r$: $d$ to $R$ instead of $R$ to $d$.)
The area for the segment is:
\begin{align} A_s &= \dfrac{r^2}{2}\left(\theta - \sin\theta\right) \\ &= \dfrac{r^2}{2}\left(2\cos^{-1}\dfrac{d}{r} - \sin\left(2\cos^{-1}\dfrac{d}{r}\right)\right) \\ &= r^2\cos^{-1}\dfrac{d}{r} - d\sqrt{r^2-d^2} \\ & \qquad\text{using $\;\sin(2\alpha)=2\sin\alpha\cos\alpha\;$ and $\;\cos(\sin^{-1}(a/b))=\sqrt{b^2-a^2}/b$}. \end{align}
Therefore the volume of a thin horizontal slice of thickness $dz\;$ is
\begin{align} dV &= A_sdz \\ &= A_s\dfrac{h}{R}dr \\ &= \dfrac{h}{R}\left(r^2\cos^{-1}\dfrac{d}{r} - d\sqrt{r^2-d^2}\right)dr. \end{align}
Therefore the required volume is:
\begin{align} V &= \int_{r=d}^{R} \dfrac{h}{R}\left(r^2\cos^{-1}\dfrac{d}{r} - d\sqrt{r^2-d^2}\right)\;dr \\ &= \dfrac{h}{3R}\left[-2dr\sqrt{r^2-d^2} + d^3\ln\left(\sqrt{r^2-d^2}+r\right) + r^3\cos^{-1}\dfrac{d}{r} \right]_{r=d}^{R} \\ & \qquad\text{using Wolfram Alpha} \\ &= \dfrac{h}{3}\left[-2d\sqrt{R^2-d^2} + \dfrac{d^3}{R}\ln\left(\sqrt{R^2-d^2}+R\right) + R^2\cos^{-1}\dfrac{d}{R} - \dfrac{d^3}{R}\ln{d} \right]. \\ \end{align}
As a sanity check, setting $d=0$ gives $V=\pi R^2h/6,$ and setting $d=R$ gives $V=0,$ as expected.
You can see here for interactive graph
https://www.geogebra.org/3d/ag7rjf5c
Best Answer
Sometimes a picture is worth a thousand words: