Tokyo university entrance exam - mathematics No.5 - (2) 
Circle $C$ which radious is 1 and center is origin on $xy$-plane, We could imagine a cone $S$ that bottom is $C$ and top is $(0,0,2)$. Point $A$ is $(1,0,2)$ and point $P$ is on $C$, $T$ is a cone that $\overline{AP}$'s set
Calculate the volume of $R$ which shares $S$ and $T$

Figure of setup
I think both $S$ and $T$'s section with $z=2-t, (0<t<2)$ is circle and they meet when $\frac{2}{3}<t<2$.
But I can't calculate area of circular sector...
I hear that answer is $\frac{2\pi}{3}+\frac{2}{3}$

My first try:
Section with $z=2-t$ and $S$ is $D_2:x^2+y^2=\frac{t^2}{4} $,
$T$ is $D_1:(x-\frac{2-t}{2})^2+y^2=\frac{t^2}{4} $
Section with $D_1, D_2$ is $E_1,E_2$
Than $\angle E_1OE_2=2\theta$
$\cos\theta=\frac{(2-t)/4}{t/2}$
$\theta=\cos^{-1}(\frac{(2-t)/4}{t/2})$
So area of circular sector is $\frac{t^2}{4}\cos^{-1}(\frac{2-t}{2t})$ and triangle $E_1OE_2$ is $\frac{(2 - t) \sqrt{3 t^2 + 4 t - 4}}{16}$...
I can't move forward after here
 A: I am going to assume that $$S = \{ (x,y,z) \in \mathbb R^3 : x^2 + y^2 \le (z/2 - 1)^2 \cap 0 \le z \le 2 \}$$ and $$T = \{ (x,y,z) \in \mathbb R^3 : (x-z/2)^2 + y^2 \le (z/2-1)^2 \cap 0 \le z \le 2 \}$$ and you want to find the common volume of intersection of $S$ and $T$.  It is immediately obvious that for a given $z_0 \in [0,2]$, the two cones intersect the plane $z = z_0$ in two disks:  $$S(z_0) : x^2 + y^2 \le (z_0/2 - 1)^2$$ and $$T(z_0) : (x - z_0/2)^2 + y^2 \le (z_0/2 - 1)^2.$$  Notably, the radii of the two disks are equal.  We can readily solve for the point of intersection of the boundary circles:  $$x^2 = (x - z_0/2)^2 = x^2 - z_0 x + \frac{z_0^2}{4},$$ hence $x = z_0/4$, and $$y = \pm\sqrt{(z_0/2-1)^2 - (z_0/4)^2} = \pm \sqrt{\frac{3z_0^2}{16} - z_0 + 1}.$$  On $z \in [0,2]$, we find that $y \in \mathbb R$ if and only if $0 \le z_0 \le 4/3$.  The area of the intersection of the disks $S(z_0) \cap T(z_0)$ is given by $$A(z_0) = 2(z_0/2 - 1)^2 \tan^{-1} \frac{\sqrt{\frac{3z_0^2}{16} - z_0 + 1}}{z_0/4} - 2 \cdot \frac{z_0}{4} \sqrt{\frac{3z_0^2}{16} - z_0 + 1}.$$  The desired volume is then given by $$V = \int_{z=0}^{4/3} A(z) \, dz = \frac{2\pi}{3} - \frac{4}{27} \left(6 + \sqrt{3} \log (2 - \sqrt{3}) \right).$$
