In the three dimensional plane, two three- dimensional regions $R_1 \ : \ x^2 + y^2 \leq z^2$ and $R_2 \ : \ x^2 + y^2 \leq z$ are plotted $\forall z \in \left [0 , 5 \right]$ . Now, a point is randomly chosen which lies inside $R_1$ . What is the probability that this point also lies inside $R_2$ ?

The answer is of the form $\dfrac {a}{b}$ where $a$ and $b$ are relatively co-prime positive integers. Submit the value of $a + b$ .

The answer is 162.

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The region $R_1 \ : \ x^2 + y^2 = z^2$ is a right circular cone $\forall z \in \left[ 0,5 \right]$ . Now, the plane $z = 5$ cuts it at $x^2 + y^2 = 25$ , a circle of radius $5$ units. So, the volume of the full cone taken into consideration = $\dfrac {1}{3} \pi r^2 × h \ = \ \dfrac {1}{3} × \pi × 5^2 × 5 \ = \ \dfrac {125 \pi}{3}$ .

The plane $z=1$ cuts the cone at circle $x^2 + y^2 = 1$ , a circle of radius $1$ . The volume of cone $\forall z \in \left(0,1\right) \ = \ \dfrac {1}{3} \pi × 1^2 × 1 \ = \ \dfrac {\pi}{3}$ .

Now, the region $R_2 \ : \ x^2 + y^2 = z$ which is a paraboloid occupies a larger region than $R_1 \ \forall z \in \left(0,1\right)$ , and lesser region than $R_1 \ \forall z \in \left(1,5\right)$ . (They intersect at $x^2 + y^2 = 1$ .

$\color{#D61F06}{\text {The volume for a paraboloid is given by}} \ \color{#3D99F6}{\dfrac {1}{2} \pi × r^2 × h}$ .

The plane $z=5$ cuts the paraboloid at $x^2 + y^2 = 5$ , a circle of radius $\sqrt {5}$ . The volume for the region $R_2 \forall z \in \left (0,5\right) \ = \ \dfrac {1}{2} \pi × {\sqrt{5}}^2 × 5 \ = \dfrac{25\pi}{2}$ .

The plane $z=1$ cuts the paraboloid at $x^2 + y^2 = 1$ , a circle of radius $1$ . The volume for the region $R_2 \forall z \in \left (0,1\right) \ = \ \dfrac {1}{2} \pi × 1^2 × 1 \ = \dfrac{\pi}{2}$ .

So, the volume of paraboloid $\forall z \in \left (1,5\right)$ $\forall z \in \left(1,5\right) \ = \ \dfrac {25\pi}{2} \ - \ \dfrac {\pi}{2} \ = 12\pi$ .

Therefore, the volume of $R_2$ inside $R_1$ = $12\pi + \dfrac {\pi}{3} \ = \ \dfrac {37\pi}{3}$ .

Hence, probability of the point chosen to be in $R_2$ also = $\dfrac{\frac {37\pi}{3}}{\frac {125\pi}{3}} \ = \ \dfrac {37}{125}$ .

$\therefore \ a \ + \ b \ = \ 37 \ + \ 125 \ = \ \color{#69047E}{\boxed{162}}$