3D probability?

Geometry Level 3

In the three dimensional plane, two three- dimensional regions R 1 : x 2 + y 2 z 2 R_1 \ : \ x^2 + y^2 \leq z^2 and R 2 : x 2 + y 2 z R_2 \ : \ x^2 + y^2 \leq z are plotted z [ 0 , 5 ] \forall z \in \left [0 , 5 \right] . Now, a point is randomly chosen which lies inside R 1 R_1 . What is the probability that this point also lies inside R 2 R_2 ?

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


The answer is 162.

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1 solution

Ashish Menon
Oct 24, 2017

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

The plane z = 1 z=1 cuts the cone at circle x 2 + y 2 = 1 x^2 + y^2 = 1 , a circle of radius 1 1 . The volume of cone z ( 0 , 1 ) = 1 3 π × 1 2 × 1 = π 3 \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 R_2 \ : \ x^2 + y^2 = z which is a paraboloid occupies a larger region than R 1 z ( 0 , 1 ) R_1 \ \forall z \in \left(0,1\right) , and lesser region than R 1 z ( 1 , 5 ) R_1 \ \forall z \in \left(1,5\right) . (They intersect at x 2 + y 2 = 1 x^2 + y^2 = 1 .

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

The plane z = 5 z=5 cuts the paraboloid at x 2 + y 2 = 5 x^2 + y^2 = 5 , a circle of radius 5 \sqrt {5} . The volume for the region R 2 z ( 0 , 5 ) = 1 2 π × 5 2 × 5 = 25 π 2 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 z=1 cuts the paraboloid at x 2 + y 2 = 1 x^2 + y^2 = 1 , a circle of radius 1 1 . The volume for the region R 2 z ( 0 , 1 ) = 1 2 π × 1 2 × 1 = π 2 R_2 \forall z \in \left (0,1\right) \ = \ \dfrac {1}{2} \pi × 1^2 × 1 \ = \dfrac{\pi}{2} .
So, the volume of paraboloid z ( 1 , 5 ) \forall z \in \left (1,5\right) z ( 1 , 5 ) = 25 π 2 π 2 = 12 π \forall z \in \left(1,5\right) \ = \ \dfrac {25\pi}{2} \ - \ \dfrac {\pi}{2} \ = 12\pi .

Therefore, the volume of R 2 R_2 inside R 1 R_1 = 12 π + π 3 = 37 π 3 12\pi + \dfrac {\pi}{3} \ = \ \dfrac {37\pi}{3} .

Hence, probability of the point chosen to be in R 2 R_2 also = 37 π 3 125 π 3 = 37 125 \dfrac{\frac {37\pi}{3}}{\frac {125\pi}{3}} \ = \ \dfrac {37}{125} .

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

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