A slipped disc

Calculus Level 4

In the x y xy -plane there is a disc of radius 1 1 centred at ( 1 , 1 ) (1,1) .

Suppose the distance from a random point on the disc to the origin is D D . What is the expected value of D 2 D^{2} ?

Inspiration


The answer is 2.5.

Brian Charlesworth by Brian Charlesworth , 55, Canada

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2 solutions

Steven Chase
Nov 16, 2017

Coordinates within the disk:

x = 1 + r c o s θ y = 1 + r s i n θ x = 1 + r \, cos\theta \\ y = 1 + r \, sin\theta

Square of distance from origin:

D 2 = x 2 + y 2 = 2 + 2 r c o s θ + 2 r s i n θ + r 2 D^2 = x^2 + y ^2 = 2 + 2r \, cos\theta + 2r \, sin\theta + r^2

Probability as a function of radius and angle:

P = d A A = r d r d θ π P = \frac{dA}{A} = \frac{r \, dr \, d\theta}{\pi}

Infinitesimal contribution to expectation:

d E = D 2 P = ( 2 + 2 r c o s θ + 2 r s i n θ + r 2 ) r d r d θ π = 1 π ( 2 r + 2 r 2 c o s θ + 2 r 2 s i n θ + r 3 ) d r d θ dE = D^2 P = (2 + 2r \, cos\theta + 2r \, sin\theta + r^2) \frac{r \, dr \, d\theta}{\pi} \\ = \frac{1}{\pi} (2r + 2r^2 \, cos\theta + 2r^2 \, sin\theta + r^3) \, dr \, d\theta

Total expected value:

E = 1 π ∫ 0 2 π ∫ 0 1 ( 2 r + 2 r 2 c o s θ + 2 r 2 s i n θ + r 3 ) d r d θ = 1 π ∫ 0 2 π ( 5 4 + 2 3 c o s θ + 2 3 s i n θ ) d θ = 1 π ( 10 4 π ) = 2.5 E = \frac{1}{\pi} \int_0^{2 \pi} \int_0^1 (2r + 2r^2 \, cos\theta + 2r^2 \, sin\theta + r^3) \, dr \, d\theta \\ = \frac{1}{\pi} \int_0^{2 \pi} (\frac{5}{4} + \frac{2}{3} cos\theta + \frac{2}{3} sin\theta) d\theta \\ = \frac{1}{\pi} (\frac{10}{4} \pi) = \boxed{2.5}

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Nice solution! Thanks for posting it. :)

I guess the logical extension is to look at the similar problem for a sphere of radius 1 1 centred at ( 1 , 1 , 1 ) (1,1,1) .

Brian Charlesworth - 3 years, 6 months ago

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I was thinking the same.

Steven Chase - 3 years, 6 months ago

That one is up now

Steven Chase - 3 years, 6 months ago

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Great! I'm glad I got that one on the first try; there were quite a few moving parts to slip up on. :)

Brian Charlesworth - 3 years, 6 months ago

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@Brian Charlesworth – Indeed. I did it analytically in spherical coordinates, and then numerically in Cartesian coordinates just to be sure.

Steven Chase - 3 years, 6 months ago

I solved it without polar coordinates (sticking to the theme), but I initially miscalculated, so my answer was wrong. Here was my approach:

If you consider a line segment in the disc, parallel to the y y -axis and at some distance x x from the rightmost point on the disc, we can calculate its length to be

2 1 − ( 1 − x ) 2 2\sqrt{1-(1-x)^2}

Now integrating this function from 0 0 to 2 2 will give the area of the circle, which is π \pi . So we can calculate a probability density function for a given line defined by some x x as

f ( x ) = 2 π 1 − ( 1 − x ) 2 f(x)=\dfrac{2}{\pi }\sqrt{1-(1-x)^2}

Now, given some point ( ( 2 − x ) , a ) ((2-x),a) on this line, we have

D 2 = ( 2 − x ) 2 + a 2 D^2=(2-x)^2+a^2

and thus the average value of the square of the distance given only x x becomes

1 2 1 − ( 1 − x ) 2 ∫ 1 − 1 − ( 1 − x ) 2 1 + 1 − ( 1 − x ) 2 ( 2 − x ) 2 + a 2 d a = 2 x 2 3 − 10 x 3 + 5 \dfrac{1}{2\sqrt{1-(1-x)^2}}\int_{1-\sqrt{1-(1-x)^2}}^{1+\sqrt{1-(1-x)^2}}(2-x)^2+a^2\; da=\dfrac{2x^2}{3}-\dfrac{10x}{3}+5

Then integrating their product gives

E [ D 2 ] = ∫ 0 2 f ( x ) × ( 2 x 2 3 − 10 x 3 + 5 ) = 2.5 \mathbb{E} \left[D^2\right]=\int_0^2f(x)\times \left(\dfrac{2x^2}{3}-\dfrac{10x}{3}+5\right)=\boxed{2.5}

Miles Koumouris - 3 years, 6 months ago
Jon Haussmann
Nov 22, 2017

More generally, consider a circle of radius R R centered at O = ( 0 , 0 ) O = (0,0) , and let P = ( a , b ) P = (a,b) .

Parametrically, we can represent a point inside the disc by ( r cos ⁡ θ , r sin ⁡ θ ) (r \cos \theta, r \sin \theta) , where 0 ≤ r ≤ R 0 \le r \le R and 0 ≤ θ ≤ 2 π 0 \le \theta \le 2 \pi . If D D is the distance from a point inside the disc to P P , then the expected value of D 2 D^2 is 1 π R 2 ∫ 0 R ∫ 0 2 π [ ( r cos ⁡ θ − a ) 2 + ( r sin ⁡ θ − b ) 2 ] r d r d θ = 1 π R 2 ∫ 0 R ∫ 0 2 π ( r 2 cos ⁡ 2 θ − 2 a r cos ⁡ θ + a 2 + r 2 sin ⁡ 2 θ − 2 b r sin ⁡ θ + b 2 ) r d r d θ = 1 π R 2 ∫ 0 R ∫ 0 2 π ( r 2 − 2 a r cos ⁡ θ − 2 b r sin ⁡ θ + a 2 + b 2 ) r d r d θ = 1 π R 2 ∫ 0 R ∫ 0 2 π r 3 − 2 a r 2 cos ⁡ θ − 2 b r 2 sin ⁡ θ + ( a 2 + b 2 ) r d r d θ = 1 π R 2 ∫ 0 R 2 π r 3 + 2 π ( a 2 + b 2 ) r d r = 1 π R 2 [ π R 4 2 + ( a 2 + b 2 ) π R 2 ] = R 2 2 + a 2 + b 2 = R 2 2 + O P 2 . \begin{aligned} &\frac{1}{\pi R^2} \int_0^R \int_0^{2 \pi} [(r \cos \theta - a)^2 + (r \sin \theta - b)^2] r \ dr \ d \theta \\ &= \frac{1}{\pi R^2} \int_0^R \int_0^{2 \pi} (r^2 \cos^2 \theta - 2ar \cos \theta + a^2 + r^2 \sin^2 \theta - 2br \sin \theta + b^2) r \ dr \ d \theta \\ &= \frac{1}{\pi R^2} \int_0^R \int_0^{2 \pi} (r^2 - 2ar \cos \theta - 2br \sin \theta + a^2 + b^2) r \ dr \ d \theta \\ &= \frac{1}{\pi R^2} \int_0^R \int_0^{2 \pi} r^3 - 2ar^2 \cos \theta - 2br^2 \sin \theta + (a^2 + b^2) r \ dr \ d \theta \\ &= \frac{1}{\pi R^2} \int_0^R 2 \pi r^3 + 2 \pi (a^2 + b^2) r \ dr \\ &= \frac{1}{\pi R^2} \left[ \frac{\pi R^4}{2} + (a^2 + b^2) \pi R^2 \right] \\ &= \frac{R^2}{2} + a^2 + b^2 = \frac{R^2}{2} + OP^2. \end{aligned}

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