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Question

The likelihood that a person with a contagious disease will infect others in a social situation may be assumed to be a function f(s)f(s) of the distance ss between individuals. Suppose contagious individuals are uniformly distributed throughout a rectangular region RR in the xyx y plane. Then the likelihood of infection for someone at the origin (0,0)(0,0) is proportional to the exposure index EE, given by the double integral

E=Rf(s)dAE=\iint_R f(s) d A

where s=x2+y2s=\sqrt{x^2+y^2} is the distance between (0,0)(0,0) and (x,y)(x, y). Find EE for the case where

f(s)=1s29f(s)=1-\frac{s^2}{9}

and RR is the square

R:2x2,2y2R:-2 \leq x \leq 2,-2 \leq y \leq 2

Solution

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The goal of this exercise is to solve for EE which is equal to

Rf(s)dA\begin{aligned} \iint_{R}f(s)dA \end{aligned}

where RR is the square R:2x2,2y2R:-2\leq x\leq 2, -2\leq y\leq 2.

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