## Related questions with answers

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

$E=\iint_R f(s) d A$

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

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

and $R$ is the square

$R:-2 \leq x \leq 2,-2 \leq y \leq 2$

Solution

VerifiedThe goal of this exercise is to solve for $E$ which is equal to

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

where $R$ is the square $R:-2\leq x\leq 2, -2\leq y\leq 2$.

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