## Spread of an Disease (Double Integral Problem)

I have yet another problem giving me trouble. Sorry to keep asking, but I am doing a little self teaching and running into some snags in this chapter

I partially answered it, but I'll go ahead and type the whole question:

When studying the spread of an epidemic, we assume that the probability of an infected individual will spread the disease to an uninfected individual is a function of the distance between them. Consider a circular city of radius 10 mi in which the population is uniformly distributed. For an uninfected individual at a fixed point A(x0, y0), assume that the probability function is given by

f(P) = (1/20)(20 - d(P, A))

where d(P, A) denotes the distance between P and A.

(a) Suppose the exposure of a person to the disease is the sum of probabilities of catching the disease from all members of the population. Assume that the infected people are uniformly distributed throughout the city, with k infected individuals per mile. Find a double integral that represents the exposure of a person residing at A.

(b) Evaluate the integral for the case in which A is the center of the city and for the case in which A is located on the edge of the city. Which location would be the safest from disease?

Okay, I did part (a):
int( int( (k/20)(20 - sqrt((x - x0)^2 + (y - y0)^2)) dA

I also solved for the center of the city in part (b):
int( int( (rk/20)(20 - r), r=0..10), theta=0..2pi) = k*200pi/3

How would I evaluate the integral for the outside? I tried using (10, 0) as A, but that gave me the wrong solution.

Any help would be appreciated.