What is the probability that $|X - Y| < 10$ if $X$ and $Y$ both $\sim \operatorname{Uniform}(0, 50)$?

Let there be two identical and independent random variables $X$ and $Y$. Both $X$ and $Y$ are uniformly distributed from $0$ to $50$. What is the probability that $|X - Y| < 10$? This is not homework. I am practicing probability.

I think if $|X - Y| < 10$, then either $X$ is greater than $Y$ or $Y$ is greater than $X$, the expected value of $A = |X - Y|$ should be $\int_0^{50} \int_0^y (y - x) \, dx \, dy + \int_0^{50} \int_0^x (x - y) \, dy \, dx = \frac{125000}{3}$, but how to do I use this value to find $P(A < 10)$?

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Think of $(X,Y)$ as being the cooordinates of a point that is uniformly chosen from the square $[0,50] \times [0,50]$. What does it mean that $|X - Y| < 10$? –  Hans Engler Nov 4 '12 at 1:17
Thanks! In that case, $Y < 10 + X$ and $Y < X - 10$... ah, so I just set up a relevant integral within the bounds of [0, 50] X [0, 50]? –  David Faux Nov 4 '12 at 1:30

We want to be between the lines $x-y=-10$ and $x-y=10$. The part of the square which is not between these lines consists of two isosceles right triangles with legs of length $40$. Their combined area is therefore $2(40)^2/2=1600$.
Thus the probability that $|X-Y|\ge 10$ is $\dfrac{1600}{2500}$. Subtract this from $1$ to get the asked for probability.