Expected value of distance

By symmetry (across the line $y = x$ ) we need only consider the average distance to the origin $O$ of a point $P(x,y)$ in the region $\{(x,y): 0 \le y \le x \le 1\}$ . As the distance from $P$ to the origin is given by $\sqrt{x^{2} + y^{2}}$ and the area of the region of integration is $1/2$ , the expected distance of $OP$ is

$\displaystyle v = \dfrac{1}{\dfrac{1}{2}} \int_{0}^{1} \int_{0}^{x} \sqrt{x^{2} + y^{2}} dydx = 2 \int_{0}^{1} \int_{0}^{x} x\sqrt{1 + \left(\dfrac{y}{x}\right)^{2}} dydx$ .

Now for the interior integral, where $y$ is the variable and $x$ is constant, let $u = \dfrac{y}{x} \Longrightarrow x du = dy$ , with $u$ going from $0$ to $1$ . We then have that

$\displaystyle v = 2\int_{0}^{1} \int_{0}^{1} x^{2}\sqrt{1 + u^{2}} dudx = 2\int_{0}^{1} \sqrt{1 + u^{2}} du \int_{0}^{1} x^{2} dx = \dfrac{2}{3} \int_{0}^{1} \sqrt{1 + u^{2}} du$ .

For this remaining integral, we can use trig substitution with $u = \tan(\theta) \Longrightarrow du = \sec^{2}(\theta)$ , with $\theta$ going from $0$ to $\pi/4$ . We then have that

$\displaystyle v = \dfrac{2}{3} \int_{0}^{\pi/4} \sec^{3}(\theta) d\theta = \dfrac{2}{3} \times \dfrac{1}{2}\left(\sec(\theta)\tan(\theta) + \ln(\sec(\theta) + \tan(\theta))\right)_{0}^{\pi/4} = \dfrac{1}{3}(\sqrt{2} + \ln(\sqrt{2} + 1))$ ,

and so $a + b + c + d = 2 + 2 + 1 + 3 = \boxed{8}$ .

Note: The integral of $\sec^{3}(\theta)$ is standard and can be solved by parts or partial fractions, as shown here .