Probability that the circumference of circle is larger than $\pi$ 
The joint probability density function of the random variables $X$ and
$Y$ is given by $f(x,y)=x+y$ for $0 \leq x,y \leq 1$ and $f(x,y)=0$
otherwise. Consider the circle centered at the origin and passing
through the point $(X,Y)$. What is the probability that the
circumference of the circle is larger that $\pi$?

try
We know circumference of circle is $2 \pi r = 2 \pi \sqrt{X^2+Y^2}$. We want $P( 2 \pi \sqrt{X^2+Y^2} > \pi ) = P(X^2+Y^2 > 1/4) = 1 - P(X^2+Y^2 \leq 1/4).$
To calculate $P(X^2+Y^2 \leq 1/4)$, we want to integrate over the intersection of the unit square (first quadrant) and the circle of radius one half. This gives
$$ \int\limits_0^{1/2} \int\limits_0^{\sqrt{1/4-x^2}} (x+y) dy dx = \int_0^{1/2} (xy+y^2/2 )_0^{\sqrt{1/4-x^2}} dy = \int_0^{1/2} x \sqrt{1/4-x^2}+ \frac{1}{2} (1/4-x^2) dx = \dfrac{1}{12}$$
Is this a correct solution?

\begin{align}
P(X^2 + Y^2 \le \frac14) &=\int_0^{\frac12} \int_0^{\frac{\pi}2}r(\cos(\theta)+\sin(\theta) \cdot r \, \, dr d\theta \\
&= \int_0^\frac12r^2 \, dr \int_0^\frac{\pi}2 \cos(\theta) + \sin(\theta) \, d\theta\\
&=\frac13\left( \frac12\right)^3 [ \sin(\theta)  - \cos(\theta)]_0^\frac{\pi}2 \\
&=\frac13\left( \frac12\right)^3 (1-(0-1))\\
&= \frac1{12}
\end{align}
Hence the probability of having a circumference that is larger than $\pi$ is $\frac{11}{12}$.
