Calculating EY with an even function Let X ∼f(x−θ), where f(x−θ) is symmetric about θ.  I want to find EY of the following function Y.  
Y= 1 for |x−θ|<2 and 0 otherwise.  
Please tell me if this interpretation of my question is correct.  We are integrating over a circle with radius 2 so our double integral would be the following:
$$ EY= \int_{0}^{2\pi} \int_{0}^{2} 1 f(x−θ) dxdθ$$
I'm not so sure how to deal with f(x−θ) given that its an even function. I mean this can be anything, and its not really a specific function.  Can anyone help me with this?  Thanks.
 A: First, let's formulate the question a little better.
(1) We have a random vector variable $\vec X=(X_x,X_y)$.
(2) The joint density of the random variables $X_x,X_y$ is $f(x-\theta_x,y-\theta_y)$.
(3) where  $\vec{\theta}=(\theta_x,\theta_y)$ is a given vector.
(4) We know that $f(x-\theta_x,y-\theta_y)$ is simmetrical around $\vec \theta$.
(5) The definition of the random variable $Y$
$$Y=\begin{cases}
1,&\text{ if } |(X_x-\theta_x,X_y-\theta_y)|<2\\
0,&\text{ otherwise. }
\end{cases},$$
(6) that is $Y$ indicates if $\vec X$ fell in the circle of radious $2$ centered at $\vec \theta$. Let the disk defined by this circle be called $\mathscr C.$
(7) The task is to calculate the following probability
$$E[Y]=P(\vec X\in \mathscr C).$$

The solution, in theory, is
$$E[Y]=\iint_{\mathscr C}f(x-\theta_x,y-\theta_y) \ dxdy.$$
Introducing polar coordinates $x=r\cos(\varphi),\ y=r\sin(x)$ we have
$$E[Y]=\int_0^{2\pi}\left[\int_0^2rf(r\cos(\varphi)-\theta_x,r\sin(\varphi)-\theta_y)\ dr\right]d\varphi$$
where $r$ in front of $f$ is the Jacobian.
