Find $E[X^2 + Y^2]$ for the given joint density function. Let X and Y have joint probability density function    
$f(x,y) = \frac{3}{2}(x^2 +y^2)$, with $0<x<1, 0<y<1$.  
Find $E[X^2+Y^2]$.  
I came up with $\frac{14}{15}$ by integrating $\int_0^1 \int_0^1 (x^2 +y^2)(x^2+y^2)$, but I've
never had  to do anything but E[X].

$\mathbf{E}[\varphi(X,Y)] = \int \int_{D} \varphi(x,y) f(x,y) dx dy = \frac{3}{2} \int _{0}^{1} \int _{0}^{1}(x^4 + 2 x^2  y^2 + y^4)dx dy = \frac{3}{2}\int_{0}^{1}[\int_{0}^{1}(x^4 + 2 x^2  y^2 + y^4)dy]dx$
Are you familiar with double integration? Can you handle from here?
