I have $f(x,y)=\frac{3}{11}(x^2+y)$ for $0<x<2$ and $0<y<1$.
I calculated $f(x)=\frac{3}{11}(x^2+1/2)$ for $0<x<2$.
I am asked to find $E(y|x>1)$.
My instinct was to first find $f(y|x>1)$ by taking $f(x>1,y)/f(x>1)$. The solutions did this, except instead of doing a double integral for $f(x>1,y)$, they only derived the original $f(x,y)$ with respect to $x$ (with $x$ going from 1 to 2). Why did they omit the $dy$ part here?
Tip: Use the definition for conditional expectation over an (non-zero probbability) event, that: $$\require{enclose}\mathsf E(Y\mid X>1) =\dfrac{\mathsf E(Y~\mathbf 1_{X>1})}{\mathsf P(X>1)} =\dfrac{\displaystyle\int_1^2\int_0^1 y~(x^2+y)~\mathsf d y~\mathsf d x}{\displaystyle\int_1^2\int_0^1 (x^2+y)~\mathsf d y~\mathsf d x}=\dfrac{\displaystyle\int_1^2\int_0^1 y~(x^2+y)~\mathsf d y~\mathsf d x}{\displaystyle\int_1^2 f_X(x)~\mathsf d x}$$