Wish to identify what I'm doing wrong when finding the E(X|Y=5) of the following:
$$f(x, y)=\begin{cases}
1/6  & \text{if } 0<x<2, 0<y<6-3x \\ 0 &  \text{otherwise}
 & 
\end{cases}.$$

My working:
$$\begin{align}\mathsf E(X\mid Y{=}5)~&=~\displaystyle\int_0^{2} x \frac{\frac{1}{6}}{f(y)}\,\mathsf d x\end{align}$$
where
$$\begin{align}\mathsf f(y)~&=~\displaystyle\int_0^{\frac{-y+6}{3}} \frac{1}{6} \,\mathsf d x = \frac{6-y}{18}\end{align}$$

Therefore, 
$$\begin{align}\mathsf E(X\mid Y{=}5)~&=~\displaystyle\int_0^{2} x \frac{\frac{1}{6}}{\frac{6-y}{18}}\,\mathsf d x \mathsf ~&=~\displaystyle\int_0^{2} x \frac{\frac{1}{6}}{\frac{6-5}{18}}\,\mathsf d x~&=~\displaystyle\int_0^{2} 3x \,\mathsf d x \end{align} = 6$$

But then when I compute Var(X|Y=5), the answer is getting negative, which is impossible. So I think that I might have done something wrong when calculating the expectation.

My working for Variance:
$$Var(X|Y=5) = E(X^2|Y=5) - (E(X|Y=5) )^2 = 8 - 6^2 = -28$$

I appreciate any help.