Wish to identify what I'm doing wrong when finding the $\operatorname E(X\mid Y=5)$ 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: $$\begin{align}\operatorname E(X\mid Y=5)~&=~\int_0^2 x \frac{\frac{1}{6}}{f(y)} \, \mathsf d x\end{align}$$ where $$\begin{align}\mathsf f(y)~&=~\int_0^{\frac{-y+6}{3}} \frac{1}{6} \,\mathsf d x = \frac{6-y}{18}\end{align}$$

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

But then when I compute $\operatorname{Var}(X\mid Y=5),$ 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: $$\operatorname{Var}(X\mid Y=5) = \operatorname E(X^2\mid Y=5) - (\operatorname E(X\mid Y=5) )^2 = 8 - 6^2 = -28$$

I appreciate any help.