Let $Y_1$ and $Y_2$ be continuous random variables with the following joint probability density function.
$f(y_1, y_2) = \left\{\begin{array}{ll} cy_1y_2 & : 0 \lt y \lt 2, \frac{y_1}{2} \lt y_2 \lt 2y_1\\\ 0 & :elsewhere\\ \end{array} \right.$
(a) Find $E(Y_1|Y_2 = y_2)$ and $E(Y_1|Y_2=2)$.
(b) Find the CDF and pdf of $ U=\frac{Y_1}{Y_2}$. Your work should include a graph that supports your computation. Specify the domain where the pdf is positive.
What I have tried so far:
I started by drawing my graph, and then solving for C. To solve for C, I simply solved the double integral: $$\int_0^2 \int_{0.5y_1}^{2y_1} cy_1y_2 \,dy_2\,dy_1$$
From this I got $C = \frac{2}{15}$
Next, I needed to get the marginal pdf of $y_2$, so I did: $$f_2(y_2) = \int_{\frac{y_2}{2}}^{2y_2} \frac{2}{15}y_1y_2 \,dy_1 + \int_{\frac{y_2}{2}}^{2} \frac{2}{15}y_1y_2 \,dy_1$$
I did this because when integrating out $y_1$, it appeared that I needed to split the graph into two regions. Though I'm not sure if I'm doing this part correctly. However, under the assumption that I was, I continued to find $f(Y_1|Y_2)$.
To do that, I simply divided the original jpdf by the $y_2$ marginal pdf, and ended up with: $$f(Y_1|Y_2) = \frac{4y_1}{7{y_2}^2 + 8}$$
Again, this part seemed kind of weird to me, but to my understanding, everything was correct. I'm not confident in this answer though.
Finally, to get $E(Y_1|Y_2=y_2)$, I did: $$\int_{\frac{y_2}{2}}^{2y_2} \frac{4y_1}{7{y_2}^2 + 8} \,dy_1 + \int_{\frac{y_2}{2}}^{2} \frac{4y_1}{7{y_2}^2 + 8} \,dy_1 $$
This gave me an answer of $E(Y_1|Y_2=y_2) = \frac{31{y_2}^3 + 32}{21{y_2}^2 + 24}$
And then to solve for $E(Y_1|Y_2=2)$, I simply plugged in $y_2 = 2$ to get $\frac{70}{27} \approx 2.59259$
I feel like the answer is wrong, but the steps I took were somewhat in the correct direction. But I'm not sure. Any assistance would be greatly appreciated.