Finding CDF and PDF from shaded area Joint PDF fx,y = $\frac{1}{a^2}$ in shaded region where a=4. If z=x+y, find Fz(z) and fz(z).

What I did so far:
$\int_0^4$$\int_0^{z-y}$ $\frac 1 {16}$ dx dy
Fz(z) =$\frac{z}{4}$-$\frac{1}{2}$ for |z|$\leq$ 4.
For the PDF, I did:
$\int_0^4$$\frac 1 {16}$ dy 
fz(z) = $\frac 1 {4}$ for |z|$\leq$ 4.
Obviously these are both wrong because if you draw them out, you end up with a CDF with negative values. However, I'm not sure where I messed up in my steps. Can someone tell me where I messed up? 
 A: Your integral $\ \int_0^4\int_0^{z-y}\frac{1}{16}\, dxdy\ $ doesn't give you $\ F_z(z)\ $, even for $\ z>0\ $, because it omits the region where $\ x< 0\ $. Also, the inner integral vanishes when $\ y>z\ $, so for $\ 0<z<4\ $ its value is $\ \int_\limits{0}^z\int_\limits{0}^{z-y}\frac{1}{16}\, dxdy=\frac{z^2}{32}\ \ $, rather than $\ \frac{z}{4}-\frac{1}{2}\ $. There are simpler ways of doing the calculation, but if you want to use the integral formula to do it, the one you need to start from is
$$
F_z(z) =\int_{-\infty}^\infty\int_{-\infty}^{z-y} f_{x,y}(x,y)
\,dxdy\ .
$$
For $\ y\not\in [-4,4]\ $, the integrand vanishes, so the formula becomes
$$
F_z(z) =\int_{-4}^4\int_{-\infty}^{z-y} f_{x,y}(x,y)\,dxdy\ .
$$
Now, for $\ -4\le y\le 0\ $ the integrand vanishes for $\ x\not\in [-4-y,0]\ $, and for $\ 0\le y\le 4\ $ it vanishes for $\ x\not\in [0,4-y]\ $, so the formula becomes
\begin{align}
F_z(z) =\int_{-4}^0&\int_{\max(-4-y,z-y)}^{\min(0,z-y)} f_{x,y}(x,y)\,dxdy\\
&+\int_0^4\int_{\max(0,z-y)}^{\min(4-y,z-y)}f_{x,y}(x,y)\,dxdy\ .
\end{align}
For $\ z\le -4\ $, the integral evaluates to $\ 0\ $, and for $\ z\ge 4\ $ it evaluates to $\ 1\ $. For $\ -4\le z \le 0\ $ the formula becomes
$$
F_z(z) =\int_{-4}^z\int_{-4-y}^0\frac{1}{16}\,dxdy+\int_z^0\int_{4-y}^{z-y}\frac{1}{16}\,dxdy\ ,
$$
and for $\ 0\le z\le4\ $ it becomes
\begin{align}
F_z(z) &=\int_{-4}^0\int_{-4-y}^0\frac{1}{16}\,dxdy+\int_0^z\int_0^{z-y}\frac{1}{16}\,dxdy\\
&=\frac{1}{2} +\int_0^z\int_0^{z-y}\frac{1}{16}\,dxdy\ .
\end{align}
