Thank you both Mr Fantastic and Matheagle for your patience! I promise I am not dumb, but I may be having a 'bad convolution' day

I earnestly wanted to try both methods - even though I think they are the same but explained in two different math 'dialects' - and the second one (change of variable language) proved to be easier for me. I did see the three different regions of integration in the W,Z coordinates and the rest was automatic.

0<z<1

$\displaystyle \int_{z=0}^1\int_{w=0}^z(wz-w^2)dwdz=\int_0^1dz[\frac{w^2}{2}z-\frac{w^3}{3}]_0^z=\int_0^1\frac{z^3}{6}dz$

which means $\displaystyle f_Z(z)=\frac{z^3}{6}, 0<z<1$

1<z<2

$\displaystyle \int_{z=1}^2\int_{w=0}^1(wz-w^2)dwdz=\int_1^2dz[\frac{w^2z}{2}-\frac{w^3}{3}]_0^1=\int_1^2\frac{z}{2}dz$

therefore density is $\displaystyle \frac{z}{2}, 1<z<2$

2<z<3

$\displaystyle \int_{z=2}^3\int_{w=z-2}^1(wz-w^2)dwdz=\int_2^3dz[\frac{w^2z}{2}-\frac{w^3}{3}]_{z-2}^{1}=\int_2^3\frac{-z^3+15z-18}{6}dz$ (or thereabouts...)

therefore density is $\displaystyle \frac{-z^3+15z-18}{6}, 2<z<3$

I hope I did get it right this time...

When I tried to do

*distribution functions* method using definition of G(z)=P(Z<z), I did get lost again, and I'd love to get to the bottom of it - Mr Fantastic, if you don't mind, could you let me know if you referred to 3D coordinates (x,y,z) or 2D coordinates (z,x) in your sentence "simple sketch graph of the line y = -x + z over the region defined by the support f_XY"?

Mr F says: It's a sketch of the line y = -x + z in the xy-plane. z is a parameter here not a variable.
Also, is it

$\displaystyle G(z)=P(X+Z\leq{z})$ or $\displaystyle P(X+Y\leq{z})$

Mr F says: By the given definition Z = X + Y so surely it must be the latter ....!