Showing a joint function is a pdf and joint MGF 

I have these two problems that are on a previous final for my class.
for Number 3, I know that the double integral from 0 to infinity of f(x+y)/x+y dxdy has to equal 1.
But I have no idea how to tackle that f(x+y).
For Number 4 I tried to do something along the lines of doing a change of variables and using
 $$\ E(e^{ua + vb}) = E(e^{a(X+Y)} + e^{b(X^2 +Y^2)})
$$
Then I expanded that out and put it in terms of U, V. It doesn't seem like it's working out for me though, I also tried completing the square, which I think is the right thing to do.
 A: 

*Let $f(\cdot)$ denote the density function of a continuous positive random variable
and define $$g(x,y) = \begin{cases}\displaystyle \frac{f(x+y)}{x+y}, & x, y > 0,\\
0, & \text{otherwise.}\end{cases}$$
We are asked to prove that $g(x,y)$ is the density function of two jointly
continuous positive random variables. It is obvious that $g(x,y) \geq 0$ for all
$x$ and $y$, and so all that remains to be proven is that the volume $V$ of the
(prismatic) solid whose base is the first quadrant and whose upper surface is
$g(x,y)$ is equal to $1$.  The standard way of doing this is via a two-dimensional
or iterated integral:
$$ V = \int_0^\infty \left[\int_0^\infty g(x,y)\,\mathrm dx \right ]\,\mathrm dy
\tag{1}$$
which can be thought of as finding the volume of a thin slice of the solid
between $y$ and $y+\Delta y$. The area of the cross-section is
$\int_0^\infty g(x,y)\,\mathrm dx$ (which is the inner integral in $(1)$), the thickness of the slice is $\Delta y$, and when
we pass from the Riemann sum to the integral, we get the outer integral in 
$(1)$.
In this instance, since $g(x,y)$ has constant value $\frac{f(z)}{z}$
on the straight-line segment with endpoints $(0,z)$ and $(z,0)$, we can instead
consider a diagonal slice whose cross-section is a rectangle of
base length $z\sqrt{2}$ and height $\frac{f(z)}{z}$ so that the area of
the cross-section is just $\sqrt{2}f(z)$. But, the thickness of the slice
(which lies between the straight-line segment with endpoints $(0,z)$ and $(z,0)$
and the straight-line segment with endpoints $(0,z+\Delta z)$ and 
$(z+\Delta z,0)$) is just $\frac{\Delta z}{\sqrt{2}}$ (think about it!)
and so the volume of the slice is just $f(z)\Delta z$. Adding up the thickness
of all such slices and passing to the limit, we find that the volume $V$ is
just
$$V = \int_0^\infty f(z)\,\mathrm dz $$
and so we get that $V = 1$ since we have been told that $f(\cdot)$ is a density with support $(0,\infty).\\$
More formally, as I suggested in a comment,
all this can be done via a change of variables corresponding
to a rotation of axes by $\pi/4$, but I can never remember the Jacobian
stuff, especially on exams.





*Your expression 
$E\left[e^{Ua + Vb}\right] = E\left[e^{a(X+Y)} + e^{b(X^2 +Y^2)}\right]$
is incorrect; you should get $E\left[e^{a(X+Y)+ b(X^2 +Y^2)}\right]$.
Then, when you set up the corresponding integral, you will need to use
a technique called completing the square; the kind of stuff that
you do when you add and subtract $b^2$ to convert
$$x^2+2bx + c = x^2 + 2bx + b^2 - b^2 + c
= (x+b)^2 - b^2 + c$$

