$O=(0,0),~ A=(3,0),~ B=(0,3)$
$I= \int_\Gamma x(a+y)(ydx+xdy)$
$I= 36$, $a \in \mathbb {R}$
The first portion of the line is $OA$, then $AB$ and finally $BO$
AB is characterized by $\{(x,y) \in \mathbb {R^2}$, $x^2 + y^2 = 9$, $x \ge 0$, $y \ge 0\}$
And I need to find a. I applied Green formula and after some calculations I got to $a=\dfrac {16}{\pi}$ which I am not sure if it is right. Sorry for format or bad english, it is not my first language.
Link to how I solved it, last integral was put into wolfram https://i.stack.imgur.com/O2dzn.jpg
EDIT: I might have gotten a better answer as both $x$ and $y$ are positive and my $AB$ might be actually a quarter of a circle so I use the constraints $0$ and $\dfrac{\pi}{2}$ for my integral with polar coordinates and it results in $a = 4$
You made a mistake in your integral. See below. Also your diagram shows $AB$ as a straight line instead of circular arc.
$a \displaystyle \int_0^{3} \bigg[\int_0^{\sqrt{9-y^2}} x \ dx \bigg] \ dy = a \displaystyle \int_0^{3} \bigg[\frac{x^2}{2} \bigg]_0^{\sqrt{9-y^2}} \ dy$
$ = \frac{a}{2} \displaystyle \int_0^{3} (9-y^2) \ dy = 9a$
So $9a = 36 \implies a = 4$.
You could also set up your integral in polar coordinates as,
$a \displaystyle \int_0^{\pi/2} \int_0^3 r^2 \cos \theta \ dr \ d\theta = 9a$