Time for me to rush to battle again! I fight constantly against the very notation "[itex]\vec{f}\cdot\vec{n}d\sigma[/itex]" because if you follow that literally, computing [itex]\vec{n}[/itex] and [itex]d\sigma[/itex] separately you wind up calculating the length of a vector twice and then watch them cancel out! (Unless you forget one of them!)
In this problem, we can write the position vector of a point on the plane as [itex]\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ (3x+ 2)\vec{k}[/itex].
Differentiating: [itex]\vec{r}_x= \vec{i}+ 3\vec{k}[/itex] and [itex]\vec{r}_y= \vec{j}[/itex]. The cross product of those two vectors, [itex]-3\vec{i}+ \vec{k}[/itex], is the "fundamental vector product" for the plane and the vector differential of area, [itex]d\vec{\sigma}[/itex] is [itex](-3\vec{i}+ \vec{k})dxdy[/itex].
Now [itex]\int\int \vec{F}\cdot\vec{d\sigma}[/itex] [itex]= \int\int (2y\vec{j}+ 3z\vec{k})\cdot(-3\vec{i}+ \vec{k})dxdy[/itex] [itex]= \int\int 3z dxdy= 3\int\int (3x+ 2)dxdy[/itex].
Integrate that over the circle of radius 2.
Obviously the way to do that is to switch to polar coordinates. We could do that right at the beginning:
write [itex]\vec{r}(r,\theta)= rcos(\theta)\vec{i}+ rsin(\theta)\vec{j}+ (3rcos(\theta)+ 2)\vec{k}[/itex], differentiate with respect to r and [itex]\theta[/itex] and the "r" you need for rdrd[itex]\theta[/itex] pops out automatically!
[Here endeth the sermon]