Let $$B=\{(x,y,z):x^2+y^2+(z-1)^2<4, \ z\geq 0\}$$ and consider the vector field $$F:(x,y,z)\mapsto(x^3,y^3,z)$$
I want to compute the flux of $F$ through $\partial B.$
We have $$\text{Div}F=3x^2+3y^2+1$$ and so by the divergence theorem we could compute the flux as $$\int_B 3x^2+3y^2+1 \ dx dy dz$$ but this triple integral does not look very friendly to me using spherical coordinates, because the sphere is centered in $(0,0,1).$
Another approach is by computing $$\partial B =\{x^2+y^2\leq 3, \ z=0\}\cup\{x^2+y^2+(z-1)^2=4, \ z>0\}$$
I can parametrize $\{x^2+y^2+(z-1)^2=4, \ z>0\}$ as $$x=2\sin \phi \cos \psi$$ $$y=2 \sin \phi \sin \psi$$ $$z= 2 \cos \phi +1$$ $$\phi\in[0,\pi], \psi \in [0,2 \pi)$$ but it does not seem to make calculations easier.
You'll want to use a change of variables, then use spherical coordinates. Starting with
$$\int_B\left(3x^2+3y^2+1\right)dx\ dy\ dz$$
we let $u=x$, $v=y$, and $w=z-1$. Let's call our new volume of integration $C$; with the new variables, $C$ is the sphere centred at the origin with radius 2:
$$C=\{(u, v, w) : u^2+v^2+w^2<4, w>-1\}$$
Noting that the Jacobian determinant for this is just 1, we get that the flux is equivalently stated as
$$\int_C\left(3u^2+3v^2+1\right)du\ dv\ dw$$
From here, you can use either spherical or cylindrical coordinates to finish the computation, and it should be a lot more pleasant this time.