Let $$D = (x,y,z) : (x^2 +y^2 \leq 1^2, 0 \leq z \leq 1)$$ Furthermore define $$\phi(x,y,z) = x^2 - y^2 + \frac{3}{2\pi}z^2$$ and $$\boldsymbol{G}(x,y,z) =\frac{3}{\pi}zy\boldsymbol{i} + \arctan(x^2)\boldsymbol{j} + x^5 y \sqrt{z^2 +1}\boldsymbol{k}$$ Let the vector field F be defined by $$\boldsymbol{F}(x,y,z) = \nabla \phi + \nabla \times \boldsymbol{G} $$ Calculate the flux of F out of the cylindrical part of the surface of D.
Im supposed to use the divergence theorem, and the identity $\nabla \cdot(\nabla \times \boldsymbol{F}) = 0 \rightarrow$ (div curl$ = 0$) The expression for F is pretty awkward as its a function + a vector. Also not sure how Im supposed to use the identity and the smartest way to approach the problem.
You have: $$\operatorname{div}(F)=\operatorname{div}(\nabla \phi+\nabla \times G)=\Delta \phi+\operatorname{div}(\nabla \times G)=\Delta \phi$$ and: $$\Delta \phi=2-2+\frac{3}{\pi}$$ all it remains is to compute: $$\int_D \operatorname{div}(F)=\int_D \frac{3}{\pi}=\frac{3}{\pi}\operatorname{Vol}(D)$$
It remains to compute the flux on the upper and lower part of the cylinder.
For $z=0$ this is: $$\int_{x^2+y^2 \leq 1} F(x,y,0) \cdot (-\mathbf{k}) dx dy$$
But $$F \cdot \mathbf{k}=\nabla \phi \cdot \mathbf{k}+(\nabla \times G) \cdot \mathbf{k}=\partial_z \phi +(\partial_x G_y-\partial_y G_x)$$ here you obtain: $$F \cdot \mathbf{k}=\frac{3}{2 \pi}\color{blue}{2} z +\frac{1}{1+x^2}-\frac{3}{\pi} z$$
the $\frac{1}{1+x^2}$ does not matter as it is independent of $z$ so it disappear when computing:
$$\int_{x^2+y^2 \leq 1} F(x,y,0) \cdot (-\mathbf{k}) dx dy+\int_{x^2+y^2 \leq 1} F(x,y,1) \cdot (\mathbf{k}) dx dy$$