Calculate the flux of the vector field $$F(x,y,z) = (\cos y \sin z + x, y^2 + \sin z, \sin z + x + y^2)$$ through the surface $S = \{ (x,y,z) : x^2 + y^2 = z + 1 \leq 1\}$.
I tried to apply the Divergence Theorem but I get stuck.
$S$ is not a closed surface, but we can build an auxiliar closed surface, let's call it D, by unioning to S the disc $S_2 = \lbrace (x,y,0) : x^2 + y^2 \leq 1\rbrace$. Then, $D = S \sqcup S_2$ and by the Gaussian Divergence Theorem, the outward flux is $$\iint_S \mathbf{F}\cdot d\mathbf{S}=\iiint_D \text{div}(\mathbf{F})\cdot dxdydz-\iint_{S_2} \mathbf{F}\cdot d\mathbf{S}$$ Let's calculate the dicervence of F $$\text{div} F = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} + \frac{\partial F}{\partial z} = 1 + 2y + \cos z$$ In order to parametrice $S_2$, we consider polar coordinates $$\Phi (r, \theta) = \begin{cases} x = r \cos \theta \\ y = r \sin \theta \\ z = 0 \ \end{cases} : \theta \in [0, 2 \pi], \ r \in (0,1)$$ Then $\Phi_r = (\cos \theta, \sin \theta, 0) \ \& \ \Phi_\theta = (-r \sin\theta, r \cos\theta, 0)$, and therefore $\Phi_r \times \Phi_\theta = (0,0, r)$. $$\iint_{S_2} \mathbf{F}\cdot d\mathbf{S} = \int_0^1 \int_0^{2\pi} (\cos y \sin z + x, y^2 + \sin z, \sin z + x + y^2) \cdot (0,0,r) d\theta dr$$ $$= \int_0^1 \int_0^{2\pi} r^3 \sin^2 (\theta ) = \frac{\pi}{4}$$
The problem now is that I don´t know how to approach the calculation of the integral of the divergence. Any help?
Yes, you are correct, $S$ is not closed, but you can get a closed surface by gluing the disc $S_2= \{ (x,y,0) : x^2 + y^2 \leq 1\}$ to $S$. Then, by the divergence theorem, the outward flux is $$\iint_S \mathbf{F}\cdot d\mathbf{S}=\iiint_D \text{div}(\mathbf{F}) \,dxdydz-\iint_{S_2} \mathbf{F}\cdot d\mathbf{S}$$ where $D= \{ (x,y,z) : x^2 + y^2 \leq z + 1 \leq 1\}$.
Can you take it from here? Please show your work just below your question.