Applying divergence theorem

Question

I am trying to solve the problem below applying the divergence theorem. I take the cylinder as a closed surface. However, the divergence of the vector field appears to be 0. And the answer I should get is PI. My guess is that I just can't apply divergence theorem here, yet I don't really understand why. I would be grateful for any help! Problem

Answer 1

The Divergence Theorem:

For a region $R \subset \Bbb R^3$ with piecewise smooth boundary $\partial R$, and $\mathbf F$ be a smooth, differentiable vector field on $R$, we have $$\iint_{\partial R} \mathbf F \cdot \mathbf n \, dS = \iiint_R \nabla \cdot \mathbf F \, dV$$ where $\mathbf n$ is the unit outward normal to $\partial R$.

An important point is that $\partial R$ (which is the $S$ in your problem) is the boundary of a 3D region $R$, i.e. it must enclose a certain volume.

As such, in order to apply the Divergence Theorem in the problem you have, you must add in something to close off $S$. This can be done, for example, by adding in the plane $z=1$ (the portion of it that is on top of $x^2+y^2 \leq 1$). So

$$\iint_S \mathbf F \cdot \mathbf n \, dS + \iint_{z=1, x^2+y^2 \leq 1}\mathbf F \cdot \mathbf n \, dS = \iiint_R \nabla \cdot \mathbf F \, dV \qquad \qquad \text{(*)}$$

where $R$ is the region enclosed by $S$ and the plane $z=1$.

As you have pointed out, $\nabla \cdot \mathbf F = 0$, so the integral on the right hand side vanishes.

On the other hand, the outward unit normal on $z=1$ is $\mathbf n = (0,0,1)$, so we may compute

\begin{align} \iint_{z=1, x^2+y^2 \leq 1}\mathbf F \cdot \mathbf n \, dS & = \iint_{z=1, x^2+y^2 \leq 1} \begin{pmatrix} e^y+x \\ e^{\sin (z)} + \sin (x) \\ -z+xy \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \, dS \\ & = \iint_{z=1, x^2+y^2 \leq 1} (-z+xy) \, dS \\ & = \int_{\theta = 0}^{\theta = 2\pi}\int_{r=0}^{r=1} \big(-1+r^2\cos(\theta)\sin(\theta) \big) \cdot rdrd\theta \\ & = -\pi \end{align}

Plugging this back into equation (*), we find that

\begin{align} & \iint_S \mathbf F \cdot \mathbf n \, dS - \pi = 0 \\ \implies & \iint_S \mathbf F \cdot \mathbf n \, dS = \pi \end{align}