Divergence Theorem: evaluate $\iint_S F \cdot dS$

Question

Use the Divergence Theorem to evaluate

$$\iint_S F \cdot dS$$

where $$F = (x^3y + z)i + xze^yj + (xy − 3z^4)k$$

and $S$ consists of the five “lower” faces of the cube $[0, 1]\times[0, 1]\times[0, 1]$. That is, the face with $z = 1$ is not a part of $S$.

(Hint: The surface $S$ is not closed so you need to close it up to apply the theorem.)

I actually have no idea what to do, since the surface is not closed and what I should do to close it.

Answer 1

The trick here is realising that the surface integral over the lower 5 faces of the cube is equal to the surface integral over all faces of the cube minus the surface integral over the top face.

Let $S_T$ denote the top face of the cube and $S_A$ denote all faces of the cube. Then we have...

$$\iint_{S} F \cdot dS = \iint_{S_A} F \cdot dS - \iint_{S_T} F \cdot dS.$$

You can then use the divergence theorem to evaluate the surface integral over all faces of the cube.

Let $V$ denote the volume enclosed by $S_A$. Then, by the divergence theorem we have...

$$ \iint_{S_A} F \cdot dS = \iint_{S_A} \nabla \cdot F dV = \frac{e - 11}{4},$$

(by standard triple integral stuff). Then we can evaluate the surface integral over the top face...

$$\iint_{S_T} F \cdot dS = \frac{-11}{4},$$

(by standard double integral stuff). Simply subtract the second from the first to obtain...

$$\iint_{S} F \cdot dS = \frac{e - 11}{4} - \frac{-11}{4} = \frac{e}{4}.$$