Calculate $\iint_S (\nabla \times F)\cdot dS= 0$ if $F= i+j+k$ where $\omega$ is the unit cube.

Question

Calculate $\iint_S (\nabla \times F)\cdot dS= 0$ if $F= i+j+k$ where $\omega$ is the unit cube. First solve directly the integral and then use the Divergence Theorem.

In the firs part I do not know how to evaluate the integral directly I am confused about because in the Divergence Theorem we use $\iiint_V\nabla\cdot F$ instead of $\iint_S (\nabla \times F)$

Any hint or idea?

Answer 1

Remember that $${\rm div}({\rm curl} \vec{F})\equiv0.$$

Using the Divergence theorem, the result follows.

Answer 2

If ${\bf F}={\bf i}+{\bf j}+{\bf k}$ then $\nabla\times{\bf F}={\bf 0}$, whatever $\nabla\times{\bf F}$ means. The $\omega$ (why didn't you use $\pi$, $e$, or $5$ to denote the unit cube?) does not occur in $\int_?\int_S (\nabla\times{\bf F})\cdot{\rm d}S$, but this does not play a rôle under the circumstances of this lousy question.

Answer 3

Since F= i+ j+ k is constant vector, $\nabla\times F$ is the 0 vector. The integral over any surface is 0.