TITLE: Volume integral of the curl of a vector field
QUESTION [17 upvotes]: I am having hard time recalling some of the theorems of vector calculus. I want to calculate the volume integral of the curl of a vector field, which would give a vector as the answer. Is there any formula? As far as I can recall, maybe I can write
$$\int \nabla \times \vec{A} \ \mathrm{d}V=\int \vec{A} \times \hat{n}\, \mathrm{d}\sigma,$$ where $d\sigma$ is the enclosing boundary. Is this right?
But more importantly, what does it follow from? Does this follow directly from one of the integral theorems, or how can it be proved independantly?

REPLY [12 votes]: You are right. The volume integral of the curl follows from the standard Gauss Theorem. This may be clearly seen if one analyzes each Cartesian component separately. To be concrete, let us consider the $z$ component: $$(\vec\nabla\times\vec F)_z= \left(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)=\vec\nabla\cdot (F_y,-F_x,0)=\vec\nabla\cdot\vec G,$$ where $\vec G=(F_y,-F_x,0)$. Since the Gauss Theorem states that $$\iiint_\Omega \vec\nabla\cdot \vec G\,dV=\iint_{\partial\Omega} \vec n\cdot \vec G\,dS$$ and $$\vec n\cdot \vec G=n_x F_y-n_y F_x=(\vec n\times \vec F)_z,$$ we obtain $$\iiint_\Omega (\vec\nabla\times\vec F)_z\,dV=\iiint_\Omega \vec\nabla\cdot \vec G\,dV=\iint_{\partial\Omega} \vec n\cdot \vec G\,dS=\iint_{\partial\Omega} (\vec n\times \vec F)_z\,dS.$$ By applying the same procedure to the $x$ and $y$ components, one arrives at the identity under discussion, i.e., $$\iiint_\Omega \vec\nabla\times\vec F\,dV=\iint_{\partial\Omega} \vec n\times \vec F\,dS.$$ It should be noted that the surface is considered to point outwards $\Omega$. If one considers it to point inwards, then the factors in the vector product of $\vec n$ and $\vec F$ should be commuted. Of course, the identity may be seen as a particular case of a general Stokes theorem, as you are already aware of.