Volume integral of the curl of a vector field

Question

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?

Answer 1

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.

Answer 2

This is one of the many guises of the fundamental theorem of calculus (aka generalized Stokes' theorem). The fundamental theorem can be written like so, in geometric calculus, on a region $\Omega$ that is $n$-dimensional in $\mathbb R^n$:

$$\int_\Omega |dV| \, \nabla A = \oint_{\partial \Omega} |dS| \, \hat n A $$

This very general result contains within it, for various kinds of fields $A$, all of the corollaries usually associated with the divergence theorem.

When $A$ is a scalar field, the result looks exactly the same as in vector calculus. When $A$ is a vector field, the equation separates into two distinct parts: the first is the scalar part.

$$\int_\Omega |dV| \, \nabla \cdot A = \oint_{d\Omega} |dS| \, \hat n \cdot A$$

This is the divergence theorem. Replace all the dot products with wedge products instead, and you get the corresponding bivector part:

$$\int_\Omega |dV| \, \nabla \wedge A = \oint_{d\Omega} |dS| \, \hat n \wedge A$$

Such is the usefulness of the geometric product, which marries the dot and wedge products together and thus contains both expressions in one; it is with this product that I wrote the fundamental theorem initially, exactly because it compactly captures all this behavior.

Wedge products are appropriate for generalization to higher dimensional spaces; they allow you to talk about planes and volumes and such in a way that the cross product is clumsy for. Still, in $\mathbb R^3$ we can convert wedges to crosses at the cost of using "Hodge duality". This, however, is the same for both sides of the integral above, so we get

$$\int_\Omega |dV| \, \nabla \times A = \oint_{d\Omega} |dS| \, \hat n \times A$$

Crucially, this means that you're off by a minus sign on one side of your proposed identity. The form I have given is in agreement with wikipedia also.

At any rate, you should see that this integral identity ultimately comes from the fundamental theorem of calculus, as so many such identities do. And given the proper tools and notation for expressing the fundamental theorem in its full generality, you no longer have to remember a whole suite of disparate theorems--there is just the one theorem, and that's all.

Answer 3

Yes, it is right except the the sign on the right hand side should be minus. It can be viewed as a Corollary to the Gauss-Green Theorem (Evans's PDE book appendix C.2 uses this name): $$ \int_{\Omega} \partial_{x_i} u\, dx = \int_{\partial \Omega} u n_i \,d\sigma. $$ The formula you gave is essentially: $$ \int_{\Omega} (\partial_{x_j} A_i - \partial_{x_i} A_j)dV = \int_{\partial \Omega} (n_j A_i - n_i A_j)d\sigma, \tag{1} $$ where $\bar{A} = (A_1,A_2,A_3)$ and $\hat{n} = (n_1,n_2,n_3)$, and (1) also applies for the $n$-dimensional curl.