Divergence Theorem for the Curl

[meteorologist1]

Hi, I'm having trouble proving the following result:

$$\int_{V} (\nabla\times\vec{A}) dV = -\int_{S} (\vec{A}\times\vec{n}) dS$$

I'm not sure how I should Stokes' and/or the Divergence Theorem in proving this, or if you should use them at all. Thanks in advance.

 

[dextercioby]

You know that the theorems of Gauss-Ostrogradski and Green are written with scalar integrands,so,to prove your statement,u'll need to pick only one component.
Chose the firstcomponent of the vectors.The LHS becomes:
$$\int\int\int (\nabla\times\vec{A})_{1} dxdydz = \int\int\int (\frac{\partial A_{z}}{\partial y} -\frac{\partial A_{y}}{\partial z}) dxdydz$$ (1)

The RHS becomes:
$$\oint\oint (\vec{n}\times\vec{A})_{1} dS = \oint\oint (A_{z}\cos\beta -A_{y}\cos\gamma) dS$$ (2)

Look at the RHS from (1).Consider the vector field:
$$\vec{B}=:(0,A_{z},-A_{y})$$ (3)

In the light of (3),the RHS from (1) becomes
$$\int\int\int (\frac{\partial 0}{\partial x}+\frac{\partial B_{y}}{\partial y} +\frac{\partial B_{z}}{\partial z})$$(4)

,which,by the virtue of the famous Gauss-Ostrogradski theorem becomes:

$$\oint\oint (0\cos\alpha+B_{y}\cos\beta+B_{z}\cos\gamma) dS$$(5)

In the light of (3),the LHS from (2) becomes:

$$\oint\oint (0\cos\alpha+B_{y}\cos\beta+B_{z}\cos\gamma) dS$$ (6)

Looking at (5) and (6),we see that they coincide,therefore the RHS from (1) coincides with the RHS from (2),and by the means of the property of the equality relation,the LHS from (1) coincides with the LHS from (2),therefore the first component from the LHS of the equality which should have been proven coincides with the RHS of the same equality.

Since our choice of components was arbitrary (we could have picked any component from the 3,we chosed the first),we can say:

Quod Erat Demonstrandum.

Daniel.

 

[meteorologist1]

Thanks. But is there a way to prove it without using the Gauss-Ostrogradski theorem? I haven't learned about it.

 

[dextercioby]

I can't see any other way to getting from a triple/volume integral to a surface one...If u've seen a method to do so,without taking into consideration the Gauss-Ostrogradski theorem/flux-divergence theorem,please let me know ASAP.
Listen,i'm not saying that i cannot be done in other way...I'm saying that i just do not see any other way...That's it...Maybe my math knowledge is not that wide as i like to think,but that's all i can do and say...

Daniel.

 

[jdavel]

dextercioby,

I'm hesitant to question anything you say regarding math, as your knowledge seems formidable!

But...as the statement to be proven is written, does it even make sense? What is the meaning of a vector integrated over a volume? Or am I having an early senior moment? :-(

Never mind!

 

[dextercioby]

It has sense...Any tensor field can be integrated.In particular R^{3} is a manifold and vectors are tensors of rank one.Tensors are integrated on manifolds.
Besides,Stokes theorem involves vectors integrated on loops and open surfaces,which are again manifolds of smaler dimension.

Daniel.

 

[Galileo]

What is the meaning of a vector integrated over a volume?

Surface and triple integrals of vector valued functions are vectors defined by integrating each component function.

 

[vincentchan]

the integral of vector is also a vector , so

$$\int_{V} \nabla\times\vec{A} dV = (\int_{V} \nabla\times\vec{A} dV)_{x}\vec{i} +(\int_{V} \nabla\times\vec{A} dV)_{y}\vec{j} +(\int_{V} \nabla\times\vec{A} dV)_{z}\vec{k}$$

my idea is prove each component of $$\vec{i},\vec{j},\vec{k}$$ in LHS is equal to that of RHS, than we can say the LHS = RHS

first, $$\nabla\times\vec{A} = (\frac{\partial A_{z}}{\partial y} -\frac{\partial A_{y}}{\partial z})\vec{i} - (\frac{\partial A_{z}}{\partial x} -\frac{\partial A_{x}}{\partial z})\vec{j} + (\frac{\partial A_{y}}{\partial x} -\frac{\partial A_{x}}{\partial y})\vec{kj}$$

so, the z component of $$\nabla\times\vec{A}$$ is $$\frac{\partial A_{y}}{\partial x} -\frac{\partial A_{x}}{\partial y}$$

$$(\int_{V} \nabla\times\vec{A} dV)_{z} = \int_{V} \frac{\partial A_{y}}{\partial x} -\frac{\partial A_{x}}{\partial y} dV=\int_{V}\frac{\partial A_{y}}{\partial x} -\frac{\partial A_{x}}{\partial y} dxdydz$$

$$= \int_{V}\frac{\partial A_{y}}{\partial x}dxdydz - \int_{V}\frac{\partial A_{x}}{\partial y} dxdydz =\int A_{y}dydz - \int A_{x} dxdz$$

The z component of LHS is $$\int A_{y}dydz - \int A_{x} dxdz$$ , now prove the z component of RHS is equal to above...

RHS = $$- \int_{S} \vec{A}\times\vec{n}dS = -\int_{S} \vec{A}\times d \vec{S}$$

what this integral do is sum up all little $$\vec{A}\times\Delta\vec{S}$$ on a closed surface, we can compute the x,y,z direction of $$\vec{A}\times\Delta\vec{S}$$ first, and put the integral signs back later..

$$\vec{A}\times\Delta\vec{S}$$ is just and vector, and the cross product of this vector is simple:

$$\vec{A}\times\Delta\vec{S} = (\vec{A}\times\Delta\vec{S})_{x} \vec{i} +(\vec{A}\times\Delta\vec{S})_{y} \vec{j} +(\vec{A}\times\Delta\vec{S})_{z} \vec{k}$$

$$= (A_{y}\Delta S_{z} - A_{z}\Delta S_{y})\vec{i} - (A_{x}\Delta S_{z} - A_{z}\Delta S_{x})\vec{j} + (A_{x}\Delta S_{y} - A_{y}\Delta S_{x}) \vec{k}$$
the z component of $$\vec{A}\times\Delta\vec{S}$$ is $$A_{x}\Delta S_{y} - A_{y} \Delta S_{x}$$

OK, $$\Delta S_{x}$$ is the surface area that point to the direction $$\vec{x}$$, that means it is parallel to the y,z plane, therefore,

$$\Delta S_{x} = \Delta y \Delta z$$, for the same reason, $$\Delta S_{y} = \Delta x \Delta z,$$ and $$\Delta S_{z} = \Delta x \Delta y$$

$$A_{x}\Delta S_{y} - A_{y} \Delta S_{x}$$ become

$$A_{x} \Delta x \Delta z - A_{y} \Delta y \Delta z$$

put the integral sign back and become:

$$- (\int A_{x} dxdz - \int A_{y} dydz) = \int A_{y}dydz - \int A_{x}dxdz = LHS$$

after you done with the z component, you can argue that the x and the y component is also equal

 

[vincentchan]

since i did a lot copy and paste stuff here, there might be typo error... too tired to check...can someone check it for me

 

[meteorologist1]

I finally figured it out... you have to use the Divergence Theorem along with the following two identities:
1) $$\vec{\nabla}\cdot (\vec{a}\times\vec{b}) = \vec{b}\cdot (\vec{\nabla}\times\vec{a}) - \vec{a}\cdot (\vec{\nabla}\times\vec{b})$$
2) $$\vec{a}\cdot (\vec{b}\times\vec{c}) = \vec{b}\cdot (\vec{c}\times\vec{a}) = \vec{c}\cdot (\vec{a}\times\vec{b})$$

Let $$\vec{B}$$ be a constant vector. From the Divergence Theorem applied to the vector field $$\vec{A}\times\vec{B}$$, we have
$$\oint_{S} (\vec{A}\times\vec{B}) \cdot \vec{n} dS = \int_{V} \vec{\nabla}\cdot (\vec{A}\times\vec{B}) dV$$

Apply the first identity to the right-hand side and the second identity to the left-hand side, and the B vectors come out of the integrals since they're constant. Since B was arbitrary, we get the desired result.

 

[jdavel]

dextercioby,

"Besides,Stokes theorem involves vectors integrated on loops and open surfaces,which are again manifolds of smaler dimension."

Yes, but in every book I've ever seen, either the integrand is dotted with a unit vector of some definition (for example curl with a unit normal to the surface of integration, or a vector integrand is dotted with vector infinitesimal of the the integration variable (for example "F dot ds" for a line integral).

However, after thinking about it and reading the posts here, I don't see any ambiguity in notation where a vector is multiplied by a scalar infinitesimal and then summed, as long as the direction of the vector is constant. And that's assured by the fact that the vector can be resolved into its components.

On the other hand (since vincentchan invited us to proofread post #8) I'm pretty sure the notation in his first line is meaningless. That first parnthesis has to be inside the integral. Probably a typo!

 

[dextercioby]

Nope,no typo.He took out from the integral the unit vectors "i,j,k",it was natural to use a paranthesis to separate the scalar.

Daniel.