Divergence Theorem: Show 0 Integral on Closed Surface

[sszabo]

In h.m. schey, div grad curl and all that, II-25:
Use the divergence theorem to show that
$$\int\int_S \hat{\mathbf{n}}\,dS=0,$$
where $$S$$ is a closed surface and
$$\hat{\mathbf{n}}$$ the unit vector
normal to the surface $$S$$.
How should I understand the l.h.s. ?
Coordinatewise? The r.h.s. is not 0, but zero vector?

 

[wofsy]

the divergence is the total flux through the region. it is not a vector.
your calculation of divergence integrates the inner product of the flow with the unit normal to the surface. this inner product is a scalar.

 

[sszabo]

the divergence is the total flux through the region. it is not a vector.
your calculation of divergence integrates the inner product of the flow with the unit normal to the surface. this inner product is a scalar.

Right. However in the l.h.s. we have a vector $$\hat{\mathbf{n}}$$ and not an inner product which would be a scalar. I can integrate only vector-variable scalar-valued functions on a surface and not a vector-valued function.
So what does it mean?

 

[slider142]

If the surface is closed, intuitively, the unit normals should "sum" to the 0 vector (generalizing the fact that on a "nice" closed curve, summing the normals should lead to no displacement). This should then be proven properly.

 

[wofsy]

Right. However in the l.h.s. we have a vector $$\hat{\mathbf{n}}$$ and not an inner product which would be a scalar. I can integrate only vector-variable scalar-valued functions on a surface and not a vector-valued function.
So what does it mean?

The integral of the unit normal is just a vector, the result of a Riemann sum of vectors.
But this is not the divergence theorem.

 

[sszabo]

If the surface is closed, intuitively, the unit normals should "sum" to the 0 vector (generalizing the fact that on a "nice" closed curve, summing the normals should lead to no displacement). This should then be proven properly.

I see. Yes, indeed, the answer -intuitively- is the 0 vector.
It means
$$\int\int_S \hat{\mathbf{n}}\,dS=\left( \int\int_S n_1\,dS,\int\int_S n_2\,dS,\int\int_S n_3\,dS\right),$$
where $$\hat{\mathbf{n}}=(n_1,n_2,n_3)$$.
Now $$n_1=\left\langle \mathbf{i},\hat{\mathbf{n}} \right\rangle$$ and here we can use the divergence theorem to obtain
$$\int\int_S n_1\,dS=\int\int_S \left\langle \mathbf{i},\hat{\mathbf{n}} \right\rangle \,dS=\int_V div\, \mathbf{i}\, dV=0,$$
because $$div\, \mathbf{i}=0$$. Nice result.
Thanks for your helps.

 

[wofsy]

I see. Yes, indeed, the answer -intuitively- is the 0 vector.
It means
$$\int\int_S \hat{\mathbf{n}}\,dS=\left( \int\int_S n_1\,dS,\int\int_S n_2\,dS,\int\int_S n_3\,dS\right),$$
where $$\hat{\mathbf{n}}=(n_1,n_2,n_3)$$.
Now $$n_1=\left\langle \mathbf{i},\hat{\mathbf{n}} \right\rangle$$ and here we can use the divergence theorem to obtain
$$\int\int_S n_1\,dS=\int\int_S \left\langle \mathbf{i},\hat{\mathbf{n}} \right\rangle \,dS=\int_V div\, \mathbf{i}\, dV=0,$$
because $$div\, \mathbf{i}=0$$. Nice result.
Thanks for your helps.

no. your your three integrals are integrals of vectors not inner products.

 

[sszabo]

no. your three integrals are integrals of vectors not inner products.

Sorry, I absolutely don't understand what you think. Please, give more details, for example point out the wrong step in the calculations. Thanks.

 

[wofsy]

I'm just saying that n1 n2 and n3 are inner products and thus are scalars. Originally you wanted to integrate a vector. Splitting into components still requires keeping the basis vectors in the integral. I don't think you did this. Instead I thing you substituted scalars.Unless I don't understand your post.

 

[sszabo]

I'm just saying that n1 n2 and n3 are inner products and thus are scalars.

Agree.

Originally you wanted to integrate a vector.

Agree.

Splitting into components still requires keeping the basis vectors in the integral.

Of course.

I don't think you did this.

Certainly a similar calculation gives
$$\int\int_S n_2\,dS=\int\int_S \left\langle \mathbf{j},\hat{\mathbf{n}} \right\rangle \,dS=\int_V div\, \mathbf{j}\, dV=0,$$
and
$$\int\int_S n_3\,dS=\int\int_S \left\langle \mathbf{k},\hat{\mathbf{n}} \right\rangle \,dS=\int_V div\, \mathbf{k}\, dV=0$$

Instead I thing you substituted scalars.Unless I don't understand your post.

The final result, the value of the original integral is $$\mathbf{0}$$.