# Divergence theorem

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?

## Answers and Replies

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.

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?

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.

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.

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.

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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.

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.

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.

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 $$\mathbf{0}$$.