Application of Stokes' theorem

[Heirot]

Homework Statement

Evaluate the following integrals

$$I_1 = \oint \vec{r} (\vec{a} \cdot \vec{n}) dS$$
$$I_2 = \oint (\vec{a} \cdot \vec{r})\vec{n} dS$$

where $$\vec{a}$$ is a constant vector, and $$\vec{n}$$ is an unit vector normal to the closed surface $$S$$.

Homework Equations

Stokes' theorem, or in this case, some sort of Gauss' divergence theorem.

The Attempt at a Solution

I'm not sure whether the Stokes's theorem can be used in this case. For instance, if we write the integrals in the component notation

$$I_1 = \oint x_i (a_j dS_j)$$
$$I_2 = \oint (a_j x_j)dS_i$$

then the Stokes' theorem would suggest the supstitution $$dS_i \to dV \partial_i$$ where $$S=\partial(V)$$. This results in

$$I_1 = \int (a_j \partial_j) x_i dV=\vec{a} V$$
$$I_2 = \int \partial_i (a_j x_j)dV=\vec{a} V$$

Is this correct?

 

[mighty2000]

Thank you for your post. I am a scientist and I would be happy to help you evaluate these integrals.

Firstly, I would like to clarify that the integrands in both integrals are vector quantities, so the result of the integration should also be a vector. This means that your results for I_1 and I_2 should be vectors, not scalars.

Now, let's look at the integrals one by one. For I_1, we can use the divergence theorem, which states that for a vector field \vec{F} and a closed surface S, the integral of the divergence of \vec{F} over S is equal to the integral of \vec{F} over the volume enclosed by S. In other words,

\int_S (\nabla \cdot \vec{F}) dS = \int_V \vec{F} \cdot d\vec{V}

In our case, \vec{F} = \vec{r}(\vec{a} \cdot \vec{n}), so the divergence of \vec{F} is given by

\nabla \cdot \vec{F} = \nabla \cdot (\vec{r}(\vec{a} \cdot \vec{n})) = \vec{a} \cdot \nabla (\vec{r} \cdot \vec{n}) + \vec{r} \cdot \nabla (\vec{a} \cdot \vec{n}) = \vec{a} \cdot \vec{n} + \vec{r} \cdot (\vec{a} \cdot \nabla \vec{n})

Since \vec{n} is a unit vector, its gradient is zero, so the second term disappears. This leaves us with

\nabla \cdot \vec{F} = \vec{a} \cdot \vec{n}

Substituting this in the divergence theorem, we get

\int_S (\vec{a} \cdot \vec{n}) dS = \int_V \vec{r} (\vec{a} \cdot \vec{n}) \cdot d\vec{V}

Note that the right-hand side is the same as your integral I_1, so we can conclude that

I_1 = \int_S (\vec{a} \cdot \vec{n}) dS = \vec{a} \cdot \