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To expand on the question, I'm trying to show that, for all smooth orientable surfaces $S$ and all constant vectors $\vec{v} = \langle a, b, c\rangle$,

$$2\iint\limits_S\vec{v}\cdot d\vec{S} = \oint\limits_{\partial S}(\vec{v}\times\vec{r})\cdot d\vec{s}$$

where $\vec{r}(x,y,z) = \langle x,y,z\rangle$, and the boundary $\partial S$ is oriented to be compatible with the orientation of $S$.

My thoughts:

Other than through Stokes' theorem, I'm not sure how I would convert the surface integral to the line integral, but Stokes' requires a curl of a vector function and I can't come up with a vector $\vec{V}$ s.t. $\vec{\nabla}\times\vec{V} = \vec{v}$. Are there any other ways to convert a surface integral to a line integral or does anybody have any other tips?

Thank you for any assistance!

Eric

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By Stokes's theorem,

$$\oint_{\partial S} (\vec{v}\times \vec{r})\cdot d\vec{s} = \iint_S \operatorname{curl}(\vec{v}\times \vec{r})\cdot d\vec{S}.$$

Since

$$\operatorname{curl}(\vec{v}\times \vec{r}) = \operatorname{curl}(bz - cy,cx - az, ay - bx) = (2a,2b,2c) = 2\vec{v},$$

we have

$$\iint_S \operatorname{curl}(\vec{v}\times \vec{r})\cdot d\vec{S} = \iint_S 2\vec{v}\cdot d\vec{S} = 2\iint_S \vec{v}\cdot d\vec{S}.$$

Therefore

$$\oint_{\partial S} (\vec{v}\times \vec{r})\cdot d\vec{s} = 2\iint_S \vec{v}\cdot d\vec{S}.$$

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