Problem: By using Stoke's Theorem, deduce that $\int_{C} \mathbf{r} ( \mathbf{r} \cdot d\mathbf{r}) = \int \int _{S} \mathbf{r} \wedge d \mathbf{S}$. Where $C$ is the simple closed curve bounding the smooth surface $S$.
My thoughts were applying the relation $$ \nabla \wedge ( \phi \mathbf{u}) = (\nabla \phi) \wedge \mathbf{u} + \phi \nabla \wedge \mathbf{u}$$ with $\mathbf{u}$ as a constant vector. Hence with Stoke's Theorem, $$ \mathbf{u} \cdot \int \int_{S} -\nabla \phi \wedge d \mathbf{S} = \mathbf{u} \cdot \int_{C} \phi d \mathbf{r} $$ So if we can some how relate $$\int \int_{S} -\nabla \phi \wedge d \mathbf{S} = \int_{C} \phi d \mathbf{r} $$ with the original question that would be pretty nice. However, I couldn't really make sense of what the question means $\mathbf{r} ( \mathbf{r} \cdot d \mathbf{r} ) $. Any ideas? Thank you so much.
Consider $\oint_C x{\bf r}\cdot d{\bf r}$. By Stoke's thm $$\oint_C x{\bf r}\cdot d{\bf r}=\iint_S (\nabla \times (x{\bf r}))\cdot d{\bf S}$$ $$=\iint_S (\nabla x \times {\bf r}+x\nabla \times {\bf r})\cdot d{\bf S}$$ $$=\iint_S (\hat{{\bf x}} \times {\bf r})\cdot d{\bf S}$$ $$=\iint_S ({\bf r}\times d{\bf S})\cdot \hat{{\bf x}}$$
Similarly, $$\oint_C y{\bf r}\cdot d{\bf r}=\iint_S ({\bf r}\times d{\bf S})\cdot \hat{{\bf y}}$$ $$\oint_C z{\bf r}\cdot d{\bf r}=\iint_S ({\bf r}\times d{\bf S})\cdot \hat{{\bf z}}$$
Hence $$\oint_C {\bf r}({\bf r}\cdot d{\bf r})=\iint_S {\bf r}\times d{\bf S}$$