Multivariable Calculus - Vector field 
For the first part I've tried multiplying both sides by a constant vector but got nowhere. I am stumped as to how to go about the second part. Any help would be appreciated. Thanks!
 A: Multiply out the norm:
$$ \lvert r-R \rvert^2 = \lvert r \rvert^2 - 2 r \cdot R + \lvert R \rvert^2. $$
The first term is independent of $R$, so its integral is a constant. The last term is independent of $r$, so one needs to compute
$$ \int_C dr = 0 $$
essentially by the Fundamental Theorem of Calculus: it's $r$ evaluated at both ends of $C$, which is the same since $C$ is closed. This leaves
$$ -2\int_C R \cdot r \, dr. $$
If we dot this with a constant vector, we have
$$ -2\int_C (R \cdot r)k \cdot dr, $$
which is in the form "vector field dot line element" that allows us to apply Stokes's Theorem. Let $S$ be the surface described in the second part, and then
$$ -2\int_C (R \cdot r)k \cdot dr = -2\iint_S (\nabla_r \times (R \cdot r)k) \cdot dS $$
Now, if $f$ is a scalar field and $k$ constant, $\nabla \times f k = (\nabla f) \times k $ (use summation convention if you're unsure), so we find
$$ -2\int_C (R \cdot r)k \cdot dr = -2\iint_S (\nabla_r(R \cdot r)) \times  k \cdot dS = -2\iint_S (R \times  k) \cdot dS = -(R \times  k) \cdot 2\iint_S  dS. $$
The order of the dot and the cross in a scalar triple product don't matter, and swapping two elements changes the sign, so we find this is
$$ -2\int_C (R \cdot r)k \cdot dr = k \cdot \left( R \times \left(2\iint_S  dS\right) \right). $$
But this is true for arbitrary $k$, hence
$$ -2\int_C (R \cdot r) \, dr = R \times \left(2\iint_S  dS\right). $$
So
$$ F(R) = R \times \left(2\iint_S  dS\right) + \int_C \lvert r \rvert^2 \, dr, $$
in the form specified.

For the second part, calculate $\nabla \times F(R)$:
$$ \nabla \times (R \times A + B) = \nabla \times (R \times A) $$
Breaking out summation convention,
$$[\nabla \times (R \times A)]_i = \epsilon_{ijk} \partial_j \epsilon_{klm} R_l A_m = (\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})\delta_{jl} A_m = (\delta_{im} -\delta_{jj}\delta_{im})A_m = -2A_i $$
since $\delta_{jj}=3$. Inserting the actual value of $A$ gives the answer.
