Let $u_0$ be a fixed vector, and let $b=u_0\times r$, where $r$ is the position vector $x\hat{i}+y\hat{j}+z\hat{k}$. What is $\int_C b.\hat{T}ds$, where $C$ is a closed curve?
Assuming $u_0=\begin{pmatrix} u\\v\\w\end{pmatrix}$, I am getting $b=(vz-wy)\hat{i}+(wx-uz)\hat{j}+(uy-vx)\hat{k}$
$\int_C b.\hat{T} ds= \int_c (vz-wy)dx + \int_C (wx-uz)dy+\int_C (uy-vx) dz=0$
Hence, the answer that I'm getting is $0$. But the book says that the answer is wrong. Where am I going wrong?
EDIT: OK I realise that my answer is incorrect as $y$ and $z$ may chance as $x$ changes along the closed curve. My book says the answer is $B(u_0.n_p)$, where $n_p$ is a normal to the plane containing $C$, and $B$ is some constant. How do I arrive upon this answer?
Using Stokes theorem: $$ \int\limits_C b \cdot dr = \int\limits_A \mbox{rot } b \cdot dA = \int\limits_A (2u, 2v, 2w) \cdot dA = 2 \int\limits_A u_0 \cdot dA = 2 A (u_0 \cdot n_A) $$ where $(\mbox{rot } u)_i = \epsilon_{ijk} \partial_j u_k$, and $C = \partial A$ ($A$ is a surface with border $C$).
Your line integral calculation was probably parameterized wrongly.