Evaluating line integrals in the form $\oint F\cdot n ds$

Question

Let $C$ be the unit circle on the plane, oriented counterclockwise, consider vector field $$F=(x^2,xy)$$ We are asked to find the flux through $C$ by evaluating the line integral $\oint_C F\cdot n ds$, but I am only familiar with line integrals of the form $\oint_C F\cdot dr$ which is against the tangent as opposed to the normal vector. How do we evaluate this other form?

Answer 1

I don't know your backgrounds, so I don't know if I should give you the full explanation, in term of differential forms...

The short answer, indeed, is what follows. Since you want to compute "how much flow" of $F$ is going trough your curve $C$ you have to integrate the inner product of the vectors $F$ with the unit normal vectors of the curve $C$. That is, you compute $C'(t)$ and take $n(t)$ a unit normal vector. The reason is that $F(C(t))\cdot n(t)$ gives you a measurement of how transversal is $F(C(t))$ to $C'(t)$. For example, if $F$ is tangent to $C$ at $C(t)$ then $F(C(t))\cdot n(t)=0$ (no flow of $F$ is going through $C$ locally at $C(t)$).

In particular, you take $C(t)=(\cos(t),\sin(t))$, and so $C'(t)=(-\sin(t),\cos(t))$ and $$ n(t)=(\cos(t),\sin(t)). $$

So finally $$ \int_C F\cdot n dt=\int_0^{2\pi} \cos^3(t)+\cos(t)\sin^2(t)dt=\int_0^{2\pi} +\cos(t)dt=... $$

Answer 2

The curve $C$ can be given by the parametric equation $r(t):=(\cos t,\sin t) $ over the interval $[0,2\pi[$, since $C$ is a circle of radius $1$.

We can show that the flux of the vector field $\vec{F}=(f,g)$ across $C$ is given by $$\int_{C}\vec{F}\cdot \vec{n}\, ds=\int_{C}f\, dy-g\, dx.$$

Thus, since $f=x^2$ and $g=xy$, then \begin{align*}\int_{C}\vec{F}\cdot \vec{n}\, ds&=\int_{C}x^2\, dy-xy\, dx\\&=\int_{0}^{2\pi}(\cos t)^2\cdot(\cos t)-(\cos t\sin t)(-\sin t)\, dt\\ &=\int_{0}^{2\pi}\cos^{3}t+\cos t\sin^{2} t\, dt\\ &=0. \end{align*}

Also, we can use Green's theorem in Normal form:

$$\int_{C}\vec{F}\cdot \vec{n}\, ds=\iint_{R}{\rm div}(\vec{F})\, dA=\iint_{R}3x\, dA=0,$$ where $R$ is the region closed by the circle $x^2+y^2=1$.

N.B: It is also interesting notice since $0=f_y\not=g_x=y$, then $\vec{F}$ is not conservative.