Flux across a plane curve

Let \(\textbf{F}(x,y)= \langle P(x,y), Q(x,y) \rangle\) be a two dimensional vector field. In this article, we talked about finding the work done by force \(F\) on a particle moving along curve \(C\):

$$W = \int_C \textbf{F} \cdot \textbf{T} ds $$

In the work done formula we use the tangent vector \(\textbf{T}\). If we replace \(\textbf{T}\) by the unit normal vector \(N\), we get the flux across the curve \(C\):

$$\text{flux across } C = \int_C \textbf{F} \cdot \textbf{N} ds $$

Let \(\textbf{n}(t) = \langle y'(t), -x'(t) \rangle\). The definition of unit normal vector is:

$$\textbf{N}(t) = \frac{\textbf{n}(t)}{\Vert \textbf{n}(t) \Vert} $$

Let \(C\) be a smooth curve with parameterization \(\textbf{r}(t) = \langle x(t), y(t) \rangle\), where \(a \le t \le b\). Using the definition of unit normal vector, we can say:

$$ \int_C \textbf{F} \cdot \textbf{N} ds = \int_c \textbf{F} \cdot \frac{\textbf{n}(t)}{\Vert \textbf{n}(t) \Vert} ds = \int^a_b \textbf{F} \cdot \frac{\textbf{n}(t)}{\Vert \textbf{n}(t) \Vert} \Vert \textbf{r}'(t) \Vert dt$$

Since \(\Vert \textbf{n}(t) \Vert = \Vert \textbf{r}'(t) \Vert\):

$$ \int_C \textbf{F} \cdot \textbf{N} ds = \int^a_b \textbf{F}(\textbf{r}(t)) \cdot \textbf{n}(t) dt$$

Since \([\textbf{n}(t) = \langle y'(t), -x'(t) \rangle]\) or \([\textbf{n}(t) = \langle dy(t)/dt, - dx(t)/dt \rangle]\), then \([\textbf{n}(t) dt = \langle dy(t), - dx(t) \rangle ]\). This means:

$$ \int_C \textbf{F} \cdot \textbf{n}(t) dt = \int_C \textbf{F} \cdot \langle dy, - dx \rangle = \int_C Pdy -Qdx$$

Flux Across A Plane Curve

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