Flux across the circle Find the flux of the fields

and 
across the circle


I'm not sure if I did this correctly, but this is what I got as the result...

 A: It appears to me that the OP, in the notes posted, has computed the circulation of these vector fields $F_1(x, y)$, $F_2(x, y)$ around the circular path 
$r(t) = (a \cos t)i + (a \sin t)j, \, 0 \le t \le 2\pi, \tag{1}$
rather than the flux of each across it.  I type these remarks based upon my scrutiny of the posted image of the notebook page which, according to my reading of it in any event, indicates that the author is wrestling with such expressions as $F_1 \cdot \frac{dr}{dt}$ which is the component of $F_1$ tangent to the circle $r(t)$ multiplied by the magnitude of the tangent vector.  Here we shall calculate the true flux vector, and that by two methods.  We recall that for any vector field $V(x, y)$ defined on $\Bbb R^2$ the flux across any closed path $\gamma(t)$ is defined to be
$\int_{\gamma(t)} (V \cdot n) ds, \tag{2}$
where $n$ is the outward pointing unit normal vector field to $\gamma$.  We note the discrepancy between $F_1(x, y) = 2i + 3j$ in the title and $F_1(x, y) = 2xi -3yj$ in the notes; here we shall stick with the vector fields as presented in the title and leave it to the reader to, if he/she so desires, perform the analogous calculations for $F_1(x, y) = 2xi - 3yj$ etc.  Thus we take
$V(x, y) = F_1(x, y) = 2xi + 3yj \tag{3}$
and observe that the outward pointing unit normal field to the circle (1) is in fact
$n(t) = \begin{pmatrix} \cos t \\ \sin t \end{pmatrix} = (\cos t) i + (\sin t) j, \tag{4}$
and converting $x$ and $y$ on the circle $r(t)$ to polar coordinates:
$x = a \cos t, \, y = a\sin t, \tag{5}$
so that on this circle
$F_1(x, y) = (2a \cos t) i + (3a \sin t) j, \tag{6}$
we see that
$F_1 \cdot n = 2a \cos^2 t + 3a \sin^2 t = 2a(\cos^2 t + \sin^2 t) + a \sin^2 t = 2a + a \sin^2 t. \tag{7}$
We now compute
$\int_{r(t)} F_1 \cdot n ds = \int_0^{2\pi} (2a + a \sin^2 t) a \, dt, \tag{8}$
where $ds = adt$ since, from (1), $ds/dt = \Vert \dot r(t) \Vert = a$; $s$ is, of course, the distance or arc-length along the circle $r(t)$.  Following through with (8):
$\int_0^{2\pi} (2a + a \sin^2 t) a dt = 4\pi a^2 + a^2\int_0^{2\pi}\sin^2 t \, dt = 5 \pi a^2, \tag{9}$
since $\int_0^{2 \pi} \sin^2 t \, dt = \pi$.  We validate the value of this flux integral by calculating it yet a second way, via the divergence theorem.  We have for any $C^1$ $V(x, y)$
$\int_\Gamma \nabla \cdot V \, dA = \int_{\gamma(t)} (V \cdot n) ds, \tag{10}$
where $\Gamma$ is a compact set bounded by $\gamma(t)$.  In the present case, $\Gamma = \{(x, y) \mid x^2 + y^2 \le a \}$ and $\nabla \cdot F_1 = 5$, so the integral on the left of (10) is simply
$\int_\Gamma \nabla \cdot F_1 \, dA = 5 \pi a^2, \tag{11}$
since $dA$ is the area element of $\Gamma$.  We see that (11) is in accord with (9).
The case
$F_2(x, y) = 2x i + (x - y)j \tag{12}$
may be similarly handled.  We have
$F_2 = 2a \cos t \, i + a(\cos t - \sin t)j \tag{13}$
on $r(t)$, so
$F_2 \cdot n = 2a \cos^2 t + a \cos t \sin t - a \sin^2 t, \tag{14}$
from which it is easily seen that 
$\int_{r(t)} F_2 \cdot n ds = a\int_0^{2\pi} (2\cos^2 t +  \cos t \sin t -  \sin^2 t)a dt= \pi a^2; \tag{15}$
on the other hand, $\nabla \cdot F_2 = 1$, so
$\int_\Gamma \nabla \cdot F_2 \, dA = \pi a^2; \tag{16}$
as expected, we see that (15) and (16) agree.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
